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

Optimal. Leaf size=65 \[ \frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^2}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )+b n\right )}{e^2}-\frac{x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.107657, antiderivative size = 74, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {43, 2351, 2314, 31, 2317, 2391} \[ \frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^2}+\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac{x \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}+\frac{b n \log (d+e x)}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.0620718, size = 71, normalized size = 1.09 \[ \frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )+\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{d \left (a+b \log \left (c x^n\right )\right )}{d+e x}-b n (\log (x)-\log (d+e x))}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.146, size = 389, normalized size = 6. \begin{align*}{\frac{b\ln \left ({x}^{n} \right ) \ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{b\ln \left ({x}^{n} \right ) d}{{e}^{2} \left ( ex+d \right ) }}-{\frac{bn\ln \left ( ex+d \right ) }{{e}^{2}}\ln \left ( -{\frac{ex}{d}} \right ) }-{\frac{bn}{{e}^{2}}{\it dilog} \left ( -{\frac{ex}{d}} \right ) }-{\frac{bn\ln \left ( ex \right ) }{{e}^{2}}}+{\frac{bn\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) d}{{e}^{2} \left ( ex+d \right ) }}+{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) d}{{e}^{2} \left ( ex+d \right ) }}+{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) \ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}d}{{e}^{2} \left ( ex+d \right ) }}+{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}d}{{e}^{2} \left ( ex+d \right ) }}-{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) \ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{b\ln \left ( c \right ) \ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{\ln \left ( c \right ) bd}{{e}^{2} \left ( ex+d \right ) }}+{\frac{a\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{ad}{{e}^{2} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*log(c*x**n))/(d + e*x)**2, 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 )} x}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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