\(\int \frac {(a+b x)^3}{c+d x^3} \, dx\) [71]

Optimal result

Integrand size = 17, antiderivative size = 222 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {b^3 x}{d}+\frac {\left (b^3 c-3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{4/3}}-\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}+\frac {a b^2 \log \left (c+d x^3\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1901, 1885, 1874, 31, 648, 631, 210, 642, 266} \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {\left (a^3 (-d)-3 a^2 b \sqrt [3]{c} d^{2/3}+b^3 c\right ) \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{4/3}}+\frac {\left (a^3 (-d)+3 a^2 b \sqrt [3]{c} d^{2/3}+b^3 c\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}-\frac {\left (a^3 (-d)+3 a^2 b \sqrt [3]{c} d^{2/3}+b^3 c\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {a b^2 \log \left (c+d x^3\right )}{d}+\frac {b^3 x}{d} \]

[In]

Int[(a + b*x)^3/(c + d*x^3),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1874

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, Dist[(-r)*((B*r - A*s)/(3*a*s)), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) +
s*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[
a/b]

Rule 1885

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1901

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b^3}{d}-\frac {b^3 c-a^3 d-3 a^2 b d x-3 a b^2 d x^2}{d \left (c+d x^3\right )}\right ) \, dx \\ & = \frac {b^3 x}{d}-\frac {\int \frac {b^3 c-a^3 d-3 a^2 b d x-3 a b^2 d x^2}{c+d x^3} \, dx}{d} \\ & = \frac {b^3 x}{d}+\left (3 a b^2\right ) \int \frac {x^2}{c+d x^3} \, dx-\frac {\int \frac {b^3 c-a^3 d-3 a^2 b d x}{c+d x^3} \, dx}{d} \\ & = \frac {b^3 x}{d}+\frac {a b^2 \log \left (c+d x^3\right )}{d}-\frac {\int \frac {\sqrt [3]{c} \left (-3 a^2 b \sqrt [3]{c} d+2 \sqrt [3]{d} \left (b^3 c-a^3 d\right )\right )+\sqrt [3]{d} \left (-3 a^2 b \sqrt [3]{c} d-\sqrt [3]{d} \left (b^3 c-a^3 d\right )\right ) x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{3 c^{2/3} d^{4/3}}-\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \int \frac {1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{3 c^{2/3} d} \\ & = \frac {b^3 x}{d}-\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {a b^2 \log \left (c+d x^3\right )}{d}-\frac {\left (b^3 c-3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \int \frac {1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{2 \sqrt [3]{c} d}+\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \int \frac {-\sqrt [3]{c} \sqrt [3]{d}+2 d^{2/3} x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{6 c^{2/3} d^{4/3}} \\ & = \frac {b^3 x}{d}-\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}+\frac {a b^2 \log \left (c+d x^3\right )}{d}-\frac {\left (b^3 c-3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{c^{2/3} d^{4/3}} \\ & = \frac {b^3 x}{d}+\frac {\left (b^3 c-3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{4/3}}-\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}+\frac {a b^2 \log \left (c+d x^3\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {6 b^3 c^{2/3} \sqrt [3]{d} x+2 \sqrt {3} \left (b^3 c-3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )-2 \left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )+\left (b^3 c+3 a^2 b \sqrt [3]{c} d^{2/3}-a^3 d\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )+6 a b^2 c^{2/3} \sqrt [3]{d} \log \left (c+d x^3\right )}{6 c^{2/3} d^{4/3}} \]

[In]

Integrate[(a + b*x)^3/(c + d*x^3),x]

[Out]

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

Maple [C] (verified)

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

Time = 1.66 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.30

method	result	size
risch	\(\frac {b^{3} x}{d}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} d +c \right )}{\sum }\frac {\left (3 \textit {\_R}^{2} a \,b^{2} d +3 a^{2} b d \textit {\_R} +a^{3} d -b^{3} c \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 d^{2}}\)	\(66\)
default	\(\frac {b^{3} x}{d}+\frac {\left (a^{3} d -b^{3} c \right ) \left (\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {2}{3}}}\right )+3 d \,a^{2} b \left (-\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )+a \,b^{2} \ln \left (d \,x^{3}+c \right )}{d}\)	\(228\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.33 (sec) , antiderivative size = 7245, normalized size of antiderivative = 32.64 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 8.53 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {b^{3} x}{d} + \operatorname {RootSum} {\left (27 t^{3} c^{2} d^{4} - 81 t^{2} a b^{2} c^{2} d^{3} + t \left (27 a^{5} b c d^{3} + 54 a^{2} b^{4} c^{2} d^{2}\right ) - a^{9} d^{3} + 3 a^{6} b^{3} c d^{2} - 3 a^{3} b^{6} c^{2} d + b^{9} c^{3}, \left ( t \mapsto t \log {\left (x + \frac {27 t^{2} a^{2} b c^{2} d^{3} + 3 t a^{6} c d^{3} - 60 t a^{3} b^{3} c^{2} d^{2} + 3 t b^{6} c^{3} d + 15 a^{7} b^{2} c d^{2} + 15 a^{4} b^{5} c^{2} d - 3 a b^{8} c^{3}}{a^{9} d^{3} + 24 a^{6} b^{3} c d^{2} + 3 a^{3} b^{6} c^{2} d - b^{9} c^{3}} \right )} \right )\right )} \]

[In]

