Integrand size = 19, antiderivative size = 206 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {b \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x}{d^3}-\frac {b^2 (b c-3 a d) x^4}{4 d^2}+\frac {b^3 x^7}{7 d}+\frac {(b c-a d)^3 \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{10/3}}-\frac {(b c-a d)^3 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{10/3}}+\frac {(b c-a d)^3 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{10/3}} \] Output:
b*(3*a^2*d^2-3*a*b*c*d+b^2*c^2)*x/d^3-1/4*b^2*(-3*a*d+b*c)*x^4/d^2+1/7*b^3 *x^7/d+1/3*(-a*d+b*c)^3*arctan(1/3*(c^(1/3)-2*d^(1/3)*x)*3^(1/2)/c^(1/3))* 3^(1/2)/c^(2/3)/d^(10/3)-1/3*(-a*d+b*c)^3*ln(c^(1/3)+d^(1/3)*x)/c^(2/3)/d^ (10/3)+1/6*(-a*d+b*c)^3*ln(c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/c^(2/3)/ d^(10/3)
Time = 0.04 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {84 b \sqrt [3]{d} \left (b^2 c^2-3 a b c d+3 a^2 d^2\right ) x-21 b^2 d^{4/3} (b c-3 a d) x^4+12 b^3 d^{7/3} x^7-\frac {28 \sqrt {3} (b c-a d)^3 \arctan \left (\frac {-\sqrt [3]{c}+2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{c^{2/3}}-\frac {28 (b c-a d)^3 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{2/3}}+\frac {14 (b c-a d)^3 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{2/3}}}{84 d^{10/3}} \] Input:
Integrate[(a + b*x^3)^3/(c + d*x^3),x]
Output:
(84*b*d^(1/3)*(b^2*c^2 - 3*a*b*c*d + 3*a^2*d^2)*x - 21*b^2*d^(4/3)*(b*c - 3*a*d)*x^4 + 12*b^3*d^(7/3)*x^7 - (28*Sqrt[3]*(b*c - a*d)^3*ArcTan[(-c^(1/ 3) + 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/c^(2/3) - (28*(b*c - a*d)^3*Log[c^(1 /3) + d^(1/3)*x])/c^(2/3) + (14*(b*c - a*d)^3*Log[c^(2/3) - c^(1/3)*d^(1/3 )*x + d^(2/3)*x^2])/c^(2/3))/(84*d^(10/3))
Time = 0.60 (sec) , antiderivative size = 206, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 915 |
| \(\displaystyle \int \left (\frac {b \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{d^3}+\frac {a^3 d^3-3 a^2 b c d^2+3 a b^2 c^2 d-b^3 c^3}{d^3 \left (c+d x^3\right )}-\frac {b^2 x^3 (b c-3 a d)}{d^2}+\frac {b^3 x^6}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b x \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{d^3}+\frac {(b c-a d)^3 \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{10/3}}-\frac {b^2 x^4 (b c-3 a d)}{4 d^2}+\frac {(b c-a d)^3 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{10/3}}-\frac {(b c-a d)^3 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{10/3}}+\frac {b^3 x^7}{7 d}\) |
Input:
Int[(a + b*x^3)^3/(c + d*x^3),x]
Output:
(b*(b^2*c^2 - 3*a*b*c*d + 3*a^2*d^2)*x)/d^3 - (b^2*(b*c - 3*a*d)*x^4)/(4*d ^2) + (b^3*x^7)/(7*d) + ((b*c - a*d)^3*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqr t[3]*c^(1/3))])/(Sqrt[3]*c^(2/3)*d^(10/3)) - ((b*c - a*d)^3*Log[c^(1/3) + d^(1/3)*x])/(3*c^(2/3)*d^(10/3)) + ((b*c - a*d)^3*Log[c^(2/3) - c^(1/3)*d^ (1/3)*x + d^(2/3)*x^2])/(6*c^(2/3)*d^(10/3))
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 = 0.92 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.64
| method | result | size |
| risch | \(\frac {b^{3} x^{7}}{7 d}+\frac {3 b^{2} a \,x^{4}}{4 d}-\frac {b^{3} c \,x^{4}}{4 d^{2}}+\frac {3 b \,a^{2} x}{d}-\frac {3 b^{2} a c x}{d^{2}}+\frac {b^{3} c^{2} x}{d^{3}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+c \right )}{\sum }\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 d^{4}}\) | \(131\) |
| default | \(\frac {b \left (\frac {1}{7} b^{2} d^{2} x^{7}+\frac {3}{4} a b \,d^{2} x^{4}-\frac {1}{4} b^{2} c d \,x^{4}+3 a^{2} d^{2} x -3 a b c d x +b^{2} c^{2} x \right )}{d^{3}}+\frac {\left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right ) \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{d^{3}}\) | \(193\) |
Input:
int((b*x^3+a)^3/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/7*b^3*x^7/d+3/4*b^2/d*a*x^4-1/4*b^3/d^2*c*x^4+3*b/d*a^2*x-3*b^2/d^2*a*c* x+b^3/d^3*c^2*x+1/3/d^4*sum((a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/ _R^2*ln(x-_R),_R=RootOf(_Z^3*d+c))
