3.257 \(\int \frac {x^3 (a+b \log (c (d+e x)^n))}{f+g x^2} \, dx\)

Optimal. Leaf size=278 \[ -\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}-\frac {b f n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 g^2}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g} \]

[Out]

1/2*b*d*n*x/e/g-1/4*b*n*x^2/g-1/2*b*d^2*n*ln(e*x+d)/e^2/g+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g-1/2*f*(a+b*ln(c*(e*x
+d)^n))*ln(e*((-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2)+d*g^(1/2)))/g^2-1/2*f*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)
+x*g^(1/2))/(e*(-f)^(1/2)-d*g^(1/2)))/g^2-1/2*b*f*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/g^2-1
/2*b*f*n*polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/g^2

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Rubi [A]  time = 0.33, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {266, 43, 2416, 2395, 260, 2394, 2393, 2391} \[ -\frac {b f n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 g^2}-\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2),x]

[Out]

(b*d*n*x)/(2*e*g) - (b*n*x^2)/(4*g) - (b*d^2*n*Log[d + e*x])/(2*e^2*g) + (x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g
) - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*g^2) - (f*(a +
b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*g^2) - (b*f*n*PolyLog[2, -(
(Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*g^2) - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] +
d*Sqrt[g])])/(2*g^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 260

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx &=\int \left (\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \left (f+g x^2\right )}\right ) \, dx\\ &=\frac {\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}-\frac {f \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx}{g}\\ &=\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {f \int \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{2 \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}\right ) \, dx}{g}-\frac {(b e n) \int \frac {x^2}{d+e x} \, dx}{2 g}\\ &=\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}-\sqrt {g} x} \, dx}{2 g^{3/2}}-\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}+\sqrt {g} x} \, dx}{2 g^{3/2}}-\frac {(b e n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx}{2 g}\\ &=\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}+\frac {(b e f n) \int \frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{d+e x} \, dx}{2 g^2}+\frac {(b e f n) \int \frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{d+e x} \, dx}{2 g^2}\\ &=\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}+\frac {(b f n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {g} x}{e \sqrt {-f}-d \sqrt {g}}\right )}{x} \, dx,x,d+e x\right )}{2 g^2}+\frac {(b f n) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {g} x}{e \sqrt {-f}+d \sqrt {g}}\right )}{x} \, dx,x,d+e x\right )}{2 g^2}\\ &=\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 243, normalized size = 0.87 \[ -\frac {2 f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+2 f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 g x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b g n \left (2 d^2 \log (d+e x)+e x (e x-2 d)\right )}{e^2}+2 b f n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+2 b f n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{4 g^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2),x]

[Out]

-1/4*((b*g*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x]))/e^2 - 2*g*x^2*(a + b*Log[c*(d + e*x)^n]) + 2*f*(a + b*Lo
g[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])] + 2*f*(a + b*Log[c*(d + e*x)^n])*Lo
g[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])] + 2*b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f]
- d*Sqrt[g]))] + 2*b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/g^2

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{3}}{g x^{2} + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x^2+f),x, algorithm="fricas")

[Out]

integral((b*x^3*log((e*x + d)^n*c) + a*x^3)/(g*x^2 + f), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x^{2} + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x^2+f),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)*x^3/(g*x^2 + f), x)