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

[Out]

b**3*x/d + RootSum(27*_t**3*c**2*d**4 - 81*_t**2*a*b**2*c**2*d**3 + _t*(27*a**5*b*c*d**3 + 54*a**2*b**4*c**2*d
**2) - a**9*d**3 + 3*a**6*b**3*c*d**2 - 3*a**3*b**6*c**2*d + b**9*c**3, Lambda(_t, _t*log(x + (27*_t**2*a**2*b
*c**2*d**3 + 3*_t*a**6*c*d**3 - 60*_t*a**3*b**3*c**2*d**2 + 3*_t*b**6*c**3*d + 15*a**7*b**2*c*d**2 + 15*a**4*b
**5*c**2*d - 3*a*b**8*c**3)/(a**9*d**3 + 24*a**6*b**3*c*d**2 + 3*a**3*b**6*c**2*d - b**9*c**3))))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {b^{3} x}{d} - \frac {\sqrt {3} {\left ({\left (b^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}} + 2 \, a b^{2}\right )} c - {\left (3 \, a^{2} b \left (\frac {c}{d}\right )^{\frac {2}{3}} + a^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}} + \frac {2 \, a b^{2} c}{d}\right )} d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, c d} + \frac {{\left (b^{3} c + {\left (6 \, a b^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} + 3 \, a^{2} b \left (\frac {c}{d}\right )^{\frac {1}{3}} - a^{3}\right )} d\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (b^{3} c - {\left (3 \, a b^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - 3 \, a^{2} b \left (\frac {c}{d}\right )^{\frac {1}{3}} + a^{3}\right )} d\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\frac {b^{3} x}{d} + \frac {a b^{2} \log \left ({\left | d x^{3} + c \right |}\right )}{d} + \frac {\sqrt {3} {\left (b^{3} c - a^{3} d + 3 \, \left (-c d^{2}\right )^{\frac {1}{3}} a^{2} b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c d^{2}\right )^{\frac {2}{3}}} + \frac {{\left (b^{3} c - a^{3} d - 3 \, \left (-c d^{2}\right )^{\frac {1}{3}} a^{2} b\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c d^{2}\right )^{\frac {2}{3}}} - \frac {{\left (3 \, a^{2} b d^{3} \left (-\frac {c}{d}\right )^{\frac {1}{3}} - b^{3} c d^{2} + a^{3} d^{3}\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.21 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x)^3}{c+d x^3} \, dx=\left (\sum _{k=1}^3\ln \left (\mathrm {root}\left (27\,c^2\,d^4\,z^3-81\,a\,b^2\,c^2\,d^3\,z^2+54\,a^2\,b^4\,c^2\,d^2\,z+27\,a^5\,b\,c\,d^3\,z+3\,a^6\,b^3\,c\,d^2-3\,a^3\,b^6\,c^2\,d+b^9\,c^3-a^9\,d^3,z,k\right )\,\left (x\,\left (3\,a^3\,d^2-3\,b^3\,c\,d\right )+\mathrm {root}\left (27\,c^2\,d^4\,z^3-81\,a\,b^2\,c^2\,d^3\,z^2+54\,a^2\,b^4\,c^2\,d^2\,z+27\,a^5\,b\,c\,d^3\,z+3\,a^6\,b^3\,c\,d^2-3\,a^3\,b^6\,c^2\,d+b^9\,c^3-a^9\,d^3,z,k\right )\,c\,d^2\,9-18\,a\,b^2\,c\,d\right )+x\,\left (6\,d\,a^4\,b^2+3\,c\,a\,b^5\right )+6\,a^2\,b^4\,c+3\,a^5\,b\,d\right )\,\mathrm {root}\left (27\,c^2\,d^4\,z^3-81\,a\,b^2\,c^2\,d^3\,z^2+54\,a^2\,b^4\,c^2\,d^2\,z+27\,a^5\,b\,c\,d^3\,z+3\,a^6\,b^3\,c\,d^2-3\,a^3\,b^6\,c^2\,d+b^9\,c^3-a^9\,d^3,z,k\right )\right )+\frac {b^3\,x}{d} \]

[In]

int((a + b*x)^3/(c + d*x^3),x)

[Out]

symsum(log(root(27*c^2*d^4*z^3 - 81*a*b^2*c^2*d^3*z^2 + 54*a^2*b^4*c^2*d^2*z + 27*a^5*b*c*d^3*z + 3*a^6*b^3*c*
d^2 - 3*a^3*b^6*c^2*d + b^9*c^3 - a^9*d^3, z, k)*(x*(3*a^3*d^2 - 3*b^3*c*d) + 9*root(27*c^2*d^4*z^3 - 81*a*b^2
*c^2*d^3*z^2 + 54*a^2*b^4*c^2*d^2*z + 27*a^5*b*c*d^3*z + 3*a^6*b^3*c*d^2 - 3*a^3*b^6*c^2*d + b^9*c^3 - a^9*d^3
, z, k)*c*d^2 - 18*a*b^2*c*d) + x*(6*a^4*b^2*d + 3*a*b^5*c) + 6*a^2*b^4*c + 3*a^5*b*d)*root(27*c^2*d^4*z^3 - 8
1*a*b^2*c^2*d^3*z^2 + 54*a^2*b^4*c^2*d^2*z + 27*a^5*b*c*d^3*z + 3*a^6*b^3*c*d^2 - 3*a^3*b^6*c^2*d + b^9*c^3 -
a^9*d^3, z, k), k, 1, 3) + (b^3*x)/d