3.266 \(\int \frac{x^5 (a+b \log (c (d+e x)^n))}{(f+g x^2)^2} \, dx\)

Optimal. Leaf size=417 \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{g^3}-\frac{b f n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{g^3}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac{f \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac{f \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (d^2 g+e^2 f\right )}+\frac{b e^2 f^2 n \log (d+e x)}{2 g^3 \left (d^2 g+e^2 f\right )}+\frac{b d e f^{3/2} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 g^{5/2} \left (d^2 g+e^2 f\right )}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2} \]

[Out]

(b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) + (b*d*e*f^(3/2)*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2*g^(5/2)*(e^2*f + d^
2*g)) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2) + (b*e^2*f^2*n*Log[d + e*x])/(2*g^3*(e^2*f + d^2*g)) + (x^2*(a + b*
Log[c*(d + e*x)^n]))/(2*g^2) - (f^2*(a + b*Log[c*(d + e*x)^n]))/(2*g^3*(f + g*x^2)) - (f*(a + b*Log[c*(d + e*x
)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/g^3 - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sq
rt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/g^3 - (b*e^2*f^2*n*Log[f + g*x^2])/(4*g^3*(e^2*f + d^2*g)) - (
b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/g^3 - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))
/(e*Sqrt[-f] + d*Sqrt[g])])/g^3

________________________________________________________________________________________

Rubi [A]  time = 0.487262, antiderivative size = 417, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.482, Rules used = {266, 43, 2416, 2395, 2413, 706, 31, 635, 205, 260, 2394, 2393, 2391} \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{g^3}-\frac{b f n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{g^3}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac{f \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac{f \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (d^2 g+e^2 f\right )}+\frac{b e^2 f^2 n \log (d+e x)}{2 g^3 \left (d^2 g+e^2 f\right )}+\frac{b d e f^{3/2} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 g^{5/2} \left (d^2 g+e^2 f\right )}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2)^2,x]

[Out]

(b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) + (b*d*e*f^(3/2)*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2*g^(5/2)*(e^2*f + d^
2*g)) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2) + (b*e^2*f^2*n*Log[d + e*x])/(2*g^3*(e^2*f + d^2*g)) + (x^2*(a + b*
Log[c*(d + e*x)^n]))/(2*g^2) - (f^2*(a + b*Log[c*(d + e*x)^n]))/(2*g^3*(f + g*x^2)) - (f*(a + b*Log[c*(d + e*x
)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/g^3 - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sq
rt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/g^3 - (b*e^2*f^2*n*Log[f + g*x^2])/(4*g^3*(e^2*f + d^2*g)) - (
b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/g^3 - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))
/(e*Sqrt[-f] + d*Sqrt[g])])/g^3

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 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 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 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 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 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{x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx &=\int \left (\frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac{f^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )^2}-\frac{2 f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \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^2}-\frac{(2 f) \int \frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx}{g^2}+\frac{f^2 \int \frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx}{g^2}\\ &=\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac{(2 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^2}+\frac{\left (b e f^2 n\right ) \int \frac{1}{(d+e x) \left (f+g x^2\right )} \, dx}{2 g^3}-\frac{(b e n) \int \frac{x^2}{d+e x} \, dx}{2 g^2}\\ &=\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}+\frac{f \int \frac{a+b \log \left (c (d+e x)^n\right )}{\sqrt{-f}-\sqrt{g} x} \, dx}{g^{5/2}}-\frac{f \int \frac{a+b \log \left (c (d+e x)^n\right )}{\sqrt{-f}+\sqrt{g} x} \, dx}{g^{5/2}}-\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^2}+\frac{\left (b e f^2 n\right ) \int \frac{d g-e g x}{f+g x^2} \, dx}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac{\left (b e^3 f^2 n\right ) \int \frac{1}{d+e x} \, dx}{2 g^3 \left (e^2 f+d^2 g\right )}\\ &=\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac{b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\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 )}{g^3}-\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 )}{g^3}+\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}{g^3}+\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}{g^3}+\frac{\left (b d e f^2 n\right ) \int \frac{1}{f+g x^2} \, dx}{2 g^2 \left (e^2 f+d^2 g\right )}-\frac{\left (b e^2 f^2 n\right ) \int \frac{x}{f+g x^2} \, dx}{2 g^2 \left (e^2 f+d^2 g\right )}\\ &=\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2}+\frac{b d e f^{3/2} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 g^{5/2} \left (e^2 f+d^2 g\right )}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac{b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\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 )}{g^3}-\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 )}{g^3}-\frac{b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (e^2 f+d^2 g\right )}+\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 )}{g^3}+\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 )}{g^3}\\ &=\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2}+\frac{b d e f^{3/2} n \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{2 g^{5/2} \left (e^2 f+d^2 g\right )}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac{b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\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 )}{g^3}-\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 )}{g^3}-\frac{b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (e^2 f+d^2 g\right )}-\frac{b f n \text{Li}_2\left (-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{g^3}-\frac{b f n \text{Li}_2\left (\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}+d \sqrt{g}}\right )}{g^3}\\ \end{align*}

