Integrand size = 19, antiderivative size = 317 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\frac {d^2 (3 b c-2 a d) x}{b^3}+\frac {d^3 x^5}{5 b^2}+\frac {(b c-a d)^3 x}{4 a b^3 \left (a+b x^4\right )}-\frac {3 (b c-a d)^2 (b c+3 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{13/4}}+\frac {3 (b c-a d)^2 (b c+3 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{13/4}}-\frac {3 (b c-a d)^2 (b c+3 a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{13/4}}+\frac {3 (b c-a d)^2 (b c+3 a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{13/4}} \]
d^2*(-2*a*d+3*b*c)*x/b^3+1/5*d^3*x^5/b^2+1/4*(-a*d+b*c)^3*x/a/b^3/(b*x^4+a )+3/16*(-a*d+b*c)^2*(3*a*d+b*c)*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/ 4)/b^(13/4)*2^(1/2)+3/16*(-a*d+b*c)^2*(3*a*d+b*c)*arctan(1+b^(1/4)*x*2^(1/ 2)/a^(1/4))/a^(7/4)/b^(13/4)*2^(1/2)-3/32*(-a*d+b*c)^2*(3*a*d+b*c)*ln(-a^( 1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))/a^(7/4)/b^(13/4)*2^(1/2)+3/32* (-a*d+b*c)^2*(3*a*d+b*c)*ln(a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2)) /a^(7/4)/b^(13/4)*2^(1/2)
Time = 0.24 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\frac {160 \sqrt [4]{b} d^2 (3 b c-2 a d) x+32 b^{5/4} d^3 x^5+\frac {40 \sqrt [4]{b} (b c-a d)^3 x}{a \left (a+b x^4\right )}-\frac {30 \sqrt {2} (b c-a d)^2 (b c+3 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{7/4}}+\frac {30 \sqrt {2} (b c-a d)^2 (b c+3 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{7/4}}-\frac {15 \sqrt {2} (b c-a d)^2 (b c+3 a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{7/4}}+\frac {15 \sqrt {2} (b c-a d)^2 (b c+3 a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{7/4}}}{160 b^{13/4}} \]
(160*b^(1/4)*d^2*(3*b*c - 2*a*d)*x + 32*b^(5/4)*d^3*x^5 + (40*b^(1/4)*(b*c - a*d)^3*x)/(a*(a + b*x^4)) - (30*Sqrt[2]*(b*c - a*d)^2*(b*c + 3*a*d)*Arc Tan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(7/4) + (30*Sqrt[2]*(b*c - a*d)^2* (b*c + 3*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(7/4) - (15*Sqrt[ 2]*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + S qrt[b]*x^2])/a^(7/4) + (15*Sqrt[2]*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/a^(7/4))/(160*b^(13/4))
Time = 0.49 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {915, 2009}
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.
| \(\displaystyle \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 915 |
| \(\displaystyle \int \left (\frac {d^2 (3 b c-2 a d)}{b^3}+\frac {3 b d x^4 (b c-a d)^2+(b c-a d)^2 (2 a d+b c)}{b^3 \left (a+b x^4\right )^2}+\frac {d^3 x^4}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) (b c-a d)^2 (3 a d+b c)}{8 \sqrt {2} a^{7/4} b^{13/4}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) (b c-a d)^2 (3 a d+b c)}{8 \sqrt {2} a^{7/4} b^{13/4}}-\frac {3 (b c-a d)^2 (3 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{13/4}}+\frac {3 (b c-a d)^2 (3 a d+b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{13/4}}+\frac {d^2 x (3 b c-2 a d)}{b^3}+\frac {x (b c-a d)^3}{4 a b^3 \left (a+b x^4\right )}+\frac {d^3 x^5}{5 b^2}\) |
