\(\int \frac {(d x)^{21/2}}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\) [473]

Optimal result

Integrand size = 28, antiderivative size = 308 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}-\frac {19 d^3 (d x)^{15/2}}{160 b^2 \left (a+b x^2\right )^4}-\frac {19 d^5 (d x)^{11/2}}{128 b^3 \left (a+b x^2\right )^3}-\frac {209 d^7 (d x)^{7/2}}{1024 b^4 \left (a+b x^2\right )^2}-\frac {1463 d^9 (d x)^{3/2}}{4096 b^5 \left (a+b x^2\right )}-\frac {4389 d^{21/2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{8192 \sqrt {2} \sqrt [4]{a} b^{23/4}}+\frac {4389 d^{21/2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{8192 \sqrt {2} \sqrt [4]{a} b^{23/4}}-\frac {4389 d^{21/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d x}}{\sqrt {d} \left (\sqrt {a}+\sqrt {b} x\right )}\right )}{8192 \sqrt {2} \sqrt [4]{a} b^{23/4}} \] Output:

-1/10*d*(d*x)^(19/2)/b/(b*x^2+a)^5-19/160*d^3*(d*x)^(15/2)/b^2/(b*x^2+a)^4 
-19/128*d^5*(d*x)^(11/2)/b^3/(b*x^2+a)^3-209/1024*d^7*(d*x)^(7/2)/b^4/(b*x 
^2+a)^2-1463/4096*d^9*(d*x)^(3/2)/b^5/(b*x^2+a)-4389/16384*d^(21/2)*arctan 
(1-2^(1/2)*b^(1/4)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*2^(1/2)/a^(1/4)/b^(23/4)+4 
389/16384*d^(21/2)*arctan(1+2^(1/2)*b^(1/4)*(d*x)^(1/2)/a^(1/4)/d^(1/2))*2 
^(1/2)/a^(1/4)/b^(23/4)-4389/16384*d^(21/2)*arctanh(2^(1/2)*a^(1/4)*b^(1/4 
)*(d*x)^(1/2)/d^(1/2)/(a^(1/2)+b^(1/2)*x))*2^(1/2)/a^(1/4)/b^(23/4)
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.67 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {d^9 (d x)^{3/2} \left (-4 \sqrt [4]{a} b^{3/4} x^{3/2} \left (7315 a^4+33440 a^3 b x^2+59470 a^2 b^2 x^4+50312 a b^3 x^6+19015 b^4 x^8\right )+21945 \sqrt {2} \left (a+b x^2\right )^5 \arctan \left (\frac {-\sqrt {a}+\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )-21945 \sqrt {2} \left (a+b x^2\right )^5 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )\right )}{81920 \sqrt [4]{a} b^{23/4} x^{3/2} \left (a+b x^2\right )^5} \] Input:

Integrate[(d*x)^(21/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

(d^9*(d*x)^(3/2)*(-4*a^(1/4)*b^(3/4)*x^(3/2)*(7315*a^4 + 33440*a^3*b*x^2 + 
 59470*a^2*b^2*x^4 + 50312*a*b^3*x^6 + 19015*b^4*x^8) + 21945*Sqrt[2]*(a + 
 b*x^2)^5*ArcTan[(-Sqrt[a] + Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] 
 - 21945*Sqrt[2]*(a + b*x^2)^5*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/( 
Sqrt[a] + Sqrt[b]*x)]))/(81920*a^(1/4)*b^(23/4)*x^(3/2)*(a + b*x^2)^5)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.43, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {1380, 27, 252, 252, 252, 252, 252, 266, 27, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx\)

