\(\int \frac {(a+b x^3)^2}{c+d x^3} \, dx\) [20]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 173 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=-\frac {b (b c-2 a d) x}{d^2}+\frac {b^2 x^4}{4 d}-\frac {(b c-a d)^2 \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{7/3}}+\frac {(b c-a d)^2 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{7/3}}-\frac {(b c-a d)^2 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{7/3}} \] Output:

-b*(-2*a*d+b*c)*x/d^2+1/4*b^2*x^4/d-1/3*(-a*d+b*c)^2*arctan(1/3*(c^(1/3)-2 
*d^(1/3)*x)*3^(1/2)/c^(1/3))*3^(1/2)/c^(2/3)/d^(7/3)+1/3*(-a*d+b*c)^2*ln(c 
^(1/3)+d^(1/3)*x)/c^(2/3)/d^(7/3)-1/6*(-a*d+b*c)^2*ln(c^(2/3)-c^(1/3)*d^(1 
/3)*x+d^(2/3)*x^2)/c^(2/3)/d^(7/3)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\frac {-12 b c^{2/3} \sqrt [3]{d} (b c-2 a d) x+3 b^2 c^{2/3} d^{4/3} x^4+4 \sqrt {3} (b c-a d)^2 \arctan \left (\frac {-\sqrt [3]{c}+2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )+4 (b c-a d)^2 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )-2 (b c-a d)^2 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{12 c^{2/3} d^{7/3}} \] Input:

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

Output:

(-12*b*c^(2/3)*d^(1/3)*(b*c - 2*a*d)*x + 3*b^2*c^(2/3)*d^(4/3)*x^4 + 4*Sqr 
t[3]*(b*c - a*d)^2*ArcTan[(-c^(1/3) + 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))] + 4* 
(b*c - a*d)^2*Log[c^(1/3) + d^(1/3)*x] - 2*(b*c - a*d)^2*Log[c^(2/3) - c^( 
1/3)*d^(1/3)*x + d^(2/3)*x^2])/(12*c^(2/3)*d^(7/3))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 173, 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.

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

\(\Big \downarrow \) 915

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {(b c-a d)^2 \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{7/3}}-\frac {(b c-a d)^2 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{7/3}}+\frac {(b c-a d)^2 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{7/3}}-\frac {b x (b c-2 a d)}{d^2}+\frac {b^2 x^4}{4 d}\)

Input:

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

Output:

-((b*(b*c - 2*a*d)*x)/d^2) + (b^2*x^4)/(4*d) - ((b*c - a*d)^2*ArcTan[(c^(1 
/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*c^(2/3)*d^(7/3)) + ((b*c - 
 a*d)^2*Log[c^(1/3) + d^(1/3)*x])/(3*c^(2/3)*d^(7/3)) - ((b*c - a*d)^2*Log 
[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(6*c^(2/3)*d^(7/3))
 

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 = 0.89 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.45

method result size
risch \(\frac {b^{2} x^{4}}{4 d}+\frac {2 b a x}{d}-\frac {b^{2} c x}{d^{2}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+c \right )}{\sum }\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 d^{3}}\) \(78\)
default \(\frac {b \left (\frac {1}{4} b d \,x^{4}+2 a d x -b c x \right )}{d^{2}}+\frac {\left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right ) \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{d^{2}}\) \(140\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 505, normalized size of antiderivative = 2.92 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\left [\frac {3 \, b^{2} c^{2} d^{2} x^{4} + 6 \, \sqrt {\frac {1}{3}} {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \log \left (\frac {2 \, c d x^{3} - 3 \, \left (c^{2} d\right )^{\frac {1}{3}} c x - c^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, c d x^{2} + \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{d x^{3} + c}\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right ) - 12 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2}\right )} x}{12 \, c^{2} d^{3}}, \frac {3 \, b^{2} c^{2} d^{2} x^{4} + 12 \, \sqrt {\frac {1}{3}} {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{c^{2}}\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (c^{2} d\right )^{\frac {2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right ) - 12 \, {\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2}\right )} x}{12 \, c^{2} d^{3}}\right ] \] Input:

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

Output:

[1/12*(3*b^2*c^2*d^2*x^4 + 6*sqrt(1/3)*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c* 
d^3)*sqrt(-(c^2*d)^(1/3)/d)*log((2*c*d*x^3 - 3*(c^2*d)^(1/3)*c*x - c^2 + 3 
*sqrt(1/3)*(2*c*d*x^2 + (c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt(-(c^2*d)^( 
1/3)/d))/(d*x^3 + c)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(c^2*d)^(2/3)*lo 
g(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) + 4*(b^2*c^2 - 2*a*b*c*d + 
a^2*d^2)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 12*(b^2*c^3*d - 2*a*b* 
c^2*d^2)*x)/(c^2*d^3), 1/12*(3*b^2*c^2*d^2*x^4 + 12*sqrt(1/3)*(b^2*c^3*d - 
 2*a*b*c^2*d^2 + a^2*c*d^3)*sqrt((c^2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(c^2 
*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt((c^2*d)^(1/3)/d)/c^2) - 2*(b^2*c^2 - 2 
*a*b*c*d + a^2*d^2)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^ 
(1/3)*c) + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(c^2*d)^(2/3)*log(c*d*x + (c^ 
2*d)^(2/3)) - 12*(b^2*c^3*d - 2*a*b*c^2*d^2)*x)/(c^2*d^3)]
 

