\(\int \frac {x^{15/2}}{(a+b x^2+c x^4)^3} \, dx\) [939]

Optimal result

Integrand size = 20, antiderivative size = 617 \[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {x^{9/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \sqrt {x} \left (a \left (b^2+12 a c\right )+b \left (b^2+4 a c\right ) x^2\right )}{16 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 \left (28 a b-\frac {b^3}{c}-\frac {b^4-30 a b^2 c-24 a^2 c^2}{c \sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{32 \sqrt [4]{2} \sqrt [4]{c} \left (b^2-4 a c\right )^2 \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}-\frac {3 \left (b^3-28 a b c-\frac {b^4-30 a b^2 c-24 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {3 \left (28 a b-\frac {b^3}{c}-\frac {b^4-30 a b^2 c-24 a^2 c^2}{c \sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{32 \sqrt [4]{2} \sqrt [4]{c} \left (b^2-4 a c\right )^2 \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}-\frac {3 \left (b^3-28 a b c-\frac {b^4-30 a b^2 c-24 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{32 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right )^2 \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \] Output:

1/4*x^(9/2)*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^2-3/16*x^(1/2)*(a*(12 
*a*c+b^2)+b*(4*a*c+b^2)*x^2)/c/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)+3/64*(28*a*b 
-b^3/c-(-24*a^2*c^2-30*a*b^2*c+b^4)/c/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c 
^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*2^(3/4)/c^(1/4)/(-4*a*c+b^2) 
^2/(-b-(-4*a*c+b^2)^(1/2))^(3/4)-3/64*(b^3-28*a*b*c-(-24*a^2*c^2-30*a*b^2* 
c+b^4)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2) 
^(1/2))^(1/4))*2^(3/4)/c^(5/4)/(-4*a*c+b^2)^2/(-b+(-4*a*c+b^2)^(1/2))^(3/4 
)+3/64*(28*a*b-b^3/c-(-24*a^2*c^2-30*a*b^2*c+b^4)/c/(-4*a*c+b^2)^(1/2))*ar 
ctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*2^(3/4)/c^(1/ 
4)/(-4*a*c+b^2)^2/(-b-(-4*a*c+b^2)^(1/2))^(3/4)-3/64*(b^3-28*a*b*c-(-24*a^ 
2*c^2-30*a*b^2*c+b^4)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/ 
(-b+(-4*a*c+b^2)^(1/2))^(1/4))*2^(3/4)/c^(5/4)/(-4*a*c+b^2)^2/(-b+(-4*a*c+ 
b^2)^(1/2))^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.86 (sec) , antiderivative size = 506, normalized size of antiderivative = 0.82 \[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 c^2 \sqrt {x} \left (36 a^3 c+b^3 x^4 \left (3 b-c x^2\right )+a b x^2 \left (6 b^2+7 b c x^2+28 c^2 x^4\right )+a^2 \left (3 b^2+48 b c x^2+68 c^2 x^4\right )\right )}{\left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )^2}+32 c \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {\log \left (\sqrt {x}-\text {$\#$1}\right )}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]+\frac {8 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {3 b^4 \log \left (\sqrt {x}-\text {$\#$1}\right )-22 a b^2 c \log \left (\sqrt {x}-\text {$\#$1}\right )+28 a^2 c^2 \log \left (\sqrt {x}-\text {$\#$1}\right )+3 b^3 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4+6 a b c^2 \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{a \left (b^2-4 a c\right )}-\frac {3 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {8 b^6 \log \left (\sqrt {x}-\text {$\#$1}\right )-80 a b^4 c \log \left (\sqrt {x}-\text {$\#$1}\right )+223 a^2 b^2 c^2 \log \left (\sqrt {x}-\text {$\#$1}\right )-140 a^3 c^3 \log \left (\sqrt {x}-\text {$\#$1}\right )+8 b^5 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4-17 a b^3 c^2 \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4-36 a^2 b c^3 \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{a \left (b^2-4 a c\right )^2}}{64 c^3} \] Input:

Integrate[x^(15/2)/(a + b*x^2 + c*x^4)^3,x]
 

Output:

((-4*c^2*Sqrt[x]*(36*a^3*c + b^3*x^4*(3*b - c*x^2) + a*b*x^2*(6*b^2 + 7*b* 
c*x^2 + 28*c^2*x^4) + a^2*(3*b^2 + 48*b*c*x^2 + 68*c^2*x^4)))/((b^2 - 4*a* 
c)^2*(a + b*x^2 + c*x^4)^2) + 32*c*RootSum[a + b*#1^4 + c*#1^8 & , Log[Sqr 
t[x] - #1]/(b*#1^3 + 2*c*#1^7) & ] + (8*RootSum[a + b*#1^4 + c*#1^8 & , (3 
*b^4*Log[Sqrt[x] - #1] - 22*a*b^2*c*Log[Sqrt[x] - #1] + 28*a^2*c^2*Log[Sqr 
t[x] - #1] + 3*b^3*c*Log[Sqrt[x] - #1]*#1^4 + 6*a*b*c^2*Log[Sqrt[x] - #1]* 
#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(a*(b^2 - 4*a*c)) - (3*RootSum[a + b*#1^4 + 
 c*#1^8 & , (8*b^6*Log[Sqrt[x] - #1] - 80*a*b^4*c*Log[Sqrt[x] - #1] + 223* 
a^2*b^2*c^2*Log[Sqrt[x] - #1] - 140*a^3*c^3*Log[Sqrt[x] - #1] + 8*b^5*c*Lo 
g[Sqrt[x] - #1]*#1^4 - 17*a*b^3*c^2*Log[Sqrt[x] - #1]*#1^4 - 36*a^2*b*c^3* 
Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(a*(b^2 - 4*a*c)^2))/(64* 
c^3)
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 525, normalized size of antiderivative = 0.85, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1435, 1701, 27, 1822, 25, 1826, 1752, 756, 218, 221}

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.

\(\displaystyle \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 1435

\(\displaystyle 2 \int \frac {x^8}{\left (c x^4+b x^2+a\right )^3}d\sqrt {x}\)

\(\Big \downarrow \) 1701

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\int \frac {3 x^4 \left (6 a-b x^2\right )}{\left (c x^4+b x^2+a\right )^2}d\sqrt {x}}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \int \frac {x^4 \left (6 a-b x^2\right )}{\left (c x^4+b x^2+a\right )^2}d\sqrt {x}}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1822

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (-\frac {\int -\frac {x^2 \left (\left (b^2+12 a c\right ) x^2+40 a b\right )}{c x^4+b x^2+a}d\sqrt {x}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\int \frac {x^2 \left (\left (b^2+12 a c\right ) x^2+40 a b\right )}{c x^4+b x^2+a}d\sqrt {x}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1826

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\frac {\sqrt {x} \left (12 a c+b^2\right )}{c}-\frac {\int \frac {b \left (b^2-28 a c\right ) x^2+a \left (b^2+12 a c\right )}{c x^4+b x^2+a}d\sqrt {x}}{c}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1752

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\frac {\sqrt {x} \left (12 a c+b^2\right )}{c}-\frac {\frac {1}{2} \left (-\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}d\sqrt {x}+\frac {1}{2} \left (\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}d\sqrt {x}}{c}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 756

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\frac {\sqrt {x} \left (12 a c+b^2\right )}{c}-\frac {\frac {1}{2} \left (\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}\right )+\frac {1}{2} \left (-\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}\right )}{c}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\frac {\sqrt {x} \left (12 a c+b^2\right )}{c}-\frac {\frac {1}{2} \left (\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (-\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{c}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle 2 \left (\frac {x^{9/2} \left (2 a+b x^2\right )}{8 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 \left (\frac {\frac {\sqrt {x} \left (12 a c+b^2\right )}{c}-\frac {\frac {1}{2} \left (\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (-\frac {-24 a^2 c^2-30 a b^2 c+b^4}{\sqrt {b^2-4 a c}}-28 a b c+b^3\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{c}}{4 \left (b^2-4 a c\right )}-\frac {x^{5/2} \left (x^2 \left (12 a c+b^2\right )+8 a b\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{8 \left (b^2-4 a c\right )}\right )\)

Input:

Int[x^(15/2)/(a + b*x^2 + c*x^4)^3,x]
 

Output:

2*((x^(9/2)*(2*a + b*x^2))/(8*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (3*(- 
1/4*(x^(5/2)*(8*a*b + (b^2 + 12*a*c)*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x 
^4)) + (((b^2 + 12*a*c)*Sqrt[x])/c - (((b^3 - 28*a*b*c + (b^4 - 30*a*b^2*c 
 - 24*a^2*c^2)/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b 
- Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4 
))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2 
^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))))/2 + ((b^3 - 28*a*b*c - (b 
^4 - 30*a*b^2*c - 24*a^2*c^2)/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4 
)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^2 
 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4* 
a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))))/2)/c)/(4*( 
b^2 - 4*a*c))))/(8*(b^2 - 4*a*c)))
 

