3.31 \(\int \frac{x^3 (a+b \log (c x^n))}{d+e x} \, dx\)

Optimal. Leaf size=148 \[ -\frac{b d^3 n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^4}-\frac{d^3 \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac{a d^2 x}{e^3}+\frac{b d^2 x \log \left (c x^n\right )}{e^3}-\frac{b d^2 n x}{e^3}+\frac{b d n x^2}{4 e^2}-\frac{b n x^3}{9 e} \]

[Out]

(a*d^2*x)/e^3 - (b*d^2*n*x)/e^3 + (b*d*n*x^2)/(4*e^2) - (b*n*x^3)/(9*e) + (b*d^2*x*Log[c*x^n])/e^3 - (d*x^2*(a
+ b*Log[c*x^n]))/(2*e^2) + (x^3*(a + b*Log[c*x^n]))/(3*e) - (d^3*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e^4 - (
b*d^3*n*PolyLog[2, -((e*x)/d)])/e^4

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Rubi [A]  time = 0.177357, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {43, 2351, 2295, 2304, 2317, 2391} \[ -\frac{b d^3 n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{e^4}-\frac{d^3 \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac{a d^2 x}{e^3}+\frac{b d^2 x \log \left (c x^n\right )}{e^3}-\frac{b d^2 n x}{e^3}+\frac{b d n x^2}{4 e^2}-\frac{b n x^3}{9 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d^2*x)/e^3 - (b*d^2*n*x)/e^3 + (b*d*n*x^2)/(4*e^2) - (b*n*x^3)/(9*e) + (b*d^2*x*Log[c*x^n])/e^3 - (d*x^2*(a
+ b*Log[c*x^n]))/(2*e^2) + (x^3*(a + b*Log[c*x^n]))/(3*e) - (d^3*(a + b*Log[c*x^n])*Log[1 + (e*x)/d])/e^4 - (
b*d^3*n*PolyLog[2, -((e*x)/d)])/e^4

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx &=\int \left (\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{d x \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{e^3 (d+e x)}\right ) \, dx\\ &=\frac{d^2 \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^3}-\frac{d^3 \int \frac{a+b \log \left (c x^n\right )}{d+e x} \, dx}{e^3}-\frac{d \int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^2}+\frac{\int x^2 \left (a+b \log \left (c x^n\right )\right ) \, dx}{e}\\ &=\frac{a d^2 x}{e^3}+\frac{b d n x^2}{4 e^2}-\frac{b n x^3}{9 e}-\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{d^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{e^4}+\frac{\left (b d^2\right ) \int \log \left (c x^n\right ) \, dx}{e^3}+\frac{\left (b d^3 n\right ) \int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{e^4}\\ &=\frac{a d^2 x}{e^3}-\frac{b d^2 n x}{e^3}+\frac{b d n x^2}{4 e^2}-\frac{b n x^3}{9 e}+\frac{b d^2 x \log \left (c x^n\right )}{e^3}-\frac{d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac{d^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{e^4}-\frac{b d^3 n \text{Li}_2\left (-\frac{e x}{d}\right )}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.0761221, size = 142, normalized size = 0.96 \[ \frac{-36 b d^3 n \text{PolyLog}\left (2,-\frac{e x}{d}\right )+36 a d^2 e x-36 a d^3 \log \left (\frac{e x}{d}+1\right )-18 a d e^2 x^2+12 a e^3 x^3+6 b \log \left (c x^n\right ) \left (e x \left (6 d^2-3 d e x+2 e^2 x^2\right )-6 d^3 \log \left (\frac{e x}{d}+1\right )\right )-36 b d^2 e n x+9 b d e^2 n x^2-4 b e^3 n x^3}{36 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*d^2*e*x - 36*b*d^2*e*n*x - 18*a*d*e^2*x^2 + 9*b*d*e^2*n*x^2 + 12*a*e^3*x^3 - 4*b*e^3*n*x^3 - 36*a*d^3*Lo
g[1 + (e*x)/d] + 6*b*Log[c*x^n]*(e*x*(6*d^2 - 3*d*e*x + 2*e^2*x^2) - 6*d^3*Log[1 + (e*x)/d]) - 36*b*d^3*n*Poly
Log[2, -((e*x)/d)])/(36*e^4)

