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

Optimal. Leaf size=72 \[ \frac{2 b n \text{PolyLog}\left (2,-\frac{e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{2 b^2 n^2 \text{PolyLog}\left (3,-\frac{e x}{d}\right )}{e}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e} \]

[Out]

((a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e + (2*b*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)])/e - (2*b^2*n^2*P
olyLog[3, -((e*x)/d)])/e

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Rubi [A]  time = 0.0610311, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2317, 2374, 6589} \[ \frac{2 b n \text{PolyLog}\left (2,-\frac{e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{2 b^2 n^2 \text{PolyLog}\left (3,-\frac{e x}{d}\right )}{e}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])^2/(d + e*x),x]

[Out]

((a + b*Log[c*x^n])^2*Log[1 + (e*x)/d])/e + (2*b*n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)])/e - (2*b^2*n^2*P
olyLog[3, -((e*x)/d)])/e

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 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx &=\frac{\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac{e x}{d}\right )}{e}-\frac{(2 b n) \int \frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{x} \, dx}{e}\\ &=\frac{\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac{e x}{d}\right )}{e}+\frac{2 b n \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2\left (-\frac{e x}{d}\right )}{e}-\frac{\left (2 b^2 n^2\right ) \int \frac{\text{Li}_2\left (-\frac{e x}{d}\right )}{x} \, dx}{e}\\ &=\frac{\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac{e x}{d}\right )}{e}+\frac{2 b n \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2\left (-\frac{e x}{d}\right )}{e}-\frac{2 b^2 n^2 \text{Li}_3\left (-\frac{e x}{d}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.0252999, size = 68, normalized size = 0.94 \[ \frac{\log \left (\frac{d+e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e}-\frac{2 b n \left (b n \text{PolyLog}\left (3,-\frac{e x}{d}\right )-\text{PolyLog}\left (2,-\frac{e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.295, size = 1412, normalized size = 19.6 \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*x^n))^2/(e*x+d),x)

[Out]

-2*b/e*n*ln(e*x+d)*ln(-e*x/d)*a+I/e*n*ln(e*x+d)*ln(-e*x/d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*ln(e*x
+d)/e*ln(c)*Pi*b^2*csgn(I*c*x^n)^3+1/2*ln(e*x+d)/e*Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^3*csgn(I*c)+1/2*ln(e*x
+d)/e*Pi^2*b^2*csgn(I*c*x^n)^5*csgn(I*c)-1/4*ln(e*x+d)/e*Pi^2*b^2*csgn(I*c*x^n)^4*csgn(I*c)^2-1/4*ln(e*x+d)/e*
Pi^2*b^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4-2*b^2*ln(e*x+d)/e*ln(x^n)*ln(x)*n+2*b^2*n*ln(x)*ln((e*x+d)/d)/e*ln(x^n)
+1/2*ln(e*x+d)/e*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5-I*ln(e*x+d)/e*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)*csg
n(I*c)-I/e*n*ln(e*x+d)*ln(-e*x/d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*ln(e*x+d)/e*Pi^2*b^2*csgn(I*x^n)^2*cs
gn(I*c*x^n)^2*csgn(I*c)^2-ln(e*x+d)/e*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^4*csgn(I*c)+I/e*n*dilog(-e*x/d)*b^2*P
i*csgn(I*c*x^n)^3-I*ln(e*x+d)/e*ln(x^n)*b^2*Pi*csgn(I*c*x^n)^3-2*b/e*n*dilog(-e*x/d)*a-I/e*n*ln(e*x+d)*ln(-e*x
/d)*b^2*Pi*csgn(I*c*x^n)^2*csgn(I*c)+ln(e*x+d)/e*ln(c)^2*b^2+I/e*n*dilog(-e*x/d)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x
^n)*csgn(I*c)-I*ln(e*x+d)/e*ln(x^n)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*ln(e*x+d)/e*Pi*a*b*csgn(I*c*x
^n)^3+1/2*ln(e*x+d)/e*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^3*csgn(I*c)^2-I/e*n*dilog(-e*x/d)*b^2*Pi*csgn(I*x^n)*
csgn(I*c*x^n)^2-I/e*n*dilog(-e*x/d)*b^2*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*ln(e*x+d)/e*ln(c)*Pi*b^2*csgn(I*c*x^n)^
2*csgn(I*c)-2/e*n*ln(e*x+d)*ln(-e*x/d)*b^2*ln(c)+b^2*ln(e*x+d)/e*ln(x^n)^2-I*ln(e*x+d)/e*Pi*a*b*csgn(I*x^n)*cs
gn(I*c*x^n)*csgn(I*c)+I*ln(e*x+d)/e*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2+I*ln(e*x+d)/e*Pi*a*b*csgn(I*c*x^n)^2*cs
gn(I*c)+I/e*n*ln(e*x+d)*ln(-e*x/d)*b^2*Pi*csgn(I*c*x^n)^3+I*ln(e*x+d)/e*ln(c)*Pi*b^2*csgn(I*x^n)*csgn(I*c*x^n)
^2+I*ln(e*x+d)/e*ln(x^n)*b^2*Pi*csgn(I*c*x^n)^2*csgn(I*c)+I*ln(e*x+d)/e*ln(x^n)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^
n)^2+b^2*n^2/e*ln(x)^2*ln(1+e*x/d)+2*b^2*n^2/e*ln(x)*polylog(2,-e*x/d)-2*b^2*n^2*dilog((e*x+d)/d)/e*ln(x)-2*b^
2*n^2*ln(x)^2*ln((e*x+d)/d)/e+a^2*ln(e*x+d)/e+b^2*ln(e*x+d)/e*n^2*ln(x)^2+2*b*ln(e*x+d)/e*ln(x^n)*a+2*b^2*n*di
log((e*x+d)/d)/e*ln(x^n)+2*ln(e*x+d)/e*ln(x^n)*b^2*ln(c)-1/4*ln(e*x+d)/e*Pi^2*b^2*csgn(I*c*x^n)^6-2/e*n*dilog(
-e*x/d)*b^2*ln(c)+2*ln(e*x+d)/e*ln(c)*a*b-2*b^2*n^2*polylog(3,-e*x/d)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*log(c*x**n))**2/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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