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

Optimal. Leaf size=151 \[ \frac {2 b n \text {Li}_2\left (-\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {2 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{d^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}+\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {d}{e x}\right )}{d^2} \]

[Out]

-e*x*(a+b*ln(c*x^n))^2/d^2/(e*x+d)-ln(1+d/e/x)*(a+b*ln(c*x^n))^2/d^2+2*b*n*(a+b*ln(c*x^n))*ln(1+e*x/d)/d^2+2*b
*n*(a+b*ln(c*x^n))*polylog(2,-d/e/x)/d^2+2*b^2*n^2*polylog(2,-e*x/d)/d^2+2*b^2*n^2*polylog(3,-d/e/x)/d^2

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Rubi [A]  time = 0.31, antiderivative size = 170, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2347, 2344, 2302, 30, 2317, 2374, 6589, 2318, 2391} \[ -\frac {2 b n \text {PolyLog}\left (2,-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {2 b^2 n^2 \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2}+\frac {2 b^2 n^2 \text {PolyLog}\left (3,-\frac {e x}{d}\right )}{d^2}-\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{d^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {2 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {\left (a+b \log \left (c x^n\right )\right )^3}{3 b d^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 2318

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

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 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 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 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 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}{x (d+e x)^2} \, dx &=\frac {\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x (d+e x)} \, dx}{d}-\frac {e \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^2} \, dx}{d}\\ &=-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {\int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{x} \, dx}{d^2}-\frac {e \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{d+e x} \, dx}{d^2}+\frac {(2 b e n) \int \frac {a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^2}\\ &=-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{d^2}+\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,a+b \log \left (c x^n\right )\right )}{b d^2 n}+\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}{d^2}-\frac {\left (2 b^2 n^2\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{d^2}\\ &=-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^3}{3 b d^2 n}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{d^2}+\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}+\frac {\left (2 b^2 n^2\right ) \int \frac {\text {Li}_2\left (-\frac {e x}{d}\right )}{x} \, dx}{d^2}\\ &=-\frac {e x \left (a+b \log \left (c x^n\right )\right )^2}{d^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^3}{3 b d^2 n}+\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^2}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{d^2}+\frac {2 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}-\frac {2 b n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {e x}{d}\right )}{d^2}+\frac {2 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 166, normalized size = 1.10 \[ \frac {-6 b n \text {Li}_2\left (-\frac {e x}{d}\right ) \left (a+b \log \left (c x^n\right )\right )-3 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2+\frac {3 d \left (a+b \log \left (c x^n\right )\right )^2}{d+e x}+6 b n \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right )^3}{b n}-3 \left (a+b \log \left (c x^n\right )\right )^2+6 b^2 n^2 \text {Li}_2\left (-\frac {e x}{d}\right )+6 b^2 n^2 \text {Li}_3\left (-\frac {e x}{d}\right )}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b*Log[c*x^n])^2 + (3*d*(a + b*Log[c*x^n])^2)/(d + e*x) + (a + b*Log[c*x^n])^3/(b*n) + 6*b*n*(a + b*Lo
g[c*x^n])*Log[1 + (e*x)/d] - 3*(a + b*Log[c*x^n])^2*Log[1 + (e*x)/d] + 6*b^2*n^2*PolyLog[2, -((e*x)/d)] - 6*b*
n*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] + 6*b^2*n^2*PolyLog[3, -((e*x)/d)])/(3*d^2)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \log \left (c x^{n}\right )^{2} + 2 \, a b \log \left (c x^{n}\right ) + a^{2}}{e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.76, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{2}}{\left (e x +d \right )^{2} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} {\left (\frac {1}{d e x + d^{2}} - \frac {\log \left (e x + d\right )}{d^{2}} + \frac {\log \relax (x)}{d^{2}}\right )} + \int \frac {b^{2} \log \relax (c)^{2} + b^{2} \log \left (x^{n}\right )^{2} + 2 \, a b \log \relax (c) + 2 \, {\left (b^{2} \log \relax (c) + a b\right )} \log \left (x^{n}\right )}{e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(1/(d*e*x + d^2) - log(e*x + d)/d^2 + log(x)/d^2) + 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^2*x^3 + 2*d*e*x^2 + d^2*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x\,{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{x \left (d + e x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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