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

Optimal. Leaf size=383 \[ -\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 f^2}-\frac{b n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 f^2}+\frac{b n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{\log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac{\log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac{\log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}-\frac{b d e \sqrt{g} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 f^{3/2} \left (d^2 g+e^2 f\right )}+\frac{b e^2 n \log \left (f+g x^2\right )}{4 f \left (d^2 g+e^2 f\right )}-\frac{b e^2 n \log (d+e x)}{2 f \left (d^2 g+e^2 f\right )} \]

[Out]

-(b*d*e*Sqrt[g]*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2*f^(3/2)*(e^2*f + d^2*g)) - (b*e^2*n*Log[d + e*x])/(2*f*(e^2*
f + d^2*g)) + (a + b*Log[c*(d + e*x)^n])/(2*f*(f + g*x^2)) + (Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^2
- ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) - ((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*e^2*n*Log[f + g*x^2])/(
4*f*(e^2*f + d^2*g)) - (b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*f^2) - (b*n*PolyLo
g[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) + (b*n*PolyLog[2, 1 + (e*x)/d])/f^2

________________________________________________________________________________________

Rubi [A]  time = 0.453394, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 13, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.482, Rules used = {266, 44, 2416, 2394, 2315, 2413, 706, 31, 635, 205, 260, 2393, 2391} \[ -\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 f^2}-\frac{b n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 f^2}+\frac{b n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{\log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac{\log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac{\log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}-\frac{b d e \sqrt{g} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 f^{3/2} \left (d^2 g+e^2 f\right )}+\frac{b e^2 n \log \left (f+g x^2\right )}{4 f \left (d^2 g+e^2 f\right )}-\frac{b e^2 n \log (d+e x)}{2 f \left (d^2 g+e^2 f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*e*Sqrt[g]*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2*f^(3/2)*(e^2*f + d^2*g)) - (b*e^2*n*Log[d + e*x])/(2*f*(e^2*
f + d^2*g)) + (a + b*Log[c*(d + e*x)^n])/(2*f*(f + g*x^2)) + (Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^2
- ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) - ((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*e^2*n*Log[f + g*x^2])/(
4*f*(e^2*f + d^2*g)) - (b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*f^2) - (b*n*PolyLo
g[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) + (b*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 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 2413

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

Rule 706

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

Rule 31

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

Rule 635

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

Rule 205

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

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

Mathematica [C]  time = 1.14459, size = 521, normalized size = 1.36 \[ \frac{b n \left (-2 \left (\text{PolyLog}\left (2,-\frac{i \sqrt{g} (d+e x)}{e \sqrt{f}-i d \sqrt{g}}\right )+\log (d+e x) \log \left (\frac{e \left (\sqrt{f}+i \sqrt{g} x\right )}{e \sqrt{f}-i d \sqrt{g}}\right )\right )-2 \left (\text{PolyLog}\left (2,\frac{i \sqrt{g} (d+e x)}{e \sqrt{f}+i d \sqrt{g}}\right )+\log (d+e x) \log \left (\frac{e \left (\sqrt{f}-i \sqrt{g} x\right )}{e \sqrt{f}+i d \sqrt{g}}\right )\right )+4 \left (\text{PolyLog}\left (2,\frac{e x}{d}+1\right )+\log \left (-\frac{e x}{d}\right ) \log (d+e x)\right )+\frac{\sqrt{f} \left (e \left (\sqrt{f}+i \sqrt{g} x\right ) \log \left (-\sqrt{g} x+i \sqrt{f}\right )-i \sqrt{g} (d+e x) \log (d+e x)\right )}{\left (\sqrt{f}+i \sqrt{g} x\right ) \left (e \sqrt{f}-i d \sqrt{g}\right )}+\frac{\sqrt{f} \left (i \sqrt{g} (d+e x) \log (d+e x)+e \left (\sqrt{f}-i \sqrt{g} x\right ) \log \left (\sqrt{g} x+i \sqrt{f}\right )\right )}{\left (\sqrt{f}-i \sqrt{g} x\right ) \left (e \sqrt{f}+i d \sqrt{g}\right )}\right )}{4 f^2}-\frac{\log \left (f+g x^2\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{2 f^2}+\frac{a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)}{2 f^2+2 f g x^2}+\frac{\log (x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.423, size = 910, normalized size = 2.4 \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/(g*x^2+f)^2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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^{2} x^{5} + 2 \, f g x^{3} + f^{2} x}, 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/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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/(g*x**2+f)**2,x)

[Out]

Timed out

________________________________________________________________________________________

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 )}^{2} x}\,{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/(g*x^2+f)^2,x, algorithm="giac")

[Out]

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