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

Optimal. Leaf size=331 \[ \frac{b g n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 f^2}+\frac{b g n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 f^2}-\frac{b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{g \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac{g \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac{b e^2 n \log (x)}{2 d^2 f}+\frac{b e^2 n \log (d+e x)}{2 d^2 f}-\frac{b e n}{2 d f x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.368075, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 44, 2416, 2395, 2394, 2315, 260, 2393, 2391} \[ \frac{b g n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 f^2}+\frac{b g n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 f^2}-\frac{b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{g \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac{g \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac{a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac{b e^2 n \log (x)}{2 d^2 f}+\frac{b e^2 n \log (d+e x)}{2 d^2 f}-\frac{b e n}{2 d f x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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]

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 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 2315

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

Rubi steps

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

Mathematica [A]  time = 0.178629, size = 279, normalized size = 0.84 \[ \frac{b g n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )+b g n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )-2 b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )+g \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+g \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-2 g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{b e f n (-e x \log (d+e x)+d+e x \log (x))}{d^2 x}}{2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.417, size = 841, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a{\left (\frac{g \log \left (g x^{2} + f\right )}{f^{2}} - \frac{2 \, g \log \left (x\right )}{f^{2}} - \frac{1}{f x^{2}}\right )} + b \int \frac{\log \left ({\left (e x + d\right )}^{n}\right ) + \log \left (c\right )}{g x^{5} + f x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(g*log(g*x^2 + f)/f^2 - 2*g*log(x)/f^2 - 1/(f*x^2)) + b*integrate((log((e*x + d)^n) + log(c))/(g*x^5 + f
*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g x^{5} + f x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((e*x + d)^n*c) + a)/(g*x^5 + f*x^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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