\(\Big \downarrow \) 1380

\(\displaystyle b^6 \int \frac {(d x)^{21/2}}{b^6 \left (b x^2+a\right )^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(d x)^{21/2}}{\left (a+b x^2\right )^6}dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {19 d^2 \int \frac {(d x)^{17/2}}{\left (b x^2+a\right )^5}dx}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \int \frac {(d x)^{13/2}}{\left (b x^2+a\right )^4}dx}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \int \frac {(d x)^{9/2}}{\left (b x^2+a\right )^3}dx}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \int \frac {(d x)^{5/2}}{\left (b x^2+a\right )^2}dx}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^2 \int \frac {\sqrt {d x}}{b x^2+a}dx}{4 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d \int \frac {d^3 x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \int \frac {d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\int \frac {\sqrt {b} x d+\sqrt {a} d}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\int \frac {1}{x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {b}}+\frac {\int \frac {1}{x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\int \frac {1}{-d x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\int \frac {1}{-d x-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a} d-\sqrt {b} d x}{b x^2 d^2+a d^2}d\sqrt {d x}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}\right )}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}\right )}{\sqrt [4]{b} \left (x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}\right )}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a} \sqrt {d}-2 \sqrt [4]{b} \sqrt {d x}}{x d+\frac {\sqrt {a} d}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b} \sqrt {d}}+\frac {\int \frac {\sqrt [4]{a} \sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{x d+\frac {\sqrt {a} d}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} \sqrt {d x} \sqrt {d}}{\sqrt [4]{b}}}d\sqrt {d x}}{2 \sqrt [4]{a} \sqrt {b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {19 d^2 \left (\frac {15 d^2 \left (\frac {11 d^2 \left (\frac {7 d^2 \left (\frac {3 d^3 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {d x}}{\sqrt [4]{a} \sqrt {d}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}+\sqrt {a} d+\sqrt {b} d x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d} \sqrt {d x}+\sqrt {a} d+\sqrt {b} d x\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {d}}}{2 \sqrt {b}}\right )}{2 b}-\frac {d (d x)^{3/2}}{2 b \left (a+b x^2\right )}\right )}{8 b}-\frac {d (d x)^{7/2}}{4 b \left (a+b x^2\right )^2}\right )}{12 b}-\frac {d (d x)^{11/2}}{6 b \left (a+b x^2\right )^3}\right )}{16 b}-\frac {d (d x)^{15/2}}{8 b \left (a+b x^2\right )^4}\right )}{20 b}-\frac {d (d x)^{19/2}}{10 b \left (a+b x^2\right )^5}\)

Input:

Int[(d*x)^(21/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

-1/10*(d*(d*x)^(19/2))/(b*(a + b*x^2)^5) + (19*d^2*(-1/8*(d*(d*x)^(15/2))/ 
(b*(a + b*x^2)^4) + (15*d^2*(-1/6*(d*(d*x)^(11/2))/(b*(a + b*x^2)^3) + (11 
*d^2*(-1/4*(d*(d*x)^(7/2))/(b*(a + b*x^2)^2) + (7*d^2*(-1/2*(d*(d*x)^(3/2) 
)/(b*(a + b*x^2)) + (3*d^3*((-(ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^( 
1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*b^ 
(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]))/(2* 
Sqrt[b]) - (-1/2*Log[Sqrt[a]*d + Sqrt[b]*d*x - Sqrt[2]*a^(1/4)*b^(1/4)*Sqr 
t[d]*Sqrt[d*x]]/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]) + Log[Sqrt[a]*d + Sqrt[b 
]*d*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d]*Sqrt[d*x]]/(2*Sqrt[2]*a^(1/4)*b^(1 
/4)*Sqrt[d]))/(2*Sqrt[b])))/(2*b)))/(8*b)))/(12*b)))/(16*b)))/(20*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 14.74 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(2 d^{11} \left (\frac {-\frac {1463 d^{8} a^{4} \left (d x \right )^{\frac {3}{2}}}{8192 b^{5}}-\frac {209 d^{6} a^{3} \left (d x \right )^{\frac {7}{2}}}{256 b^{4}}-\frac {5947 a^{2} d^{4} \left (d x \right )^{\frac {11}{2}}}{4096 b^{3}}-\frac {6289 a \,d^{2} \left (d x \right )^{\frac {15}{2}}}{5120 b^{2}}-\frac {3803 \left (d x \right )^{\frac {19}{2}}}{8192 b}}{\left (b \,d^{2} x^{2}+a \,d^{2}\right )^{5}}+\frac {4389 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{65536 b^{6} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\)	\(235\)
default	\(2 d^{11} \left (\frac {-\frac {1463 d^{8} a^{4} \left (d x \right )^{\frac {3}{2}}}{8192 b^{5}}-\frac {209 d^{6} a^{3} \left (d x \right )^{\frac {7}{2}}}{256 b^{4}}-\frac {5947 a^{2} d^{4} \left (d x \right )^{\frac {11}{2}}}{4096 b^{3}}-\frac {6289 a \,d^{2} \left (d x \right )^{\frac {15}{2}}}{5120 b^{2}}-\frac {3803 \left (d x \right )^{\frac {19}{2}}}{8192 b}}{\left (b \,d^{2} x^{2}+a \,d^{2}\right )^{5}}+\frac {4389 \sqrt {2}\, \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{65536 b^{6} \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\)	\(235\)
pseudoelliptic	\(\frac {1463 d^{10} \left (-8 \sqrt {d x}\, b x \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \left (\frac {3803}{1463} b^{4} x^{8}+\frac {2648}{385} a \,b^{3} x^{6}+\frac {626}{77} a^{2} b^{2} x^{4}+\frac {32}{7} a^{3} b \,x^{2}+a^{4}\right )+3 d \sqrt {2}\, \left (b \,x^{2}+a \right )^{5} \left (\ln \left (\frac {d x -\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}{d x +\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {a \,d^{2}}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}-\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}+\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}{\left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}}}\right )\right )\right )}{32768 \left (\frac {a \,d^{2}}{b}\right )^{\frac {1}{4}} b^{6} \left (b \,x^{2}+a \right )^{5}}\)	\(246\)

Input:

int((d*x)^(21/2)/(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*d^11*((-1463/8192/b^5*d^8*a^4*(d*x)^(3/2)-209/256/b^4*d^6*a^3*(d*x)^(7/2 
)-5947/4096*a^2*d^4/b^3*(d*x)^(11/2)-6289/5120*a*d^2/b^2*(d*x)^(15/2)-3803 
/8192/b*(d*x)^(19/2))/(b*d^2*x^2+a*d^2)^5+4389/65536/b^6/(a*d^2/b)^(1/4)*2 
^(1/2)*(ln((d*x-(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2))/(d*x+ 
(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))+2*arctan(2^(1/2)/(a* 
d^2/b)^(1/4)*(d*x)^(1/2)+1)+2*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1 
)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.77 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {21945 \, {\left (b^{10} x^{10} + 5 \, a b^{9} x^{8} + 10 \, a^{2} b^{8} x^{6} + 10 \, a^{3} b^{7} x^{4} + 5 \, a^{4} b^{6} x^{2} + a^{5} b^{5}\right )} \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {1}{4}} \log \left (84546715869 \, \sqrt {d x} d^{31} + 84546715869 \, \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {3}{4}} a b^{17}\right ) - 21945 \, {\left (i \, b^{10} x^{10} + 5 i \, a b^{9} x^{8} + 10 i \, a^{2} b^{8} x^{6} + 10 i \, a^{3} b^{7} x^{4} + 5 i \, a^{4} b^{6} x^{2} + i \, a^{5} b^{5}\right )} \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {1}{4}} \log \left (84546715869 \, \sqrt {d x} d^{31} + 84546715869 i \, \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {3}{4}} a b^{17}\right ) - 21945 \, {\left (-i \, b^{10} x^{10} - 5 i \, a b^{9} x^{8} - 10 i \, a^{2} b^{8} x^{6} - 10 i \, a^{3} b^{7} x^{4} - 5 i \, a^{4} b^{6} x^{2} - i \, a^{5} b^{5}\right )} \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {1}{4}} \log \left (84546715869 \, \sqrt {d x} d^{31} - 84546715869 i \, \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {3}{4}} a b^{17}\right ) - 21945 \, {\left (b^{10} x^{10} + 5 \, a b^{9} x^{8} + 10 \, a^{2} b^{8} x^{6} + 10 \, a^{3} b^{7} x^{4} + 5 \, a^{4} b^{6} x^{2} + a^{5} b^{5}\right )} \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {1}{4}} \log \left (84546715869 \, \sqrt {d x} d^{31} - 84546715869 \, \left (-\frac {d^{42}}{a b^{23}}\right )^{\frac {3}{4}} a b^{17}\right ) - 4 \, {\left (19015 \, b^{4} d^{10} x^{9} + 50312 \, a b^{3} d^{10} x^{7} + 59470 \, a^{2} b^{2} d^{10} x^{5} + 33440 \, a^{3} b d^{10} x^{3} + 7315 \, a^{4} d^{10} x\right )} \sqrt {d x}}{81920 \, {\left (b^{10} x^{10} + 5 \, a b^{9} x^{8} + 10 \, a^{2} b^{8} x^{6} + 10 \, a^{3} b^{7} x^{4} + 5 \, a^{4} b^{6} x^{2} + a^{5} b^{5}\right )}} \] Input:

integrate((d*x)^(21/2)/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")
 

Output:

1/81920*(21945*(b^10*x^10 + 5*a*b^9*x^8 + 10*a^2*b^8*x^6 + 10*a^3*b^7*x^4 
+ 5*a^4*b^6*x^2 + a^5*b^5)*(-d^42/(a*b^23))^(1/4)*log(84546715869*sqrt(d*x 
)*d^31 + 84546715869*(-d^42/(a*b^23))^(3/4)*a*b^17) - 21945*(I*b^10*x^10 + 
 5*I*a*b^9*x^8 + 10*I*a^2*b^8*x^6 + 10*I*a^3*b^7*x^4 + 5*I*a^4*b^6*x^2 + I 
*a^5*b^5)*(-d^42/(a*b^23))^(1/4)*log(84546715869*sqrt(d*x)*d^31 + 84546715 
869*I*(-d^42/(a*b^23))^(3/4)*a*b^17) - 21945*(-I*b^10*x^10 - 5*I*a*b^9*x^8 
 - 10*I*a^2*b^8*x^6 - 10*I*a^3*b^7*x^4 - 5*I*a^4*b^6*x^2 - I*a^5*b^5)*(-d^ 
42/(a*b^23))^(1/4)*log(84546715869*sqrt(d*x)*d^31 - 84546715869*I*(-d^42/( 
a*b^23))^(3/4)*a*b^17) - 21945*(b^10*x^10 + 5*a*b^9*x^8 + 10*a^2*b^8*x^6 + 
 10*a^3*b^7*x^4 + 5*a^4*b^6*x^2 + a^5*b^5)*(-d^42/(a*b^23))^(1/4)*log(8454 
6715869*sqrt(d*x)*d^31 - 84546715869*(-d^42/(a*b^23))^(3/4)*a*b^17) - 4*(1 
9015*b^4*d^10*x^9 + 50312*a*b^3*d^10*x^7 + 59470*a^2*b^2*d^10*x^5 + 33440* 
a^3*b*d^10*x^3 + 7315*a^4*d^10*x)*sqrt(d*x))/(b^10*x^10 + 5*a*b^9*x^8 + 10 
*a^2*b^8*x^6 + 10*a^3*b^7*x^4 + 5*a^4*b^6*x^2 + a^5*b^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate((d*x)**(21/2)/(b**2*x**4+2*a*b*x**2+a**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.22 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {21945 \, d^{12} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {d x} \sqrt {b}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b} d}}\right )}{\sqrt {\sqrt {a} \sqrt {b} d} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} d x + \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} d x - \sqrt {2} \left (a d^{2}\right )^{\frac {1}{4}} \sqrt {d x} b^{\frac {1}{4}} + \sqrt {a} d\right )}{\left (a d^{2}\right )^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{b^{5}} - \frac {8 \, {\left (19015 \, \left (d x\right )^{\frac {19}{2}} b^{4} d^{12} + 50312 \, \left (d x\right )^{\frac {15}{2}} a b^{3} d^{14} + 59470 \, \left (d x\right )^{\frac {11}{2}} a^{2} b^{2} d^{16} + 33440 \, \left (d x\right )^{\frac {7}{2}} a^{3} b d^{18} + 7315 \, \left (d x\right )^{\frac {3}{2}} a^{4} d^{20}\right )}}{b^{10} d^{10} x^{10} + 5 \, a b^{9} d^{10} x^{8} + 10 \, a^{2} b^{8} d^{10} x^{6} + 10 \, a^{3} b^{7} d^{10} x^{4} + 5 \, a^{4} b^{6} d^{10} x^{2} + a^{5} b^{5} d^{10}}}{163840 \, d} \] Input:

integrate((d*x)^(21/2)/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")
 

Output:

1/163840*(21945*d^12*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)* 
b^(1/4) + 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt 
(b)*d)*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2)^(1/4)*b^( 
1/4) - 2*sqrt(d*x)*sqrt(b))/sqrt(sqrt(a)*sqrt(b)*d))/(sqrt(sqrt(a)*sqrt(b) 
*d)*sqrt(b)) - sqrt(2)*log(sqrt(b)*d*x + sqrt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b 
^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) + sqrt(2)*log(sqrt(b)*d*x - sq 
rt(2)*(a*d^2)^(1/4)*sqrt(d*x)*b^(1/4) + sqrt(a)*d)/((a*d^2)^(1/4)*b^(3/4)) 
)/b^5 - 8*(19015*(d*x)^(19/2)*b^4*d^12 + 50312*(d*x)^(15/2)*a*b^3*d^14 + 5 
9470*(d*x)^(11/2)*a^2*b^2*d^16 + 33440*(d*x)^(7/2)*a^3*b*d^18 + 7315*(d*x) 
^(3/2)*a^4*d^20)/(b^10*d^10*x^10 + 5*a*b^9*d^10*x^8 + 10*a^2*b^8*d^10*x^6 
+ 10*a^3*b^7*d^10*x^4 + 5*a^4*b^6*d^10*x^2 + a^5*b^5*d^10))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.14 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {1}{163840} \, d^{10} {\left (\frac {43890 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{8} d} + \frac {43890 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {d x}\right )}}{2 \, \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}}}\right )}{a b^{8} d} - \frac {21945 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x + \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{8} d} + \frac {21945 \, \sqrt {2} \left (a b^{3} d^{2}\right )^{\frac {3}{4}} \log \left (d x - \sqrt {2} \left (\frac {a d^{2}}{b}\right )^{\frac {1}{4}} \sqrt {d x} + \sqrt {\frac {a d^{2}}{b}}\right )}{a b^{8} d} - \frac {8 \, {\left (19015 \, \sqrt {d x} b^{4} d^{10} x^{9} + 50312 \, \sqrt {d x} a b^{3} d^{10} x^{7} + 59470 \, \sqrt {d x} a^{2} b^{2} d^{10} x^{5} + 33440 \, \sqrt {d x} a^{3} b d^{10} x^{3} + 7315 \, \sqrt {d x} a^{4} d^{10} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{5} b^{5}}\right )} \] Input:

integrate((d*x)^(21/2)/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

1/163840*d^10*(43890*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2) 
*(a*d^2/b)^(1/4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(a*b^8*d) + 43890*sqrt(2) 
*(a*b^3*d^2)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) - 2*sqrt(d 
*x))/(a*d^2/b)^(1/4))/(a*b^8*d) - 21945*sqrt(2)*(a*b^3*d^2)^(3/4)*log(d*x 
+ sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a*b^8*d) + 21945*sqr 
t(2)*(a*b^3*d^2)^(3/4)*log(d*x - sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + sqrt( 
a*d^2/b))/(a*b^8*d) - 8*(19015*sqrt(d*x)*b^4*d^10*x^9 + 50312*sqrt(d*x)*a* 
b^3*d^10*x^7 + 59470*sqrt(d*x)*a^2*b^2*d^10*x^5 + 33440*sqrt(d*x)*a^3*b*d^ 
10*x^3 + 7315*sqrt(d*x)*a^4*d^10*x)/((b*d^2*x^2 + a*d^2)^5*b^5))
 