(d^2*(3*b*c - 2*a*d)*x)/b^3 + (d^3*x^5)/(5*b^2) + ((b*c - a*d)^3*x)/(4*a*b ^3*(a + b*x^4)) - (3*(b*c - a*d)^2*(b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/ 4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*b^(13/4)) + (3*(b*c - a*d)^2*(b*c + 3*a *d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*b^(13/4)) - (3*(b*c - a*d)^2*(b*c + 3*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(16*Sqrt[2]*a^(7/4)*b^(13/4)) + (3*(b*c - a*d)^2*(b*c + 3*a *d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(16*Sqrt[2]*a^ (7/4)*b^(13/4))
3.2.68.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a , b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.01 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.48
| method | result | size |
| risch | \(\frac {d^{3} x^{5}}{5 b^{2}}-\frac {2 d^{3} a x}{b^{3}}+\frac {3 d^{2} c x}{b^{2}}-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{4 a \,b^{3} \left (b \,x^{4}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (3 a^{3} d^{3}-5 a^{2} b c \,d^{2}+a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16 b^{4} a}\) | \(151\) |
| default | \(-\frac {d^{2} \left (-\frac {1}{5} b d \,x^{5}+2 a d x -3 b c x \right )}{b^{3}}+\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{4 a \left (b \,x^{4}+a \right )}+\frac {3 \left (3 a^{3} d^{3}-5 a^{2} b c \,d^{2}+a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2}}}{b^{3}}\) | \(220\) |
1/5*d^3*x^5/b^2-2*d^3/b^3*a*x+3*d^2/b^2*c*x-1/4*(a^3*d^3-3*a^2*b*c*d^2+3*a *b^2*c^2*d-b^3*c^3)/a*x/b^3/(b*x^4+a)+3/16/b^4/a*sum((3*a^3*d^3-5*a^2*b*c* d^2+a*b^2*c^2*d+b^3*c^3)/_R^3*ln(x-_R),_R=RootOf(_Z^4*b+a))
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 1741, normalized size of antiderivative = 5.49 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\text {Too large to display} \]
1/80*(16*a*b^2*d^3*x^9 + 48*(5*a*b^2*c*d^2 - 3*a^2*b*d^3)*x^5 + 15*(a*b^4* x^4 + a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44* a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6* c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3*c^3*d^ 9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7*b^13)) ^(1/4)*log(3*a^2*b^3*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^ 6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932*a^9*b^3* c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^12)/(a^7* b^13))^(1/4) + 3*(b^3*c^3 + a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*x) - 15*(-I*a*b^4*x^4 - I*a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14*a^2*b^10 *c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^8 - 1932 *a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81*a^12*d^ 12)/(a^7*b^13))^(1/4)*log(3*I*a^2*b^3*(-(b^12*c^12 + 4*a*b^11*c^11*d - 14* a^2*b^10*c^10*d^2 - 44*a^3*b^9*c^9*d^3 + 127*a^4*b^8*c^8*d^4 + 136*a^5*b^7 *c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 328*a^7*b^5*c^5*d^7 + 1039*a^8*b^4*c^4*d^ 8 - 1932*a^9*b^3*c^3*d^9 + 1458*a^10*b^2*c^2*d^10 - 540*a^11*b*c*d^11 + 81 *a^12*d^12)/(a^7*b^13))^(1/4) + 3*(b^3*c^3 + a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*x) - 15*(I*a*b^4*x^4 + I*a^2*b^3)*(-(b^12*c^12 + 4*a*b^11*c...
