\(\int \frac {(a+b x^4)^3}{(c+d x^4)^2} \, dx\) [215]

Optimal result

Integrand size = 19, antiderivative size = 249 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=-\frac {b^2 (2 b c-3 a d) x}{d^3}+\frac {b^3 x^5}{5 d^2}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^4\right )}-\frac {3 (b c-a d)^2 (3 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{13/4}}+\frac {3 (b c-a d)^2 (3 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{13/4}}+\frac {3 (b c-a d)^2 (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{8 \sqrt {2} c^{7/4} d^{13/4}} \] Output:

-b^2*(-3*a*d+2*b*c)*x/d^3+1/5*b^3*x^5/d^2-1/4*(-a*d+b*c)^3*x/c/d^3/(d*x^4+ 
c)+3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arctan(-1+2^(1/2)*d^(1/4)*x/c^(1/4))*2^(1 
/2)/c^(7/4)/d^(13/4)+3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arctan(1+2^(1/2)*d^(1/4 
)*x/c^(1/4))*2^(1/2)/c^(7/4)/d^(13/4)+3/16*(-a*d+b*c)^2*(a*d+3*b*c)*arctan 
h(2^(1/2)*c^(1/4)*d^(1/4)*x/(c^(1/2)+d^(1/2)*x^2))*2^(1/2)/c^(7/4)/d^(13/4 
)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\frac {-160 b^2 \sqrt [4]{d} (2 b c-3 a d) x+32 b^3 d^{5/4} x^5+\frac {40 \sqrt [4]{d} (-b c+a d)^3 x}{c \left (c+d x^4\right )}-\frac {30 \sqrt {2} (b c-a d)^2 (3 b c+a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{7/4}}+\frac {30 \sqrt {2} (b c-a d)^2 (3 b c+a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{7/4}}-\frac {15 \sqrt {2} (b c-a d)^2 (3 b c+a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{7/4}}+\frac {15 \sqrt {2} (b c-a d)^2 (3 b c+a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{c^{7/4}}}{160 d^{13/4}} \] Input:

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

Output:

(-160*b^2*d^(1/4)*(2*b*c - 3*a*d)*x + 32*b^3*d^(5/4)*x^5 + (40*d^(1/4)*(-( 
b*c) + a*d)^3*x)/(c*(c + d*x^4)) - (30*Sqrt[2]*(b*c - a*d)^2*(3*b*c + a*d) 
*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(7/4) + (30*Sqrt[2]*(b*c - a*d 
)^2*(3*b*c + a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(7/4) - (15*S 
qrt[2]*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x 
 + Sqrt[d]*x^2])/c^(7/4) + (15*Sqrt[2]*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqr 
t[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/c^(7/4))/(160*d^(13/4))
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.28, 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.

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

\(\Big \downarrow \) 915

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 (b c-a d)^2 (a d+3 b c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt {2} c^{7/4} d^{13/4}}+\frac {3 (b c-a d)^2 (a d+3 b c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{8 \sqrt {2} c^{7/4} d^{13/4}}-\frac {b^2 x (2 b c-3 a d)}{d^3}-\frac {3 (b c-a d)^2 (a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{13/4}}+\frac {3 (b c-a d)^2 (a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{16 \sqrt {2} c^{7/4} d^{13/4}}-\frac {x (b c-a d)^3}{4 c d^3 \left (c+d x^4\right )}+\frac {b^3 x^5}{5 d^2}\)

Input:

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

Output:

-((b^2*(2*b*c - 3*a*d)*x)/d^3) + (b^3*x^5)/(5*d^2) - ((b*c - a*d)^3*x)/(4* 
c*d^3*(c + d*x^4)) - (3*(b*c - a*d)^2*(3*b*c + a*d)*ArcTan[1 - (Sqrt[2]*d^ 
(1/4)*x)/c^(1/4)])/(8*Sqrt[2]*c^(7/4)*d^(13/4)) + (3*(b*c - a*d)^2*(3*b*c 
+ a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(8*Sqrt[2]*c^(7/4)*d^(13/4 
)) - (3*(b*c - a*d)^2*(3*b*c + a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)* 
x + Sqrt[d]*x^2])/(16*Sqrt[2]*c^(7/4)*d^(13/4)) + (3*(b*c - a*d)^2*(3*b*c 
+ a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(16*Sqrt[2] 
*c^(7/4)*d^(13/4))
 