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.69 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {4389\,d^{21/2}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{8192\,{\left (-a\right )}^{1/4}\,b^{23/4}}-\frac {\frac {3803\,d^{11}\,{\left (d\,x\right )}^{19/2}}{4096\,b}+\frac {5947\,a^2\,d^{15}\,{\left (d\,x\right )}^{11/2}}{2048\,b^3}+\frac {209\,a^3\,d^{17}\,{\left (d\,x\right )}^{7/2}}{128\,b^4}+\frac {1463\,a^4\,d^{19}\,{\left (d\,x\right )}^{3/2}}{4096\,b^5}+\frac {6289\,a\,d^{13}\,{\left (d\,x\right )}^{15/2}}{2560\,b^2}}{a^5\,d^{10}+5\,a^4\,b\,d^{10}\,x^2+10\,a^3\,b^2\,d^{10}\,x^4+10\,a^2\,b^3\,d^{10}\,x^6+5\,a\,b^4\,d^{10}\,x^8+b^5\,d^{10}\,x^{10}}-\frac {4389\,d^{21/2}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {d\,x}}{{\left (-a\right )}^{1/4}\,\sqrt {d}}\right )}{8192\,{\left (-a\right )}^{1/4}\,b^{23/4}} \] Input:

int((d*x)^(21/2)/(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)
 

Output:

(4389*d^(21/2)*atan((b^(1/4)*(d*x)^(1/2))/((-a)^(1/4)*d^(1/2))))/(8192*(-a 
)^(1/4)*b^(23/4)) - ((3803*d^11*(d*x)^(19/2))/(4096*b) + (5947*a^2*d^15*(d 
*x)^(11/2))/(2048*b^3) + (209*a^3*d^17*(d*x)^(7/2))/(128*b^4) + (1463*a^4* 
d^19*(d*x)^(3/2))/(4096*b^5) + (6289*a*d^13*(d*x)^(15/2))/(2560*b^2))/(a^5 
*d^10 + b^5*d^10*x^10 + 5*a^4*b*d^10*x^2 + 5*a*b^4*d^10*x^8 + 10*a^3*b^2*d 
^10*x^4 + 10*a^2*b^3*d^10*x^6) - (4389*d^(21/2)*atanh((b^(1/4)*(d*x)^(1/2) 
)/((-a)^(1/4)*d^(1/2))))/(8192*(-a)^(1/4)*b^(23/4))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 996, normalized size of antiderivative = 3.23 \[ \int \frac {(d x)^{21/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int((d*x)^(21/2)/(b^2*x^4+2*a*b*x^2+a^2)^3,x)
 

Output:

(sqrt(d)*d**10*( - 43890*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4) 
*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**5 - 219450*b 
**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt( 
b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**4*b*x**2 - 438900*b**(1/4)*a**(3/4)*sq 
rt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1 
/4)*sqrt(2)))*a**3*b**2*x**4 - 438900*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**( 
1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a* 
*2*b**3*x**6 - 219450*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sq 
rt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**4*x**8 - 4389 
0*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sq 
rt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*b**5*x**10 + 43890*b**(1/4)*a**(3/4)*s 
qrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**( 
1/4)*sqrt(2)))*a**5 + 219450*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**( 
1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**4*b*x**2 
 + 438900*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sq 
rt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**3*b**2*x**4 + 438900*b**(1/ 
4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/( 
b**(1/4)*a**(1/4)*sqrt(2)))*a**2*b**3*x**6 + 219450*b**(1/4)*a**(3/4)*sqrt 
(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4 
)*sqrt(2)))*a*b**4*x**8 + 43890*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4...