Time = 86.25 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=x \left (- \frac {2 a d^{3}}{b^{3}} + \frac {3 c d^{2}}{b^{2}}\right ) + \frac {x \left (- a^{3} d^{3} + 3 a^{2} b c d^{2} - 3 a b^{2} c^{2} d + b^{3} c^{3}\right )}{4 a^{2} b^{3} + 4 a b^{4} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{7} b^{13} + 6561 a^{12} d^{12} - 43740 a^{11} b c d^{11} + 118098 a^{10} b^{2} c^{2} d^{10} - 156492 a^{9} b^{3} c^{3} d^{9} + 84159 a^{8} b^{4} c^{4} d^{8} + 26568 a^{7} b^{5} c^{5} d^{7} - 52164 a^{6} b^{6} c^{6} d^{6} + 11016 a^{5} b^{7} c^{7} d^{5} + 10287 a^{4} b^{8} c^{8} d^{4} - 3564 a^{3} b^{9} c^{9} d^{3} - 1134 a^{2} b^{10} c^{10} d^{2} + 324 a b^{11} c^{11} d + 81 b^{12} c^{12}, \left ( t \mapsto t \log {\left (\frac {16 t a^{2} b^{3}}{9 a^{3} d^{3} - 15 a^{2} b c d^{2} + 3 a b^{2} c^{2} d + 3 b^{3} c^{3}} + x \right )} \right )\right )} + \frac {d^{3} x^{5}}{5 b^{2}} \]
x*(-2*a*d**3/b**3 + 3*c*d**2/b**2) + x*(-a**3*d**3 + 3*a**2*b*c*d**2 - 3*a *b**2*c**2*d + b**3*c**3)/(4*a**2*b**3 + 4*a*b**4*x**4) + RootSum(65536*_t **4*a**7*b**13 + 6561*a**12*d**12 - 43740*a**11*b*c*d**11 + 118098*a**10*b **2*c**2*d**10 - 156492*a**9*b**3*c**3*d**9 + 84159*a**8*b**4*c**4*d**8 + 26568*a**7*b**5*c**5*d**7 - 52164*a**6*b**6*c**6*d**6 + 11016*a**5*b**7*c* *7*d**5 + 10287*a**4*b**8*c**8*d**4 - 3564*a**3*b**9*c**9*d**3 - 1134*a**2 *b**10*c**10*d**2 + 324*a*b**11*c**11*d + 81*b**12*c**12, Lambda(_t, _t*lo g(16*_t*a**2*b**3/(9*a**3*d**3 - 15*a**2*b*c*d**2 + 3*a*b**2*c**2*d + 3*b* *3*c**3) + x))) + d**3*x**5/(5*b**2)
Time = 0.29 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.28 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x}{4 \, {\left (a b^{4} x^{4} + a^{2} b^{3}\right )}} + \frac {b d^{3} x^{5} + 5 \, {\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} x}{5 \, b^{3}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (b^{3} c^{3} + a b^{2} c^{2} d - 5 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (b^{3} c^{3} + a b^{2} c^{2} d - 5 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (b^{3} c^{3} + a b^{2} c^{2} d - 5 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b^{3} c^{3} + a b^{2} c^{2} d - 5 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )}}{32 \, a b^{3}} \]
1/4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x/(a*b^4*x^4 + a^2 *b^3) + 1/5*(b*d^3*x^5 + 5*(3*b*c*d^2 - 2*a*d^3)*x)/b^3 + 3/32*(2*sqrt(2)* (b^3*c^3 + a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*arctan(1/2*sqrt(2)*(2* sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt( sqrt(a)*sqrt(b))) + 2*sqrt(2)*(b^3*c^3 + a*b^2*c^2*d - 5*a^2*b*c*d^2 + 3*a ^3*d^3)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sq rt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + sqrt(2)*(b^3*c^3 + a*b^2 *c^2*d - 5*a^2*b*c*d^2 + 3*a^3*d^3)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1 /4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - sqrt(2)*(b^3*c^3 + a*b^2*c^2*d - 5*a^ 2*b*c*d^2 + 3*a^3*d^3)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt( a))/(a^(3/4)*b^(1/4)))/(a*b^3)