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maple [C]  time = 0.26, size = 631, normalized size = 2.27 \[ -\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{4 g}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4 g}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4 g}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{4 g}+\frac {i \pi b f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{4 g^{2}}-\frac {i \pi b f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 g^{2}}-\frac {i \pi b f \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 g^{2}}+\frac {i \pi b f \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g \,x^{2}+f \right )}{4 g^{2}}-\frac {b f n \ln \left (\frac {d g +\sqrt {-f g}\, e -\left (e x +d \right ) g}{d g +\sqrt {-f g}\, e}\right ) \ln \left (e x +d \right )}{2 g^{2}}-\frac {b f n \ln \left (\frac {-d g +\sqrt {-f g}\, e +\left (e x +d \right ) g}{-d g +\sqrt {-f g}\, e}\right ) \ln \left (e x +d \right )}{2 g^{2}}+\frac {b f n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}-\frac {b n \,x^{2}}{4 g}+\frac {b \,x^{2} \ln \relax (c )}{2 g}+\frac {b \,x^{2} \ln \left (\left (e x +d \right )^{n}\right )}{2 g}+\frac {a \,x^{2}}{2 g}-\frac {b \,d^{2} n \ln \left (e x +d \right )}{2 e^{2} g}+\frac {b d n x}{2 e g}-\frac {b f n \dilog \left (\frac {d g +\sqrt {-f g}\, e -\left (e x +d \right ) g}{d g +\sqrt {-f g}\, e}\right )}{2 g^{2}}-\frac {b f n \dilog \left (\frac {-d g +\sqrt {-f g}\, e +\left (e x +d \right ) g}{-d g +\sqrt {-f g}\, e}\right )}{2 g^{2}}-\frac {b f \ln \relax (c ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}-\frac {b f \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}-\frac {a f \ln \left (g \,x^{2}+f \right )}{2 g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*ln(c*(e*x+d)^n)+a)/(g*x^2+f),x)

[Out]

1/2*b/g*x^2*ln((e*x+d)^n)-1/2*b*ln((e*x+d)^n)*f/g^2*ln(g*x^2+f)-1/4*b/g*n*x^2+1/2*b*d/e/g*n*x-1/2*b*d^2*n*ln(e
*x+d)/e^2/g+1/2*b*n*f/g^2*ln(e*x+d)*ln(g*x^2+f)-1/2*b*n*f/g^2*ln(e*x+d)*ln((d*g+(-f*g)^(1/2)*e-(e*x+d)*g)/(d*g
+(-f*g)^(1/2)*e))-1/2*b*n*f/g^2*ln(e*x+d)*ln((-d*g+(-f*g)^(1/2)*e+(e*x+d)*g)/(-d*g+(-f*g)^(1/2)*e))-1/2*b*n*f/
g^2*dilog((d*g+(-f*g)^(1/2)*e-(e*x+d)*g)/(d*g+(-f*g)^(1/2)*e))-1/2*b*n*f/g^2*dilog((-d*g+(-f*g)^(1/2)*e+(e*x+d
)*g)/(-d*g+(-f*g)^(1/2)*e))+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3*f/g^2*ln(g*x^2+f)-1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(
e*x+d)^n)^2*f/g^2*ln(g*x^2+f)+1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*f/g^2*ln(g*x^2+f)+1/4
*I*Pi*b/g*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/4*I*Pi*b/g*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*P
i*b/g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2*f
/g^2*ln(g*x^2+f)-1/4*I*Pi*b/g*x^2*csgn(I*c*(e*x+d)^n)^3+1/2*b/g*x^2*ln(c)-1/2*b*ln(c)*f/g^2*ln(g*x^2+f)+1/2*a/
g*x^2-1/2*a*f/g^2*ln(g*x^2+f)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {x^{2}}{g} - \frac {f \log \left (g x^{2} + f\right )}{g^{2}}\right )} + b \int \frac {x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{3} \log \relax (c)}{g x^{2} + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x^2+f),x, algorithm="maxima")

[Out]

1/2*a*(x^2/g - f*log(g*x^2 + f)/g^2) + b*integrate((x^3*log((e*x + d)^n) + x^3*log(c))/(g*x^2 + f), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{g\,x^2+f} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*log(c*(d + e*x)^n)))/(f + g*x^2),x)

[Out]

int((x^3*(a + b*log(c*(d + e*x)^n)))/(f + g*x^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*(e*x+d)**n))/(g*x**2+f),x)

[Out]

Timed out

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