Defintions of rubi rules used

rule 915

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {b^{3} x^{5}}{5 d^{2}}+\frac {3 b^{2} a x}{d^{2}}-\frac {2 b^{3} c x}{d^{3}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{4 c \,d^{3} \left (d \,x^{4}+c \right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{4}+c \right )}{\sum }\frac {\left (a^{3} d^{3}+a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +3 b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{16 d^{4} c}\)	\(151\)
default	\(\frac {b^{2} \left (\frac {1}{5} x^{5} b d +3 a d x -2 c b x \right )}{d^{3}}+\frac {\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{4 c \left (d \,x^{4}+c \right )}+\frac {3 \left (a^{3} d^{3}+a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +3 b^{3} c^{3}\right ) \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c^{2}}}{d^{3}}\)	\(219\)

Input:

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

Output:

1/5*b^3*x^5/d^2+3*b^2/d^2*a*x-2*b^3/d^3*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)/c*x/d^3/(d*x^4+c)+3/16/d^4/c*sum((a^3*d^3+a^2*b*c*d^2- 
5*a*b^2*c^2*d+3*b^3*c^3)/_R^3*ln(x-_R),_R=RootOf(_Z^4*d+c))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 1742, normalized size of antiderivative = 7.00 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/80*(16*b^3*c*d^2*x^9 - 48*(3*b^3*c^2*d - 5*a*b^2*c*d^2)*x^5 + 15*(c*d^4* 
x^4 + c^2*d^3)*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c^10*d^ 
2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^5 - 64 
4*a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44*a^9*b^3 
*c^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^12)/(c^7*d^13)) 
^(1/4)*log(3*c^2*d^3*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a^2*b^10*c 
^10*d^2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b^7*c^7*d^ 
5 - 644*a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d^8 - 44*a 
^9*b^3*c^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^12)/(c^7* 
d^13))^(1/4) + 3*(3*b^3*c^3 - 5*a*b^2*c^2*d + a^2*b*c*d^2 + a^3*d^3)*x) - 
15*(-I*c*d^4*x^4 - I*c^2*d^3)*(-(81*b^12*c^12 - 540*a*b^11*c^11*d + 1458*a 
^2*b^10*c^10*d^2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 328*a^5*b 
^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b^4*c^4*d 
^8 - 44*a^9*b^3*c^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + a^12*d^ 
12)/(c^7*d^13))^(1/4)*log(3*I*c^2*d^3*(-(81*b^12*c^12 - 540*a*b^11*c^11*d 
+ 1458*a^2*b^10*c^10*d^2 - 1932*a^3*b^9*c^9*d^3 + 1039*a^4*b^8*c^8*d^4 + 3 
28*a^5*b^7*c^7*d^5 - 644*a^6*b^6*c^6*d^6 + 136*a^7*b^5*c^5*d^7 + 127*a^8*b 
^4*c^4*d^8 - 44*a^9*b^3*c^3*d^9 - 14*a^10*b^2*c^2*d^10 + 4*a^11*b*c*d^11 + 
 a^12*d^12)/(c^7*d^13))^(1/4) + 3*(3*b^3*c^3 - 5*a*b^2*c^2*d + a^2*b*c*d^2 
 + a^3*d^3)*x) - 15*(I*c*d^4*x^4 + I*c^2*d^3)*(-(81*b^12*c^12 - 540*a*b...
 

Sympy [A] (verification not implemented)

Time = 10.40 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\frac {b^{3} x^{5}}{5 d^{2}} + x \left (\frac {3 a b^{2}}{d^{2}} - \frac {2 b^{3} c}{d^{3}}\right ) + \frac {x \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{4 c^{2} d^{3} + 4 c d^{4} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} c^{7} d^{13} + 81 a^{12} d^{12} + 324 a^{11} b c d^{11} - 1134 a^{10} b^{2} c^{2} d^{10} - 3564 a^{9} b^{3} c^{3} d^{9} + 10287 a^{8} b^{4} c^{4} d^{8} + 11016 a^{7} b^{5} c^{5} d^{7} - 52164 a^{6} b^{6} c^{6} d^{6} + 26568 a^{5} b^{7} c^{7} d^{5} + 84159 a^{4} b^{8} c^{8} d^{4} - 156492 a^{3} b^{9} c^{9} d^{3} + 118098 a^{2} b^{10} c^{10} d^{2} - 43740 a b^{11} c^{11} d + 6561 b^{12} c^{12}, \left ( t \mapsto t \log {\left (\frac {16 t c^{2} d^{3}}{3 a^{3} d^{3} + 3 a^{2} b c d^{2} - 15 a b^{2} c^{2} d + 9 b^{3} c^{3}} + x \right )} \right )\right )} \] Input:

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

Output:

b**3*x**5/(5*d**2) + x*(3*a*b**2/d**2 - 2*b**3*c/d**3) + x*(a**3*d**3 - 3* 
a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(4*c**2*d**3 + 4*c*d**4*x**4) 
 + RootSum(65536*_t**4*c**7*d**13 + 81*a**12*d**12 + 324*a**11*b*c*d**11 - 
 1134*a**10*b**2*c**2*d**10 - 3564*a**9*b**3*c**3*d**9 + 10287*a**8*b**4*c 
**4*d**8 + 11016*a**7*b**5*c**5*d**7 - 52164*a**6*b**6*c**6*d**6 + 26568*a 
**5*b**7*c**7*d**5 + 84159*a**4*b**8*c**8*d**4 - 156492*a**3*b**9*c**9*d** 
3 + 118098*a**2*b**10*c**10*d**2 - 43740*a*b**11*c**11*d + 6561*b**12*c**1 
2, Lambda(_t, _t*log(16*_t*c**2*d**3/(3*a**3*d**3 + 3*a**2*b*c*d**2 - 15*a 
*b**2*c**2*d + 9*b**3*c**3) + x)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (198) = 396\).

Time = 0.12 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=-\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x}{4 \, {\left (c d^{4} x^{4} + c^{2} d^{3}\right )}} + \frac {b^{3} d x^{5} - 5 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x}{5 \, d^{3}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (3 \, b^{3} c^{3} - 5 \, a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (3 \, b^{3} c^{3} - 5 \, a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (3 \, b^{3} c^{3} - 5 \, a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, b^{3} c^{3} - 5 \, a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}\right )}}{32 \, c d^{3}} \] Input:

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

Output:

-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/(c*d^4*x^4 + c^ 
2*d^3) + 1/5*(b^3*d*x^5 - 5*(2*b^3*c - 3*a*b^2*d)*x)/d^3 + 3/32*(2*sqrt(2) 
*(3*b^3*c^3 - 5*a*b^2*c^2*d + a^2*b*c*d^2 + a^3*d^3)*arctan(1/2*sqrt(2)*(2 
*sqrt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt 
(sqrt(c)*sqrt(d))) + 2*sqrt(2)*(3*b^3*c^3 - 5*a*b^2*c^2*d + a^2*b*c*d^2 + 
a^3*d^3)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x - sqrt(2)*c^(1/4)*d^(1/4))/sqrt(s 
qrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(3*b^3*c^3 - 5* 
a*b^2*c^2*d + a^2*b*c*d^2 + a^3*d^3)*log(sqrt(d)*x^2 + sqrt(2)*c^(1/4)*d^( 
1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4)) - sqrt(2)*(3*b^3*c^3 - 5*a*b^2*c^2*d + 
 a^2*b*c*d^2 + a^3*d^3)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt 
(c))/(c^(3/4)*d^(1/4)))/(c*d^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (198) = 396\).

Time = 0.13 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{4}} + \frac {3 \, \sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{16 \, c^{2} d^{4}} + \frac {3 \, \sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{4}} - \frac {3 \, \sqrt {2} {\left (3 \, \left (c d^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 5 \, \left (c d^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + \left (c d^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} + \left (c d^{3}\right )^{\frac {1}{4}} a^{3} d^{3}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{32 \, c^{2} d^{4}} - \frac {b^{3} c^{3} x - 3 \, a b^{2} c^{2} d x + 3 \, a^{2} b c d^{2} x - a^{3} d^{3} x}{4 \, {\left (d x^{4} + c\right )} c d^{3}} + \frac {b^{3} d^{8} x^{5} - 10 \, b^{3} c d^{7} x + 15 \, a b^{2} d^{8} x}{5 \, d^{10}} \] Input:

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

Output:

3/16*sqrt(2)*(3*(c*d^3)^(1/4)*b^3*c^3 - 5*(c*d^3)^(1/4)*a*b^2*c^2*d + (c*d 
^3)^(1/4)*a^2*b*c*d^2 + (c*d^3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x + s 
qrt(2)*(c/d)^(1/4))/(c/d)^(1/4))/(c^2*d^4) + 3/16*sqrt(2)*(3*(c*d^3)^(1/4) 
*b^3*c^3 - 5*(c*d^3)^(1/4)*a*b^2*c^2*d + (c*d^3)^(1/4)*a^2*b*c*d^2 + (c*d^ 
3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(c/d)^(1/4))/(c/d)^(1/ 
4))/(c^2*d^4) + 3/32*sqrt(2)*(3*(c*d^3)^(1/4)*b^3*c^3 - 5*(c*d^3)^(1/4)*a* 
b^2*c^2*d + (c*d^3)^(1/4)*a^2*b*c*d^2 + (c*d^3)^(1/4)*a^3*d^3)*log(x^2 + s 
qrt(2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^4) - 3/32*sqrt(2)*(3*(c*d^3)^(1/4 
)*b^3*c^3 - 5*(c*d^3)^(1/4)*a*b^2*c^2*d + (c*d^3)^(1/4)*a^2*b*c*d^2 + (c*d 
^3)^(1/4)*a^3*d^3)*log(x^2 - sqrt(2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^2*d^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)/((d*x^4 
+ c)*c*d^3) + 1/5*(b^3*d^8*x^5 - 10*b^3*c*d^7*x + 15*a*b^2*d^8*x)/d^10
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1615, normalized size of antiderivative = 6.49 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