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 218

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

rule 221

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

rule 756

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

rule 1435

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> With[{k = Denominator[m]}, Simp[k/d   Subst[Int[x^(k*(m + 1) - 1)*(a + b 
*(x^(2*k)/d^2) + c*(x^(4*k)/d^4))^p, x], x, (d*x)^(1/k)], x]] /; FreeQ[{a, 
b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
 

rule 1701

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x 
_Symbol] :> Simp[(-d^(2*n - 1))*(d*x)^(m - 2*n + 1)*(2*a + b*x^n)*((a + b*x 
^n + c*x^(2*n))^(p + 1)/(n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[d^(2*n)/(n*(p 
 + 1)*(b^2 - 4*a*c))   Int[(d*x)^(m - 2*n)*(2*a*(m - 2*n + 1) + b*(m + n*(2 
*p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c 
, d}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && ILtQ[p, -1 
] && GtQ[m, 2*n - 1]
 

rule 1752

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x 
_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) 
   Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   I 
nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 
, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 
 - 4*a*c] ||  !IGtQ[n/2, 0])
 

rule 1822

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( 
c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Simp[f^(n - 1)*(f*x)^(m - n + 1)*(a + 
b*x^n + c*x^(2*n))^(p + 1)*((b*d - 2*a*e - (b*e - 2*c*d)*x^n)/(n*(p + 1)*(b 
^2 - 4*a*c))), x] + Simp[f^n/(n*(p + 1)*(b^2 - 4*a*c))   Int[(f*x)^(m - n)* 
(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[(n - m - 1)*(b*d - 2*a*e) + (2*n*p + 2 
*n + m + 1)*(b*e - 2*c*d)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] & 
& EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m, 
 n - 1] && IntegerQ[p]
 

rule 1826

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( 
c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[e*f^(n - 1)*(f*x)^(m - n + 1)*((a 
+ b*x^n + c*x^(2*n))^(p + 1)/(c*(m + n*(2*p + 1) + 1))), x] - Simp[f^n/(c*( 
m + n*(2*p + 1) + 1))   Int[(f*x)^(m - n)*(a + b*x^n + c*x^(2*n))^p*Simp[a* 
e*(m - n + 1) + (b*e*(m + n*p + 1) - c*d*(m + n*(2*p + 1) + 1))*x^n, x], x] 
, x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 
 0] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*(2*p + 1) + 1, 0] && Intege 
rQ[p]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.52 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.45

method	result	size
derivativedivides	\(\frac {-\frac {3 a^{2} \left (12 a c +b^{2}\right ) \sqrt {x}}{16 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {3 a b \left (8 a c +b^{2}\right ) x^{\frac {5}{2}}}{8 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (68 a^{2} c^{2}+7 a \,b^{2} c +3 b^{4}\right ) x^{\frac {9}{2}}}{16 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {b \left (28 a c -b^{2}\right ) x^{\frac {13}{2}}}{16 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (b \left (-28 a c +b^{2}\right ) \textit {\_R}^{4}+12 c \,a^{2}+b^{2} a \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{64 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(275\)
default	\(\frac {-\frac {3 a^{2} \left (12 a c +b^{2}\right ) \sqrt {x}}{16 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {3 a b \left (8 a c +b^{2}\right ) x^{\frac {5}{2}}}{8 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (68 a^{2} c^{2}+7 a \,b^{2} c +3 b^{4}\right ) x^{\frac {9}{2}}}{16 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {b \left (28 a c -b^{2}\right ) x^{\frac {13}{2}}}{16 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (b \left (-28 a c +b^{2}\right ) \textit {\_R}^{4}+12 c \,a^{2}+b^{2} a \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{64 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(275\)

Input:

int(x^(15/2)/(c*x^4+b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

2*(-3/32*a^2*(12*a*c+b^2)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(1/2)-3/16/c*a*b* 
(8*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(5/2)-1/32*(68*a^2*c^2+7*a*b^2*c+ 
3*b^4)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^(9/2)-1/32*b*(28*a*c-b^2)/(16*a^2*c^ 
2-8*a*b^2*c+b^4)*x^(13/2))/(c*x^4+b*x^2+a)^2+3/64/c/(16*a^2*c^2-8*a*b^2*c+ 
b^4)*sum((b*(-28*a*c+b^2)*_R^4+12*c*a^2+b^2*a)/(2*_R^7*c+_R^3*b)*ln(x^(1/2 
)-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 14789 vs. \(2 (521) = 1042\).

