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

Optimal. Leaf size=156 \[ \frac{b e^2 g n^2 \text{PolyLog}\left (2,\frac{d}{d+e x}\right )}{d^2}-\frac{e^2 n \log \left (1-\frac{d}{d+e x}\right ) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 d^2}-\frac{e n (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 d^2 x}-\frac{\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )}{2 x^2}+\frac{b e^2 g n^2 \log (x)}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.556868, antiderivative size = 265, normalized size of antiderivative = 1.7, number of steps used = 17, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {2439, 2411, 2347, 2344, 2301, 2317, 2391, 2314, 31} \[ -\frac{b e^2 g n^2 \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{d^2}+\frac{e^2 g \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{4 b d^2}-\frac{e^2 g n \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 d^2}-\frac{e g n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 d^2 x}-\frac{\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )}{2 x^2}+\frac{b e^2 \left (g \log \left (c (d+e x)^n\right )+f\right )^2}{4 d^2 g}-\frac{b e^2 n \log \left (-\frac{e x}{d}\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )}{2 d^2}-\frac{b e n (d+e x) \left (g \log \left (c (d+e x)^n\right )+f\right )}{2 d^2 x}+\frac{b e^2 g n^2 \log (x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -1]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2344

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

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

Rule 2314

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

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.128616, size = 254, normalized size = 1.63 \[ b e g n \left (-\frac{e n \text{PolyLog}\left (2,\frac{d+e x}{d}\right )}{d^2}+\frac{e \log ^2\left (c (d+e x)^n\right )}{2 d^2 n}-\frac{e \log \left (-\frac{e x}{d}\right ) \log \left (c (d+e x)^n\right )}{d^2}-\frac{\log \left (c (d+e x)^n\right )}{d x}+\frac{e n \left (\frac{\log (x)}{d}-\frac{\log (d+e x)}{d}\right )}{d}\right )-\frac{a g \log \left (c (d+e x)^n\right )}{2 x^2}+\frac{1}{2} a e g n \left (-\frac{e \log (x)}{d^2}+\frac{e \log (d+e x)}{d^2}-\frac{1}{d x}\right )-\frac{a f}{2 x^2}-\frac{b f \log \left (c (d+e x)^n\right )}{2 x^2}-\frac{b g \log ^2\left (c (d+e x)^n\right )}{2 x^2}+\frac{1}{2} b e f n \left (-\frac{e \log (x)}{d^2}+\frac{e \log (d+e x)}{d^2}-\frac{1}{d x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.593, size = 1201, normalized size = 7.7 \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))*(f+g*ln(c*(e*x+d)^n))/x^3,x)

[Out]

1/2*I*e^2*n/d^2*ln(x)*Pi*b*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/2*I*e*n/d/x*Pi*b*g*csgn(I*(e*x+
d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*e^2*n/d^2*ln(x)*Pi*b*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I*e^2*n/d^2*ln(x)
*Pi*b*g*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/2*I*e^2*n/d^2*ln(e*x+d)*Pi*b*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)
^2+1/2*I*e*n/d/x*Pi*b*g*csgn(I*c*(e*x+d)^n)^3+1/2*I*ln((e*x+d)^n)/x^2*Pi*b*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(
I*c*(e*x+d)^n)+1/2*I*e^2*n/d^2*ln(e*x+d)*Pi*b*g*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*e*n/d/x*Pi*b*g*c
sgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I*ln((e*x+d)^n)/x^2*Pi*b*g*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*ln
((e*x+d)^n)/x^2*Pi*b*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I*e^2*n/d^2*ln(e*x+d)*Pi*b*g*csgn(I*c*(e*x+d)^n)^3+
1/2*I*e^2*n/d^2*ln(x)*Pi*b*g*csgn(I*c*(e*x+d)^n)^3-ln((e*x+d)^n)/x^2*ln(c)*b*g-1/2*ln((e*x+d)^n)/x^2*a*g-1/2*l
n((e*x+d)^n)/x^2*b*f-1/2*b*g/x^2*ln((e*x+d)^n)^2-1/2*e*n/d/x*a*g-1/2*e*n/d/x*b*f+1/2*e^2*n/d^2*ln(e*x+d)*b*f-1
/2*e^2*n/d^2*ln(x)*a*g-1/2*e^2*n/d^2*ln(x)*b*f+1/2*e^2*n/d^2*ln(e*x+d)*a*g-1/8*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+
d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^
2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*a)*(-I*g*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*g*Pi*
csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*g*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*g*Pi*csgn(I*c*(e*x+d)^n)^3+2*
g*ln(c)+2*f)/x^2-1/2*b*g*e^2*n^2/d^2*ln(e*x+d)^2-b*g*e^2*n^2/d^2*ln(e*x+d)+b*g*e^2*n^2/d^2*dilog((e*x+d)/d)+1/
2*I*ln((e*x+d)^n)/x^2*Pi*b*g*csgn(I*c*(e*x+d)^n)^3+b*g*e^2*n*ln((e*x+d)^n)/d^2*ln(e*x+d)-b*g*e*n*ln((e*x+d)^n)
/d/x-b*g*e^2*n*ln((e*x+d)^n)/d^2*ln(x)+b*e^2*g*n^2*ln(x)/d^2-1/2*I*e^2*n/d^2*ln(e*x+d)*Pi*b*g*csgn(I*c)*csgn(I
*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*e*n/d/x*Pi*b*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+e^2*n/d^2
*ln(e*x+d)*ln(c)*b*g-e^2*n/d^2*ln(x)*ln(c)*b*g+b*g*e^2*n^2/d^2*ln(x)*ln((e*x+d)/d)-e*n/d/x*ln(c)*b*g

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b g \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + a f +{\left (b f + a g\right )} \log \left ({\left (e x + d\right )}^{n} c\right )}{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))*(f+g*log(c*(e*x+d)^n))/x^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}{\left (g \log \left ({\left (e x + d\right )}^{n} c\right ) + 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))*(f+g*log(c*(e*x+d)^n))/x^3,x, algorithm="giac")

[Out]

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