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

Optimal. Leaf size=69 \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^2}-\frac{d \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{a x}{e}+\frac{b x \log \left (c x^n\right )}{e}-\frac{b n x}{e} \]

[Out]

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

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Rubi [A]  time = 0.0939894, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {43, 2351, 2295, 2317, 2391} \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^2}-\frac{d \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{a x}{e}+\frac{b x \log \left (c x^n\right )}{e}-\frac{b n x}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/e - (b*n*x)/e + (b*x*Log[c*x^n])/e - (d*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e^2 - (b*d*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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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]

Rubi steps

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

Mathematica [A]  time = 0.0322178, size = 66, normalized size = 0.96 \[ \frac{-b d n \text{PolyLog}\left (2,-\frac{e x}{d}\right )-a d \log \left (\frac{e x}{d}+1\right )+a e x+b \log \left (c x^n\right ) \left (e x-d \log \left (\frac{e x}{d}+1\right )\right )-b e n x}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.22, size = 343, normalized size = 5. \begin{align*}{\frac{bx\ln \left ({x}^{n} \right ) }{e}}-{\frac{b\ln \left ({x}^{n} \right ) d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{bnx}{e}}-{\frac{bdn}{{e}^{2}}}+{\frac{bdn\ln \left ( ex+d \right ) }{{e}^{2}}\ln \left ( -{\frac{ex}{d}} \right ) }+{\frac{bdn}{{e}^{2}}{\it dilog} \left ( -{\frac{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}x}{e}}+{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) x}{e}}-{\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 ) x}{e}}+{\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\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}d\ln \left ( ex+d \right ) }{{e}^{2}}}-{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}x}{e}}-{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) d\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{\ln \left ( c \right ) bx}{e}}-{\frac{\ln \left ( c \right ) bd\ln \left ( ex+d \right ) }{{e}^{2}}}+{\frac{ax}{e}}-{\frac{ad\ln \left ( ex+d \right ) }{{e}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

a*(x/e - d*log(e*x + d)/e^2) + b*integrate((x*log(c) + x*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 x \log \left (c x^{n}\right ) + a x}{e x + d}, 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 41.5534, size = 144, normalized size = 2.09 \begin{align*} - \frac{a d \left (\begin{cases} \frac{x}{d} & \text{for}\: e = 0 \\\frac{\log{\left (d + e x \right )}}{e} & \text{otherwise} \end{cases}\right )}{e} + \frac{a x}{e} + \frac{b d n \left (\begin{cases} \frac{x}{d} & \text{for}\: e = 0 \\\frac{\begin{cases} \log{\left (d \right )} \log{\left (x \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{d}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (d \right )} \log{\left (\frac{1}{x} \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{d}\right ) & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (d \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (d \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{d}\right ) & \text{otherwise} \end{cases}}{e} & \text{otherwise} \end{cases}\right )}{e} - \frac{b d \left (\begin{cases} \frac{x}{d} & \text{for}\: e = 0 \\\frac{\log{\left (d + e x \right )}}{e} & \text{otherwise} \end{cases}\right ) \log{\left (c x^{n} \right )}}{e} - \frac{b n x}{e} + \frac{b x \log{\left (c x^{n} \right )}}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e + a*x/e + b*d*n*Piecewise((x/d, Eq(e, 0)), (Piecewis
e((log(d)*log(x) - polylog(2, e*x*exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_pola
r(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)
), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e - b*d*Piecewise((x/d, Eq(e, 0)), (log(d +
e*x)/e, True))*log(c*x**n)/e - b*n*x/e + b*x*log(c*x**n)/e

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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}{e x + d}\,{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),x, algorithm="giac")

[Out]

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