Time = 8.92 (sec) , antiderivative size = 14789, normalized size of antiderivative = 23.97 \[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(15/2)/(c*x**4+b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\int { \frac {x^{\frac {15}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{3}} \,d x } \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")
 

Output:

1/16*(3*(b^2*c + 12*a*c^2)*x^(17/2) + (7*b^3 + 44*a*b*c)*x^(13/2) + 24*a^2 
*b*x^(5/2) + (35*a*b^2 + 4*a^2*c)*x^(9/2))/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^ 
2*c^4)*x^8 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^6 + a^2*b^4 - 8*a^3* 
b^2*c + 16*a^4*c^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^4 + 2*(a*b^5 - 8*a^2 
*b^3*c + 16*a^3*b*c^2)*x^2) - integrate(3/32*((b^2 + 12*a*c)*x^(7/2) + 40* 
a*b*x^(3/2))/(a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16 
*a^2*c^3)*x^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*x^2), x)
 

Giac [F]

\[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\int { \frac {x^{\frac {15}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{3}} \,d x } \] Input:

integrate(x^(15/2)/(c*x^4+b*x^2+a)^3,x, algorithm="giac")
 

Output:

integrate(x^(15/2)/(c*x^4 + b*x^2 + a)^3, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 22.57 (sec) , antiderivative size = 50970, normalized size of antiderivative = 82.61 \[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

int(x^(15/2)/(a + b*x^2 + c*x^4)^3,x)
 

Output:

atan(((((3*(3159*a^3*b^14 - 20155392*a^10*c^7 - 367497*a^4*b^12*c + 159002 
19*a^5*b^10*c^2 - 299549340*a^6*b^8*c^3 + 1945179360*a^7*b^6*c^4 + 2840323 
968*a^8*b^4*c^5 + 164042496*a^9*b^2*c^6))/(65536*(b^18*c - 262144*a^9*c^10 
 - 36*a*b^16*c^2 + 576*a^2*b^14*c^3 - 5376*a^3*b^12*c^4 + 32256*a^4*b^10*c 
^5 - 129024*a^5*b^8*c^6 + 344064*a^6*b^6*c^7 - 589824*a^7*b^4*c^8 + 589824 
*a^8*b^2*c^9)) + ((3*(-(81*(b^33 + b^8*(-(4*a*c - b^2)^25)^(1/2) - 4711042 
25280*a^16*b*c^16 + 10509*a^2*b^29*c^2 - 394248*a^3*b^27*c^3 + 9219696*a^4 
*b^25*c^4 - 140233728*a^5*b^23*c^5 + 1424368896*a^6*b^21*c^6 - 9732052992* 
a^7*b^19*c^7 + 43376799744*a^8*b^17*c^8 - 108493078528*a^9*b^15*c^9 + 1315 
1174656*a^10*b^13*c^10 + 986354024448*a^11*b^11*c^11 - 3840358219776*a^12* 
b^9*c^12 + 7562531438592*a^13*b^7*c^13 - 8212262682624*a^14*b^5*c^14 + 421 
3765570560*a^15*b^3*c^15 + 1296*a^4*c^4*(-(4*a*c - b^2)^25)^(1/2) - 157*a* 
b^31*c + 4009*a^2*b^4*c^2*(-(4*a*c - b^2)^25)^(1/2) - 54648*a^3*b^2*c^3*(- 
(4*a*c - b^2)^25)^(1/2) - 107*a*b^6*c*(-(4*a*c - b^2)^25)^(1/2)))/(3355443 
2*(1099511627776*a^20*c^25 + b^40*c^5 - 80*a*b^38*c^6 + 3040*a^2*b^36*c^7 
- 72960*a^3*b^34*c^8 + 1240320*a^4*b^32*c^9 - 15876096*a^5*b^30*c^10 + 158 
760960*a^6*b^28*c^11 - 1270087680*a^7*b^26*c^12 + 8255569920*a^8*b^24*c^13 
 - 44029706240*a^9*b^22*c^14 + 193730707456*a^10*b^20*c^15 - 704475299840* 
a^11*b^18*c^16 + 2113425899520*a^12*b^16*c^17 - 5202279137280*a^13*b^14*c^ 
18 + 10404558274560*a^14*b^12*c^19 - 16647293239296*a^15*b^10*c^20 + 20...
 

Reduce [F]

\[ \int \frac {x^{15/2}}{\left (a+b x^2+c x^4\right )^3} \, dx=\int \frac {x^{\frac {15}{2}}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{3}}d x \] Input:

int(x^(15/2)/(c*x^4+b*x^2+a)^3,x)
 

Output:

int(x^(15/2)/(c*x^4+b*x^2+a)^3,x)