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Maple [C]  time = 0.217, size = 693, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))/(e*x+d),x)

[Out]

-1/2*b*ln(c)/e^2*x^2*d+b*ln(c)/e^3*x*d^2-b*ln(c)*d^3/e^4*ln(e*x+d)+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e*x^
3+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^3*x*d^2-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^2*x^2*d-1/2*I*b*Pi
*csgn(I*c*x^n)^2*csgn(I*c)*d^3/e^4*ln(e*x+d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^3/e^4*ln(e*x+d)-1/4*I*b*
Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^2*x^2*d+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^3*x*d^2-1/6*I*b*Pi*csgn(I*x^n)*c
sgn(I*c*x^n)*csgn(I*c)/e*x^3+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d^3/e^4*ln(e*x+d)+1/4*I*b*Pi*csgn(
I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^2*x^2*d-1/6*I*b*Pi*csgn(I*c*x^n)^3/e*x^3+1/3*b*ln(x^n)/e*x^3+1/3*b*ln(c)/e*x^
3-a*d^3/e^4*ln(e*x+d)-1/2*a/e^2*x^2*d-49/36*b*n*d^3/e^4+b*n*d^3/e^4*ln(e*x+d)*ln(-e*x/d)-1/2*b*ln(x^n)/e^2*x^2
*d+b*ln(x^n)/e^3*x*d^2-b*ln(x^n)*d^3/e^4*ln(e*x+d)+1/6*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e*x^3+1/4*I*b*Pi*csgn(
I*c*x^n)^3/e^2*x^2*d+1/3*a/e*x^3-1/2*I*b*Pi*csgn(I*c*x^n)^3/e^3*x*d^2+1/2*I*b*Pi*csgn(I*c*x^n)^3*d^3/e^4*ln(e*
x+d)+b*n*d^3/e^4*dilog(-e*x/d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^3*x*d^2+a*d^2*x/e^3-b*d^2*n*x/
e^3-1/9*b*n*x^3/e+1/4*b*d*n*x^2/e^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \, a{\left (\frac{6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac{2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} + b \int \frac{x^{3} \log \left (c\right ) + x^{3} \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^3*(a+b*log(c*x^n))/(e*x+d),x, algorithm="maxima")

[Out]

-1/6*a*(6*d^3*log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^2*x)/e^3) + b*integrate((x^3*log(c) + x^3*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^{3} \log \left (c x^{n}\right ) + a x^{3}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 41.4876, size = 248, normalized size = 1.68 \begin{align*} - \frac{a d^{3} \left (\begin{cases} \frac{x}{d} & \text{for}\: e = 0 \\\frac{\log{\left (d + e x \right )}}{e} & \text{otherwise} \end{cases}\right )}{e^{3}} + \frac{a d^{2} x}{e^{3}} - \frac{a d x^{2}}{2 e^{2}} + \frac{a x^{3}}{3 e} + \frac{b d^{3} 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^{3}} - \frac{b d^{3} \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^{3}} - \frac{b d^{2} n x}{e^{3}} + \frac{b d^{2} x \log{\left (c x^{n} \right )}}{e^{3}} + \frac{b d n x^{2}}{4 e^{2}} - \frac{b d x^{2} \log{\left (c x^{n} \right )}}{2 e^{2}} - \frac{b n x^{3}}{9 e} + \frac{b x^{3} \log{\left (c x^{n} \right )}}{3 e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x+d),x)

[Out]

-a*d**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + a*d**2*x/e**3 - a*d*x**2/(2*e**2) + a*x**3/(
3*e) + b*d**3*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((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_polar(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**3 - b*d**3*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3 - b*d**2*n*x/e**3 +
b*d**2*x*log(c*x**n)/e**3 + b*d*n*x**2/(4*e**2) - b*d*x**2*log(c*x**n)/(2*e**2) - b*n*x**3/(9*e) + b*x**3*log
(c*x**n)/(3*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^{3}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x+d),x, algorithm="giac")

[Out]

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