Time = 0.27 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.56 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{4}} + \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{4}} + \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{4}} - \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} + \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{4}} + \frac {b^{3} c^{3} x - 3 \, a b^{2} c^{2} d x + 3 \, a^{2} b c d^{2} x - a^{3} d^{3} x}{4 \, {\left (b x^{4} + a\right )} a b^{3}} + \frac {b^{8} d^{3} x^{5} + 15 \, b^{8} c d^{2} x - 10 \, a b^{7} d^{3} x}{5 \, b^{10}} \]
3/16*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3 )^(1/4)*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x + s qrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^2*b^4) + 3/16*sqrt(2)*((a*b^3)^(1/4)*b ^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b^ 3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/ 4))/(a^2*b^4) + 3/32*sqrt(2)*((a*b^3)^(1/4)*b^3*c^3 + (a*b^3)^(1/4)*a*b^2* c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b^3)^(1/4)*a^3*d^3)*log(x^2 + s qrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^4) - 3/32*sqrt(2)*((a*b^3)^(1/4)* b^3*c^3 + (a*b^3)^(1/4)*a*b^2*c^2*d - 5*(a*b^3)^(1/4)*a^2*b*c*d^2 + 3*(a*b ^3)^(1/4)*a^3*d^3)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^4) + 1/4*(b^3*c^3*x - 3*a*b^2*c^2*d*x + 3*a^2*b*c*d^2*x - a^3*d^3*x)/((b*x^4 + a)*a*b^3) + 1/5*(b^8*d^3*x^5 + 15*b^8*c*d^2*x - 10*a*b^7*d^3*x)/b^10
Time = 5.72 (sec) , antiderivative size = 1616, normalized size of antiderivative = 5.10 \[ \int \frac {\left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx=\text {Too large to display} \]
(d^3*x^5)/(5*b^2) - x*((2*a*d^3)/b^3 - (3*c*d^2)/b^2) - (x*(a^3*d^3 - b^3* c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(4*a*(a*b^3 + b^4*x^4)) + (atan((((a *d - b*c)^2*(3*a*d + b*c)*((9*x*(9*a^6*d^6 + b^6*c^6 - 9*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 + 31*a^4*b^2*c^2*d^4 + 2*a*b^5*c^5*d - 30*a^5*b*c*d^5)) /(4*a^2*b^3) - (3*(a*d - b*c)^2*(3*a*d + b*c)*(36*a^3*d^3 + 12*b^3*c^3 + 1 2*a*b^2*c^2*d - 60*a^2*b*c*d^2))/(16*(-a)^(7/4)*b^(13/4)))*3i)/(16*(-a)^(7 /4)*b^(13/4)) + ((a*d - b*c)^2*(3*a*d + b*c)*((9*x*(9*a^6*d^6 + b^6*c^6 - 9*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 + 31*a^4*b^2*c^2*d^4 + 2*a*b^5*c^5*d - 30*a^5*b*c*d^5))/(4*a^2*b^3) + (3*(a*d - b*c)^2*(3*a*d + b*c)*(36*a^3*d ^3 + 12*b^3*c^3 + 12*a*b^2*c^2*d - 60*a^2*b*c*d^2))/(16*(-a)^(7/4)*b^(13/4 )))*3i)/(16*(-a)^(7/4)*b^(13/4)))/((3*(a*d - b*c)^2*(3*a*d + b*c)*((9*x*(9 *a^6*d^6 + b^6*c^6 - 9*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 + 31*a^4*b^2*c^ 2*d^4 + 2*a*b^5*c^5*d - 30*a^5*b*c*d^5))/(4*a^2*b^3) - (3*(a*d - b*c)^2*(3 *a*d + b*c)*(36*a^3*d^3 + 12*b^3*c^3 + 12*a*b^2*c^2*d - 60*a^2*b*c*d^2))/( 16*(-a)^(7/4)*b^(13/4))))/(16*(-a)^(7/4)*b^(13/4)) - (3*(a*d - b*c)^2*(3*a *d + b*c)*((9*x*(9*a^6*d^6 + b^6*c^6 - 9*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d ^3 + 31*a^4*b^2*c^2*d^4 + 2*a*b^5*c^5*d - 30*a^5*b*c*d^5))/(4*a^2*b^3) + ( 3*(a*d - b*c)^2*(3*a*d + b*c)*(36*a^3*d^3 + 12*b^3*c^3 + 12*a*b^2*c^2*d - 60*a^2*b*c*d^2))/(16*(-a)^(7/4)*b^(13/4))))/(16*(-a)^(7/4)*b^(13/4))))*(a* d - b*c)^2*(3*a*d + b*c)*3i)/(8*(-a)^(7/4)*b^(13/4)) + (3*atan(((3*(a*d...