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

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

[Out]

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

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Rubi [A]  time = 0.168242, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {263, 43, 2351, 2295, 2304, 2317, 2391} \[ -\frac{b e^3 n \text{PolyLog}\left (2,-\frac{d x}{e}\right )}{d^4}-\frac{e^3 \log \left (\frac{d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac{e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}+\frac{a e^2 x}{d^3}+\frac{b e^2 x \log \left (c x^n\right )}{d^3}-\frac{b e^2 n x}{d^3}+\frac{b e n x^2}{4 d^2}-\frac{b n x^3}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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

Mathematica [A]  time = 0.0694653, size = 142, normalized size = 0.96 \[ \frac{-36 b e^3 n \text{PolyLog}\left (2,-\frac{d x}{e}\right )-18 a d^2 e x^2+12 a d^3 x^3+36 a d e^2 x-36 a e^3 \log \left (\frac{d x}{e}+1\right )+6 b \log \left (c x^n\right ) \left (d x \left (2 d^2 x^2-3 d e x+6 e^2\right )-6 e^3 \log \left (\frac{d x}{e}+1\right )\right )+9 b d^2 e n x^2-4 b d^3 n x^3-36 b d e^2 n x}{36 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.174, 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^2*(a+b*ln(c*x^n))/(d+e/x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 109.785, size = 248, normalized size = 1.68 \begin{align*} \frac{a x^{3}}{3 d} - \frac{a e x^{2}}{2 d^{2}} - \frac{a e^{3} \left (\begin{cases} \frac{x}{e} & \text{for}\: d = 0 \\\frac{\log{\left (d x + e \right )}}{d} & \text{otherwise} \end{cases}\right )}{d^{3}} + \frac{a e^{2} x}{d^{3}} - \frac{b n x^{3}}{9 d} + \frac{b x^{3} \log{\left (c x^{n} \right )}}{3 d} + \frac{b e n x^{2}}{4 d^{2}} - \frac{b e x^{2} \log{\left (c x^{n} \right )}}{2 d^{2}} + \frac{b e^{3} n \left (\begin{cases} \frac{x}{e} & \text{for}\: d = 0 \\\frac{\begin{cases} \log{\left (e \right )} \log{\left (x \right )} - \operatorname{Li}_{2}\left (\frac{d x e^{i \pi }}{e}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (e \right )} \log{\left (\frac{1}{x} \right )} - \operatorname{Li}_{2}\left (\frac{d x e^{i \pi }}{e}\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 (e \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (e \right )} - \operatorname{Li}_{2}\left (\frac{d x e^{i \pi }}{e}\right ) & \text{otherwise} \end{cases}}{d} & \text{otherwise} \end{cases}\right )}{d^{3}} - \frac{b e^{3} \left (\begin{cases} \frac{x}{e} & \text{for}\: d = 0 \\\frac{\log{\left (d x + e \right )}}{d} & \text{otherwise} \end{cases}\right ) \log{\left (c x^{n} \right )}}{d^{3}} - \frac{b e^{2} n x}{d^{3}} + \frac{b e^{2} x \log{\left (c x^{n} \right )}}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/(3*d) - a*e*x**2/(2*d**2) - a*e**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d**3 + a*e**2*x/d
**3 - b*n*x**3/(9*d) + b*x**3*log(c*x**n)/(3*d) + b*e*n*x**2/(4*d**2) - b*e*x**2*log(c*x**n)/(2*d**2) + b*e**3
*n*Piecewise((x/e, Eq(d, 0)), (Piecewise((log(e)*log(x) - polylog(2, d*x*exp_polar(I*pi)/e), Abs(x) < 1), (-lo
g(e)*log(1/x) - polylog(2, d*x*exp_polar(I*pi)/e), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log
(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e), True))/d, True))/d**3
- b*e**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x**n)/d**3 - b*e**2*n*x/d**3 + b*e**2*x*log
(c*x**n)/d**3

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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^{2}}{d + \frac{e}{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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