Time = 0.10 (sec) , antiderivative size = 679, normalized size of antiderivative = 3.30 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx =\text {Too large to display} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c),x, algorithm="fricas")
Output:
[1/84*(12*b^3*c^2*d^3*x^7 - 21*(b^3*c^3*d^2 - 3*a*b^2*c^2*d^3)*x^4 - 42*sq rt(1/3)*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 - a^3*c*d^4)*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)) + 14*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(c^2*d )^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) - 28*(b^3*c^3 - 3 *a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^ (2/3)) + 84*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3)*x)/(c^2*d^4), 1/84*(12*b^3*c^2*d^3*x^7 - 21*(b^3*c^3*d^2 - 3*a*b^2*c^2*d^3)*x^4 - 84*sqr t(1/3)*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 - a^3*c*d^4)*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) + 14*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3 *d^3)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) - 28* (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(c^2*d)^(2/3)*log(c*d* x + (c^2*d)^(2/3)) + 84*(b^3*c^4*d - 3*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3)*x) /(c^2*d^4)]
Time = 0.60 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {b^{3} x^{7}}{7 d} + x^{4} \cdot \left (\frac {3 a b^{2}}{4 d} - \frac {b^{3} c}{4 d^{2}}\right ) + x \left (\frac {3 a^{2} b}{d} - \frac {3 a b^{2} c}{d^{2}} + \frac {b^{3} c^{2}}{d^{3}}\right ) + \operatorname {RootSum} {\left (27 t^{3} c^{2} d^{10} - a^{9} d^{9} + 9 a^{8} b c d^{8} - 36 a^{7} b^{2} c^{2} d^{7} + 84 a^{6} b^{3} c^{3} d^{6} - 126 a^{5} b^{4} c^{4} d^{5} + 126 a^{4} b^{5} c^{5} d^{4} - 84 a^{3} b^{6} c^{6} d^{3} + 36 a^{2} b^{7} c^{7} d^{2} - 9 a b^{8} c^{8} d + b^{9} c^{9}, \left ( t \mapsto t \log {\left (\frac {3 t c d^{3}}{a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}} + x \right )} \right )\right )} \] Input:
integrate((b*x**3+a)**3/(d*x**3+c),x)
Output:
b**3*x**7/(7*d) + x**4*(3*a*b**2/(4*d) - b**3*c/(4*d**2)) + x*(3*a**2*b/d - 3*a*b**2*c/d**2 + b**3*c**2/d**3) + RootSum(27*_t**3*c**2*d**10 - a**9*d **9 + 9*a**8*b*c*d**8 - 36*a**7*b**2*c**2*d**7 + 84*a**6*b**3*c**3*d**6 - 126*a**5*b**4*c**4*d**5 + 126*a**4*b**5*c**5*d**4 - 84*a**3*b**6*c**6*d**3 + 36*a**2*b**7*c**7*d**2 - 9*a*b**8*c**8*d + b**9*c**9, Lambda(_t, _t*log (3*_t*c*d**3/(a**3*d**3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3) + x)))
Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {4 \, b^{3} d^{2} x^{7} - 7 \, {\left (b^{3} c d - 3 \, a b^{2} d^{2}\right )} x^{4} + 28 \, {\left (b^{3} c^{2} - 3 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x}{28 \, d^{3}} - \frac {\sqrt {3} {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, d^{4} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c),x, algorithm="maxima")
Output:
1/28*(4*b^3*d^2*x^7 - 7*(b^3*c*d - 3*a*b^2*d^2)*x^4 + 28*(b^3*c^2 - 3*a*b^ 2*c*d + 3*a^2*b*d^2)*x)/d^3 - 1/3*sqrt(3)*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2 *b*c*d^2 - a^3*d^3)*arctan(1/3*sqrt(3)*(2*x - (c/d)^(1/3))/(c/d)^(1/3))/(d ^4*(c/d)^(2/3)) + 1/6*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)* log(x^2 - x*(c/d)^(1/3) + (c/d)^(2/3))/(d^4*(c/d)^(2/3)) - 1/3*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(x + (c/d)^(1/3))/(d^4*(c/d)^( 2/3))
Time = 0.13 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {\sqrt {3} {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c d^{2}\right )^{\frac {2}{3}} d^{2}} + \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c d^{2}\right )^{\frac {2}{3}} d^{2}} + \frac {{\left (b^{3} c^{3} d^{4} - 3 \, a b^{2} c^{2} d^{5} + 3 \, a^{2} b c d^{6} - a^{3} d^{7}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d^{7}} + \frac {4 \, b^{3} d^{6} x^{7} - 7 \, b^{3} c d^{5} x^{4} + 21 \, a b^{2} d^{6} x^{4} + 28 \, b^{3} c^{2} d^{4} x - 84 \, a b^{2} c d^{5} x + 84 \, a^{2} b d^{6} x}{28 \, d^{7}} \] Input:
integrate((b*x^3+a)^3/(d*x^3+c),x, algorithm="giac")
Output:
1/3*sqrt(3)*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(1/3 *sqrt(3)*(2*x + (-c/d)^(1/3))/(-c/d)^(1/3))/((-c*d^2)^(2/3)*d^2) + 1/6*(b^ 3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/((-c*d^2)^(2/3)*d^2) + 1/3*(b^3*c^3*d^4 - 3*a*b^2*c^2*d^5 + 3*a^2*b*c*d^6 - a^3*d^7)*(-c/d)^(1/3)*log(abs(x - (-c/d)^(1/3)))/(c*d^7) + 1/28*(4*b^3*d^6*x^7 - 7*b^3*c*d^5*x^4 + 21*a*b^2*d^6*x^4 + 28*b^3*c^2*d ^4*x - 84*a*b^2*c*d^5*x + 84*a^2*b*d^6*x)/d^7
Time = 0.68 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=x^4\,\left (\frac {3\,a\,b^2}{4\,d}-\frac {b^3\,c}{4\,d^2}\right )+x\,\left (\frac {3\,a^2\,b}{d}-\frac {c\,\left (\frac {3\,a\,b^2}{d}-\frac {b^3\,c}{d^2}\right )}{d}\right )+\frac {b^3\,x^7}{7\,d}+\frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,{\left (a\,d-b\,c\right )}^3}{3\,c^{2/3}\,d^{10/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^3}{c^{2/3}\,d^{10/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^3}{3\,c^{2/3}\,d^{10/3}} \] Input:
int((a + b*x^3)^3/(c + d*x^3),x)
Output:
x^4*((3*a*b^2)/(4*d) - (b^3*c)/(4*d^2)) + x*((3*a^2*b)/d - (c*((3*a*b^2)/d - (b^3*c)/d^2))/d) + (b^3*x^7)/(7*d) + (log(d^(1/3)*x + c^(1/3))*(a*d - b *c)^3)/(3*c^(2/3)*d^(10/3)) + (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)^3)/(c^(2/3)*d^(10/3)) - (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)^ 3)/(3*c^(2/3)*d^(10/3))
Time = 0.23 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^3\right )^3}{c+d x^3} \, dx=\frac {-28 c^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a^{3} d^{3}+84 c^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a^{2} b \,d^{2}-84 c^{\frac {7}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) a \,b^{2} d +28 c^{\frac {10}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) b^{3}-14 c^{\frac {1}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{3} d^{3}+42 c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a^{2} b \,d^{2}-42 c^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) a \,b^{2} d +14 c^{\frac {10}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) b^{3}+28 c^{\frac {1}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{3} d^{3}-84 c^{\frac {4}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a^{2} b \,d^{2}+84 c^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) a \,b^{2} d -28 c^{\frac {10}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) b^{3}+252 d^{\frac {7}{3}} a^{2} b c x -252 d^{\frac {4}{3}} a \,b^{2} c^{2} x +63 d^{\frac {7}{3}} a \,b^{2} c \,x^{4}+84 d^{\frac {1}{3}} b^{3} c^{3} x -21 d^{\frac {4}{3}} b^{3} c^{2} x^{4}+12 d^{\frac {7}{3}} b^{3} c \,x^{7}}{84 d^{\frac {10}{3}} c} \] Input:
int((b*x^3+a)^3/(d*x^3+c),x)
Output:
( - 28*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3))) *a**3*d**3 + 84*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)* sqrt(3)))*a**2*b*c*d**2 - 84*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)* x)/(c**(1/3)*sqrt(3)))*a*b**2*c**2*d + 28*c**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**(1/3)*x)/(c**(1/3)*sqrt(3)))*b**3*c**3 - 14*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**3*d**3 + 42*c**(1/3)*log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a**2*b*c*d**2 - 42*c**(1/3)*log(c* *(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*a*b**2*c**2*d + 14*c**(1/3)* log(c**(2/3) - d**(1/3)*c**(1/3)*x + d**(2/3)*x**2)*b**3*c**3 + 28*c**(1/3 )*log(c**(1/3) + d**(1/3)*x)*a**3*d**3 - 84*c**(1/3)*log(c**(1/3) + d**(1/ 3)*x)*a**2*b*c*d**2 + 84*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*a*b**2*c**2*d - 28*c**(1/3)*log(c**(1/3) + d**(1/3)*x)*b**3*c**3 + 252*d**(1/3)*a**2*b* c*d**2*x - 252*d**(1/3)*a*b**2*c**2*d*x + 63*d**(1/3)*a*b**2*c*d**2*x**4 + 84*d**(1/3)*b**3*c**3*x - 21*d**(1/3)*b**3*c**2*d*x**4 + 12*d**(1/3)*b**3 *c*d**2*x**7)/(84*d**(1/3)*c*d**3)