x*((3*a*b^2)/d^2 - (2*b^3*c)/d^3) + (b^3*x^5)/(5*d^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*c*(c*d^3 + d^4*x^4)) + (atan((((a 
*d - b*c)^2*(a*d + 3*b*c)*((9*x*(a^6*d^6 + 9*b^6*c^6 + 31*a^2*b^4*c^4*d^2 
- 4*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 - 30*a*b^5*c^5*d + 2*a^5*b*c*d^5)) 
/(4*c^2*d^3) - (3*(a*d - b*c)^2*(a*d + 3*b*c)*(12*a^3*d^3 + 36*b^3*c^3 - 6 
0*a*b^2*c^2*d + 12*a^2*b*c*d^2))/(16*(-c)^(7/4)*d^(13/4)))*3i)/(16*(-c)^(7 
/4)*d^(13/4)) + ((a*d - b*c)^2*(a*d + 3*b*c)*((9*x*(a^6*d^6 + 9*b^6*c^6 + 
31*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 - 30*a*b^5*c^5* 
d + 2*a^5*b*c*d^5))/(4*c^2*d^3) + (3*(a*d - b*c)^2*(a*d + 3*b*c)*(12*a^3*d 
^3 + 36*b^3*c^3 - 60*a*b^2*c^2*d + 12*a^2*b*c*d^2))/(16*(-c)^(7/4)*d^(13/4 
)))*3i)/(16*(-c)^(7/4)*d^(13/4)))/((3*(a*d - b*c)^2*(a*d + 3*b*c)*((9*x*(a 
^6*d^6 + 9*b^6*c^6 + 31*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^ 
2*d^4 - 30*a*b^5*c^5*d + 2*a^5*b*c*d^5))/(4*c^2*d^3) - (3*(a*d - b*c)^2*(a 
*d + 3*b*c)*(12*a^3*d^3 + 36*b^3*c^3 - 60*a*b^2*c^2*d + 12*a^2*b*c*d^2))/( 
16*(-c)^(7/4)*d^(13/4))))/(16*(-c)^(7/4)*d^(13/4)) - (3*(a*d - b*c)^2*(a*d 
 + 3*b*c)*((9*x*(a^6*d^6 + 9*b^6*c^6 + 31*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3* 
d^3 - 9*a^4*b^2*c^2*d^4 - 30*a*b^5*c^5*d + 2*a^5*b*c*d^5))/(4*c^2*d^3) + ( 
3*(a*d - b*c)^2*(a*d + 3*b*c)*(12*a^3*d^3 + 36*b^3*c^3 - 60*a*b^2*c^2*d + 
12*a^2*b*c*d^2))/(16*(-c)^(7/4)*d^(13/4))))/(16*(-c)^(7/4)*d^(13/4))))*(a* 
d - b*c)^2*(a*d + 3*b*c)*3i)/(8*(-c)^(7/4)*d^(13/4)) + (3*atan(((3*(a*d...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 1347, normalized size of antiderivative = 5.41 \[ \int \frac {\left (a+b x^4\right )^3}{\left (c+d x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 30*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt( 
d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*c*d**3 - 30*d**(3/4)*c**(1/4)*sqrt 
(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt 
(2)))*a**3*d**4*x**4 - 30*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4 
)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c**2*d**2 - 3 
0*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x) 
/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c*d**3*x**4 + 150*d**(3/4)*c**(1/4)*s 
qrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*s 
qrt(2)))*a*b**2*c**3*d + 150*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**( 
1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a*b**2*c**2*d**2* 
x**4 - 90*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sq 
rt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*b**3*c**4 - 90*d**(3/4)*c**(1/4)*sqr 
t(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqr 
t(2)))*b**3*c**3*d*x**4 + 30*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**( 
1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**3*c*d**3 + 30* 
d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/( 
d**(1/4)*c**(1/4)*sqrt(2)))*a**3*d**4*x**4 + 30*d**(3/4)*c**(1/4)*sqrt(2)* 
atan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)) 
)*a**2*b*c**2*d**2 + 30*d**(3/4)*c**(1/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)* 
sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*a**2*b*c*d**3*x**4 ...