3.1.6 \(\int \frac {a+b x^3+c x^6}{d+e x^3} \, dx\) [6]

Optimal. Leaf size=188 \[ -\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}-\frac {\left (c d^2-b d e+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt {3} d^{2/3} e^{7/3}}+\frac {\left (c d^2-b d e+a e^2\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} e^{7/3}}-\frac {\left (c d^2-b d e+a e^2\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{6 d^{2/3} e^{7/3}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1425, 396, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right ) \left (a e^2-b d e+c d^2\right )}{\sqrt {3} d^{2/3} e^{7/3}}-\frac {\log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right ) \left (a e^2-b d e+c d^2\right )}{6 d^{2/3} e^{7/3}}+\frac {\log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \left (a e^2-b d e+c d^2\right )}{3 d^{2/3} e^{7/3}}-\frac {x (c d-b e)}{e^2}+\frac {c x^4}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3 + c*x^6)/(d + e*x^3),x]

[Out]

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

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], 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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 1425

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Simp[c*x^(n + 1)*(
(d + e*x^n)^(q + 1)/(e*(n*(q + 2) + 1))), x] + Dist[1/(e*(n*(q + 2) + 1)), Int[(d + e*x^n)^q*(a*e*(n*(q + 2) +
1) - (c*d*(n + 1) - b*e*(n*(q + 2) + 1))*x^n), x], x] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rubi steps

\begin {align*} \int \frac {a+b x^3+c x^6}{d+e x^3} \, dx &=\frac {c x^4}{4 e}+\frac {\int \frac {4 a e-(4 c d-4 b e) x^3}{d+e x^3} \, dx}{4 e}\\ &=-\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}-\left (-a-\frac {d (c d-b e)}{e^2}\right ) \int \frac {1}{d+e x^3} \, dx\\ &=-\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}+\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \int \frac {1}{\sqrt [3]{d}+\sqrt [3]{e} x} \, dx}{3 d^{2/3}}+\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \int \frac {2 \sqrt [3]{d}-\sqrt [3]{e} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx}{3 d^{2/3}}\\ &=-\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}+\frac {\left (c d^2-b d e+a e^2\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} e^{7/3}}-\frac {\left (c d^2-b d e+a e^2\right ) \int \frac {-\sqrt [3]{d} \sqrt [3]{e}+2 e^{2/3} x}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx}{6 d^{2/3} e^{7/3}}+\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \int \frac {1}{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2} \, dx}{2 \sqrt [3]{d}}\\ &=-\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}+\frac {\left (c d^2-b d e+a e^2\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} e^{7/3}}-\frac {\left (c d^2-b d e+a e^2\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{6 d^{2/3} e^{7/3}}+\frac {\left (c d^2-b d e+a e^2\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}\right )}{d^{2/3} e^{7/3}}\\ &=-\frac {(c d-b e) x}{e^2}+\frac {c x^4}{4 e}-\frac {\left (c d^2-b d e+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt [3]{d}-2 \sqrt [3]{e} x}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt {3} d^{2/3} e^{7/3}}+\frac {\left (c d^2-b d e+a e^2\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{3 d^{2/3} e^{7/3}}-\frac {\left (c d^2-b d e+a e^2\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{6 d^{2/3} e^{7/3}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 176, normalized size = 0.94 \begin {gather*} \frac {12 \sqrt [3]{e} (-c d+b e) x+3 c e^{4/3} x^4-\frac {4 \sqrt {3} \left (c d^2+e (-b d+a e)\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{e} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )}{d^{2/3}}+\frac {4 \left (c d^2+e (-b d+a e)\right ) \log \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{d^{2/3}}-\frac {2 \left (c d^2+e (-b d+a e)\right ) \log \left (d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2\right )}{d^{2/3}}}{12 e^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3 + c*x^6)/(d + e*x^3),x]

[Out]

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

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Maple [A]
time = 0.19, size = 133, normalized size = 0.71