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\frac {b^{2} x^{4}}{4 d} + x \left (\frac {2 a b}{d} - \frac {b^{2} c}{d^{2}}\right ) + \operatorname {RootSum} {\left (27 t^{3} c^{2} d^{7} - a^{6} d^{6} + 6 a^{5} b c d^{5} - 15 a^{4} b^{2} c^{2} d^{4} + 20 a^{3} b^{3} c^{3} d^{3} - 15 a^{2} b^{4} c^{4} d^{2} + 6 a b^{5} c^{5} d - b^{6} c^{6}, \left ( t \mapsto t \log {\left (\frac {3 t c d^{2}}{a^{2} d^{2} - 2 a b c d + b^{2} c^{2}} + x \right )} \right )\right )} \] Input:

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

Output:

b**2*x**4/(4*d) + x*(2*a*b/d - b**2*c/d**2) + RootSum(27*_t**3*c**2*d**7 - 
 a**6*d**6 + 6*a**5*b*c*d**5 - 15*a**4*b**2*c**2*d**4 + 20*a**3*b**3*c**3* 
d**3 - 15*a**2*b**4*c**4*d**2 + 6*a*b**5*c**5*d - b**6*c**6, Lambda(_t, _t 
*log(3*_t*c*d**2/(a**2*d**2 - 2*a*b*c*d + b**2*c**2) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\frac {b^{2} d x^{4} - 4 \, {\left (b^{2} c - 2 \, a b d\right )} x}{4 \, d^{2}} + \frac {\sqrt {3} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{3} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{3} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, d^{3} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \] Input:

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

Output:

1/4*(b^2*d*x^4 - 4*(b^2*c - 2*a*b*d)*x)/d^2 + 1/3*sqrt(3)*(b^2*c^2 - 2*a*b 
*c*d + a^2*d^2)*arctan(1/3*sqrt(3)*(2*x - (c/d)^(1/3))/(c/d)^(1/3))/(d^3*( 
c/d)^(2/3)) - 1/6*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(x^2 - x*(c/d)^(1/3) 
+ (c/d)^(2/3))/(d^3*(c/d)^(2/3)) + 1/3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log 
(x + (c/d)^(1/3))/(d^3*(c/d)^(2/3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=-\frac {\sqrt {3} {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c d^{2}\right )^{\frac {2}{3}} d} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c d^{2}\right )^{\frac {2}{3}} d} - \frac {{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d^{4}} + \frac {b^{2} d^{3} x^{4} - 4 \, b^{2} c d^{2} x + 8 \, a b d^{3} x}{4 \, d^{4}} \] Input:

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

Output:

-1/3*sqrt(3)*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(1/3*sqrt(3)*(2*x + (-c 
/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2/3)*d) - 1/6*(b^2*c^2 - 2*a*b*c*d + a 
^2*d^2)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*d) - 1/3* 
(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(-c/d)^(1/3)*log(abs(x - (-c/d)^(1/3 
)))/(c*d^4) + 1/4*(b^2*d^3*x^4 - 4*b^2*c*d^2*x + 8*a*b*d^3*x)/d^4
 

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\frac {b^2\,x^4}{4\,d}-x\,\left (\frac {b^2\,c}{d^2}-\frac {2\,a\,b}{d}\right )+\frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,{\left (a\,d-b\,c\right )}^2}{3\,c^{2/3}\,d^{7/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^2}{c^{2/3}\,d^{7/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^2}{3\,c^{2/3}\,d^{7/3}} \] Input:

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

Output:

(b^2*x^4)/(4*d) - x*((b^2*c)/d^2 - (2*a*b)/d) + (log(d^(1/3)*x + c^(1/3))* 
(a*d - b*c)^2)/(3*c^(2/3)*d^(7/3)) + (log(3^(1/2)*c^(1/3)*1i + 2*d^(1/3)*x 
 - c^(1/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^2)/(c^(2/3)*d^(7/3)) - (log 
(3^(1/2)*c^(1/3)*1i - 2*d^(1/3)*x + c^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - 
 b*c)^2)/(3*c^(2/3)*d^(7/3))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a+b x^3\right )^2}{c+d x^3} \, dx=\frac {-4 c^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a^{2} d^{2}+8 c^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a b d -4 c^{\frac {7}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) b^{2}-2 c^{\frac {1}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{2} d^{2}+4 c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a b d -2 c^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) b^{2}+4 c^{\frac {1}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{2} d^{2}-8 c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a b d +4 c^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) b^{2}+24 d^{\frac {4}{3}} a b c x -12 d^{\frac {1}{3}} b^{2} c^{2} x +3 d^{\frac {4}{3}} b^{2} c \,x^{4}}{12 d^{\frac {7}{3}} c} \] Input:

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

Output:

( - 4*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))* 
a**2*d**2 + 8*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sq 
rt(3)))*a*b*c*d - 4*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1 
/3)*sqrt(3)))*b**2*c**2 - 2*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + 
d**(2/3)*x**2)*a**2*d**2 + 4*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + 
 d**(2/3)*x**2)*a*b*c*d - 2*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + 
d**(2/3)*x**2)*b**2*c**2 + 4*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*a**2*d**2 
 - 8*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*a*b*c*d + 4*c**(1/3)*log(c**(1/3) 
 + d**(1/3)*x)*b**2*c**2 + 24*d**(1/3)*a*b*c*d*x - 12*d**(1/3)*b**2*c**2*x 
 + 3*d**(1/3)*b**2*c*d*x**4)/(12*d**(1/3)*c*d**2)