3.1.14 \(\int \frac {(a+b \log (c x^n))^2 \log (1+e x)}{x} \, dx\) [14]

Optimal. Leaf size=55 \[ -\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)+2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(-e x)-2 b^2 n^2 \text {Li}_4(-e x) \]

[Out]

-(a+b*ln(c*x^n))^2*polylog(2,-e*x)+2*b*n*(a+b*ln(c*x^n))*polylog(3,-e*x)-2*b^2*n^2*polylog(4,-e*x)

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Rubi [A]
time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2421, 2430, 6724} \begin {gather*} 2 b n \text {PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )-\text {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2-2 b^2 n^2 \text {PolyLog}(4,-e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])^2*Log[1 + e*x])/x,x]

[Out]

-((a + b*Log[c*x^n])^2*PolyLog[2, -(e*x)]) + 2*b*n*(a + b*Log[c*x^n])*PolyLog[3, -(e*x)] - 2*b^2*n^2*PolyLog[4
, -(e*x)]

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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 2430

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[PolyLo
g[k + 1, e*x^q]*((a + b*Log[c*x^n])^p/q), x] - Dist[b*n*(p/q), Int[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^(
p - 1)/x), x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

Rule 6724

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 \log (1+e x)}{x} \, dx &=-\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)+(2 b n) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)+2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(-e x)-\left (2 b^2 n^2\right ) \int \frac {\text {Li}_3(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)+2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(-e x)-2 b^2 n^2 \text {Li}_4(-e x)\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 53, normalized size = 0.96 \begin {gather*} -\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)+2 b n \left (\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(-e x)-b n \text {Li}_4(-e x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])^2*Log[1 + e*x])/x,x]

[Out]

-((a + b*Log[c*x^n])^2*PolyLog[2, -(e*x)]) + 2*b*n*((a + b*Log[c*x^n])*PolyLog[3, -(e*x)] - b*n*PolyLog[4, -(e
*x)])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 835, normalized size = 15.18

method	result	size
risch	\(\text {Expression too large to display}\)	\(835\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))^2*ln(e*x+1)/x,x,method=_RETURNVERBOSE)

[Out]

-I*ln(x^n)*Pi*dilog(e*x+1)*b^2*csgn(I*x^n)*csgn(I*c*x^n)^2-I*ln(x)*Pi*polylog(2,-e*x)*b^2*n*csgn(I*x^n)*csgn(I
*c*x^n)^2-I*Pi*polylog(3,-e*x)*b^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*ln(x)*Pi*polylog(2,-e*x)*b^2*n*csgn
(I*c)*csgn(I*c*x^n)^2+I*ln(x^n)*Pi*dilog(e*x+1)*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*ln(x^n)*Pi*dilog(e*x
+1)*b^2*csgn(I*c)*csgn(I*c*x^n)^2+I*ln(x)*Pi*dilog(e*x+1)*b^2*n*csgn(I*c)*csgn(I*c*x^n)^2+2*polylog(3,-e*x)*a*
b*n+2*ln(c)*polylog(3,-e*x)*b^2*n-ln(x^n)^2*dilog(e*x+1)*b^2+2*ln(x)*ln(x^n)*dilog(e*x+1)*b^2*n+I*ln(x)*Pi*pol
ylog(2,-e*x)*b^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*ln(x)*Pi*dilog(e*x+1)*b^2*n*csgn(I*x^n)*csgn(I*c*x^n)
^2-2*ln(x^n)*dilog(e*x+1)*a*b-2*ln(c)*ln(x^n)*dilog(e*x+1)*b^2-I*ln(x)*Pi*dilog(e*x+1)*b^2*n*csgn(I*c*x^n)^3+I
*Pi*polylog(3,-e*x)*b^2*n*csgn(I*c)*csgn(I*c*x^n)^2+I*Pi*polylog(3,-e*x)*b^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2+I*l
n(x)*Pi*polylog(2,-e*x)*b^2*n*csgn(I*c*x^n)^3-2*ln(c)*ln(x)*polylog(2,-e*x)*b^2*n+2*ln(c)*ln(x)*dilog(e*x+1)*b
^2*n+ln(x)^2*polylog(2,-e*x)*b^2*n^2-ln(x)^2*dilog(e*x+1)*b^2*n^2+2*ln(x^n)*polylog(3,-e*x)*b^2*n-I*ln(x)*Pi*d
ilog(e*x+1)*b^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2*ln(x)*ln(x^n)*polylog(2,-e*x)*b^2*n+I*ln(x^n)*Pi*dilog
(e*x+1)*b^2*csgn(I*c*x^n)^3-I*Pi*polylog(3,-e*x)*b^2*n*csgn(I*c*x^n)^3-1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)
+2*a)^2*dilog(e*x+1)-2*ln(x)*polylog(2,-e*x)*a*b*n+2*ln(x)*dilog(e*x+1)*a*b*n-2*b^2*n^2*polylog(4,-e*x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="maxima")

[Out]

integrate((b*log(c*x^n) + a)^2*log(x*e + 1)/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="fricas")

[Out]

integral((b^2*log(c*x^n)^2*log(x*e + 1) + 2*a*b*log(c*x^n)*log(x*e + 1) + a^2*log(x*e + 1))/x, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))**2*ln(e*x+1)/x,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)^2*log(x*e + 1)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\ln \left (e\,x+1\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(e*x + 1)*(a + b*log(c*x^n))^2)/x,x)

[Out]

int((log(e*x + 1)*(a + b*log(c*x^n))^2)/x, x)

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