method	result	size
risch	\(\frac {c \,x^{4}}{4 e}+\frac {b x}{e}-\frac {c d x}{e^{2}}+\frac {\munderset {\textit {\_R} =\RootOf \left (e \,\textit {\_Z}^{3}+d \right )}{\sum }\frac {\left (a \,e^{2}-d e b +c \,d^{2}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}}{3 e^{3}}\)	\(67\)
default	\(\frac {\frac {1}{4} c e \,x^{4}+e b x -c d x}{e^{2}}+\frac {\left (\frac {\ln \left (x +\left (\frac {d}{e}\right )^{\frac {1}{3}}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {d}{e}\right )^{\frac {1}{3}} x +\left (\frac {d}{e}\right )^{\frac {2}{3}}\right )}{6 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{e}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 e \left (\frac {d}{e}\right )^{\frac {2}{3}}}\right ) \left (a \,e^{2}-d e b +c \,d^{2}\right )}{e^{2}}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^6+b*x^3+a)/(e*x^3+d),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 146, normalized size = 0.78 \begin {gather*} \frac {\sqrt {3} {\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (-\frac {\sqrt {3} {\left (d^{\frac {1}{3}} e^{\left (-\frac {1}{3}\right )} - 2 \, x\right )} e^{\frac {1}{3}}}{3 \, d^{\frac {1}{3}}}\right ) e^{\left (-\frac {7}{3}\right )}}{3 \, d^{\frac {2}{3}}} - \frac {{\left (c d^{2} - b d e + a e^{2}\right )} e^{\left (-\frac {7}{3}\right )} \log \left (-d^{\frac {1}{3}} x e^{\left (-\frac {1}{3}\right )} + x^{2} + d^{\frac {2}{3}} e^{\left (-\frac {2}{3}\right )}\right )}{6 \, d^{\frac {2}{3}}} + \frac {{\left (c d^{2} - b d e + a e^{2}\right )} e^{\left (-\frac {7}{3}\right )} \log \left (d^{\frac {1}{3}} e^{\left (-\frac {1}{3}\right )} + x\right )}{3 \, d^{\frac {2}{3}}} + \frac {1}{4} \, {\left (c x^{4} e - 4 \, {\left (c d - b e\right )} x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 199, normalized size = 1.06 \begin {gather*} -\frac {{\left (12 \, c d^{3} x e - 12 \, \sqrt {\frac {1}{3}} {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} {\left (d^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (d^{2}\right )}^{\frac {2}{3}} x e^{\frac {2}{3}} - {\left (d^{2}\right )}^{\frac {1}{3}} d e^{\frac {1}{3}}\right )} {\left (d^{2}\right )}^{\frac {1}{6}} e^{\left (-\frac {1}{3}\right )}}{d^{2}}\right ) e^{\left (-\frac {1}{3}\right )} + 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} {\left (d^{2}\right )}^{\frac {2}{3}} e^{\frac {2}{3}} \log \left (d x^{2} e - {\left (d^{2}\right )}^{\frac {2}{3}} x e^{\frac {2}{3}} + {\left (d^{2}\right )}^{\frac {1}{3}} d e^{\frac {1}{3}}\right ) - 4 \, {\left (c d^{2} - b d e + a e^{2}\right )} {\left (d^{2}\right )}^{\frac {2}{3}} e^{\frac {2}{3}} \log \left (d x e + {\left (d^{2}\right )}^{\frac {2}{3}} e^{\frac {2}{3}}\right ) - 3 \, {\left (c d^{2} x^{4} + 4 \, b d^{2} x\right )} e^{2}\right )} e^{\left (-3\right )}}{12 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d),x, algorithm="fricas")

[Out]

-1/12*(12*c*d^3*x*e - 12*sqrt(1/3)*(c*d^3*e - b*d^2*e^2 + a*d*e^3)*(d^2)^(1/6)*arctan(sqrt(1/3)*(2*(d^2)^(2/3)
*x*e^(2/3) - (d^2)^(1/3)*d*e^(1/3))*(d^2)^(1/6)*e^(-1/3)/d^2)*e^(-1/3) + 2*(c*d^2 - b*d*e + a*e^2)*(d^2)^(2/3)
*e^(2/3)*log(d*x^2*e - (d^2)^(2/3)*x*e^(2/3) + (d^2)^(1/3)*d*e^(1/3)) - 4*(c*d^2 - b*d*e + a*e^2)*(d^2)^(2/3)*
e^(2/3)*log(d*x*e + (d^2)^(2/3)*e^(2/3)) - 3*(c*d^2*x^4 + 4*b*d^2*x)*e^2)*e^(-3)/d^2

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Sympy [A]
time = 0.45, size = 175, normalized size = 0.93 \begin {gather*} \frac {c x^{4}}{4 e} + x \left (\frac {b}{e} - \frac {c d}{e^{2}}\right ) + \operatorname {RootSum} {\left (27 t^{3} d^{2} e^{7} - a^{3} e^{6} + 3 a^{2} b d e^{5} - 3 a^{2} c d^{2} e^{4} - 3 a b^{2} d^{2} e^{4} + 6 a b c d^{3} e^{3} - 3 a c^{2} d^{4} e^{2} + b^{3} d^{3} e^{3} - 3 b^{2} c d^{4} e^{2} + 3 b c^{2} d^{5} e - c^{3} d^{6}, \left ( t \mapsto t \log {\left (\frac {3 t d e^{2}}{a e^{2} - b d e + c d^{2}} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**6+b*x**3+a)/(e*x**3+d),x)

[Out]

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

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Giac [A]
time = 4.15, size = 173, normalized size = 0.92 \begin {gather*} -\frac {\sqrt {3} {\left (c d^{2} - b d e + a e^{2}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-d e^{\left (-1\right )}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-d e^{\left (-1\right )}\right )^{\frac {1}{3}}}\right ) e^{\left (-1\right )}}{3 \, \left (-d e^{2}\right )^{\frac {2}{3}}} - \frac {{\left (c d^{2} - b d e + a e^{2}\right )} e^{\left (-1\right )} \log \left (x^{2} + \left (-d e^{\left (-1\right )}\right )^{\frac {1}{3}} x + \left (-d e^{\left (-1\right )}\right )^{\frac {2}{3}}\right )}{6 \, \left (-d e^{2}\right )^{\frac {2}{3}}} - \frac {{\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} \left (-d e^{\left (-1\right )}\right )^{\frac {1}{3}} e^{\left (-4\right )} \log \left ({\left | x - \left (-d e^{\left (-1\right )}\right )^{\frac {1}{3}} \right |}\right )}{3 \, d} + \frac {1}{4} \, {\left (c x^{4} e^{3} - 4 \, c d x e^{2} + 4 \, b x e^{3}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^6+b*x^3+a)/(e*x^3+d),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.27, size = 165, normalized size = 0.88 \begin {gather*} x\,\left (\frac {b}{e}-\frac {c\,d}{e^2}\right )+\frac {c\,x^4}{4\,e}+\frac {\ln \left (e^{1/3}\,x+d^{1/3}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{3\,d^{2/3}\,e^{7/3}}+\frac {\ln \left (2\,e^{1/3}\,x-d^{1/3}+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{3\,d^{2/3}\,e^{7/3}}-\frac {\ln \left (d^{1/3}-2\,e^{1/3}\,x+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{3\,d^{2/3}\,e^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3 + c*x^6)/(d + e*x^3),x)

[Out]

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

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