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

Optimal. Leaf size=136 \[ -\frac{b e^3 \text{PolyLog}\left (2,-\frac{d x}{e}\right )}{d^4}-\frac{e^3 \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))}{d^4}-\frac{e x^2 (a+b \log (c x))}{2 d^2}+\frac{x^3 (a+b \log (c x))}{3 d}+\frac{a e^2 x}{d^3}+\frac{b e^2 x \log (c x)}{d^3}-\frac{b e^2 x}{d^3}+\frac{b e x^2}{4 d^2}-\frac{b x^3}{9 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.151627, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {263, 43, 2351, 2295, 2304, 2317, 2391} \[ -\frac{b e^3 \text{PolyLog}\left (2,-\frac{d x}{e}\right )}{d^4}-\frac{e^3 \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))}{d^4}-\frac{e x^2 (a+b \log (c x))}{2 d^2}+\frac{x^3 (a+b \log (c x))}{3 d}+\frac{a e^2 x}{d^3}+\frac{b e^2 x \log (c x)}{d^3}-\frac{b e^2 x}{d^3}+\frac{b e x^2}{4 d^2}-\frac{b x^3}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0658093, size = 125, normalized size = 0.92 \[ \frac{-36 b e^3 \text{PolyLog}\left (2,-\frac{d x}{e}\right )-18 d^2 e x^2 (a+b \log (c x))+12 d^3 x^3 (a+b \log (c x))-36 e^3 \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))+36 a d e^2 x+36 b d e^2 x \log (c x)+9 b d^2 e x^2-4 b d^3 x^3-36 b d e^2 x}{36 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.043, size = 171, normalized size = 1.3 \begin{align*}{\frac{a{x}^{3}}{3\,d}}-{\frac{ae{x}^{2}}{2\,{d}^{2}}}+{\frac{a{e}^{2}x}{{d}^{3}}}-{\frac{a{e}^{3}\ln \left ( cdx+ce \right ) }{{d}^{4}}}+{\frac{b{x}^{3}\ln \left ( cx \right ) }{3\,d}}-{\frac{b{x}^{3}}{9\,d}}-{\frac{eb{x}^{2}\ln \left ( cx \right ) }{2\,{d}^{2}}}+{\frac{eb{x}^{2}}{4\,{d}^{2}}}+{\frac{b{e}^{2}x\ln \left ( cx \right ) }{{d}^{3}}}-{\frac{b{e}^{2}x}{{d}^{3}}}-{\frac{b{e}^{3}}{{d}^{4}}{\it dilog} \left ({\frac{cdx+ce}{ce}} \right ) }-{\frac{b{e}^{3}\ln \left ( cx \right ) }{{d}^{4}}\ln \left ({\frac{cdx+ce}{ce}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a/d*x^3-1/2*a/d^2*e*x^2+a*e^2*x/d^3-a*e^3/d^4*ln(c*d*x+c*e)+1/3*b/d*x^3*ln(c*x)-1/9*b*x^3/d-1/2*b/d^2*e*x^
2*ln(c*x)+1/4*b*e*x^2/d^2+b*e^2*x*ln(c*x)/d^3-b*e^2*x/d^3-b*e^3/d^4*dilog((c*d*x+c*e)/c/e)-b*e^3/d^4*ln(c*x)*l
n((c*d*x+c*e)/c/e)

________________________________________________________________________________________

Maxima [A]  time = 1.60506, size = 221, normalized size = 1.62 \begin{align*} -\frac{{\left (\log \left (\frac{d x}{e} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{d x}{e}\right )\right )} b e^{3}}{d^{4}} + \frac{4 \,{\left (3 \, a d^{2} +{\left (3 \, d^{2} \log \left (c\right ) - d^{2}\right )} b\right )} x^{3} - 9 \,{\left (2 \, a d e +{\left (2 \, d e \log \left (c\right ) - d e\right )} b\right )} x^{2} + 36 \,{\left (a e^{2} +{\left (e^{2} \log \left (c\right ) - e^{2}\right )} b\right )} x + 6 \,{\left (2 \, b d^{2} x^{3} - 3 \, b d e x^{2} + 6 \, b e^{2} x\right )} \log \left (x\right )}{36 \, d^{3}} - \frac{{\left (b e^{3} \log \left (c\right ) + a e^{3}\right )} \log \left (d x + e\right )}{d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b*e^3/d^4 + 1/36*(4*(3*a*d^2 + (3*d^2*log(c) - d^2)*b)*x^3 - 9*(2*a*d
*e + (2*d*e*log(c) - d*e)*b)*x^2 + 36*(a*e^2 + (e^2*log(c) - e^2)*b)*x + 6*(2*b*d^2*x^3 - 3*b*d*e*x^2 + 6*b*e^
2*x)*log(x))/d^3 - (b*e^3*log(c) + a*e^3)*log(d*x + e)/d^4

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \log \left (c x\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))/(d+e/x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 108.016, size = 235, normalized size = 1.73 \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 x^{3} \log{\left (c x \right )}}{3 d} - \frac{b x^{3}}{9 d} - \frac{b e x^{2} \log{\left (c x \right )}}{2 d^{2}} + \frac{b e x^{2}}{4 d^{2}} + \frac{b e^{3} \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 \right )}}{d^{3}} + \frac{b e^{2} x \log{\left (c x \right )}}{d^{3}} - \frac{b e^{2} x}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*ln(c*x))/(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*x**3*log(c*x)/(3*d) - b*x**3/(9*d) - b*e*x**2*log(c*x)/(2*d**2) + b*e*x**2/(4*d**2) + b*e**3*Piecewise
((x/e, Eq(d, 0)), (Piecewise((log(e)*log(x) - polylog(2, d*x*exp_polar(I*pi)/e), Abs(x) < 1), (-log(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) + meijer
g(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e), True))/d, True))/d**3 - b*e**3*Pi
ecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/d**3 + b*e**2*x*log(c*x)/d**3 - b*e**2*x/d**3

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x\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))/(d+e/x),x, algorithm="giac")

[Out]

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