Mathematica [C]  time = 1.18919, size = 530, normalized size = 1.27 \[ \frac{b n \left (-4 f \left (\text{PolyLog}\left (2,-\frac{i \sqrt{g} (d+e x)}{e \sqrt{f}-i d \sqrt{g}}\right )+\log (d+e x) \log \left (\frac{e \left (\sqrt{f}+i \sqrt{g} x\right )}{e \sqrt{f}-i d \sqrt{g}}\right )\right )-4 f \left (\text{PolyLog}\left (2,\frac{i \sqrt{g} (d+e x)}{e \sqrt{f}+i d \sqrt{g}}\right )+\log (d+e x) \log \left (\frac{e \left (\sqrt{f}-i \sqrt{g} x\right )}{e \sqrt{f}+i d \sqrt{g}}\right )\right )+\frac{g \left (e x (2 d-e x)-2 \left (d^2-e^2 x^2\right ) \log (d+e x)\right )}{e^2}+\frac{f^{3/2} \left (i \sqrt{g} (d+e x) \log (d+e x)-e \left (\sqrt{f}+i \sqrt{g} x\right ) \log \left (-\sqrt{g} x+i \sqrt{f}\right )\right )}{\left (\sqrt{f}+i \sqrt{g} x\right ) \left (e \sqrt{f}-i d \sqrt{g}\right )}+\frac{i f^{3/2} \left (-\sqrt{g} (d+e x) \log (d+e x)+e \left (\sqrt{g} x+i \sqrt{f}\right ) \log \left (\sqrt{g} x+i \sqrt{f}\right )\right )}{\left (\sqrt{f}-i \sqrt{g} x\right ) \left (e \sqrt{f}+i d \sqrt{g}\right )}\right )-\frac{2 f^2 \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{f+g x^2}-4 f \log \left (f+g x^2\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+2 g x^2 \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{4 g^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2)^2,x]

[Out]

(2*g*x^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]) - (2*f^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))
/(f + g*x^2) - 4*f*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*Log[f + g*x^2] + b*n*((g*(e*x*(2*d - e*x) - 2
*(d^2 - e^2*x^2)*Log[d + e*x]))/e^2 + (f^(3/2)*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] - e*(Sqrt[f] + I*Sqrt[g]*x)*L
og[I*Sqrt[f] - Sqrt[g]*x]))/((e*Sqrt[f] - I*d*Sqrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (I*f^(3/2)*(-(Sqrt[g]*(d + e
*x)*Log[d + e*x]) + e*(I*Sqrt[f] + Sqrt[g]*x)*Log[I*Sqrt[f] + Sqrt[g]*x]))/((e*Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f]
- I*Sqrt[g]*x)) - 4*f*(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*f*(Log[d + e*x]*Log[(e*(Sqrt[f] - I*Sqrt[g]*x))/(e*Sqr
t[f] + I*d*Sqrt[g])] + PolyLog[2, (I*Sqrt[g]*(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g])])))/(4*g^3)

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Maple [C]  time = 0.462, size = 1008, 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(x^5*(a+b*ln(c*(e*x+d)^n))/(g*x^2+f)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{f^{2}}{g^{4} x^{2} + f g^{3}} - \frac{x^{2}}{g^{2}} + \frac{2 \, f \log \left (g x^{2} + f\right )}{g^{3}}\right )} + b \int \frac{x^{5} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{5} \log \left (c\right )}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

-1/2*a*(f^2/(g^4*x^2 + f*g^3) - x^2/g^2 + 2*f*log(g*x^2 + f)/g^3) + b*integrate((x^5*log((e*x + d)^n) + x^5*lo
g(c))/(g^2*x^4 + 2*f*g*x^2 + f^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{5} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{5}}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral((b*x^5*log((e*x + d)^n*c) + a*x^5)/(g^2*x^4 + 2*f*g*x^2 + f^2), 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(x**5*(a+b*ln(c*(e*x+d)**n))/(g*x**2+f)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="giac")

[Out]

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