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

Optimal. Leaf size=136 \[ -\frac {e^3 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^4}-\frac {e x^2 (a+b \log (c x))}{2 d^2}+\frac {x^3 (a+b \log (c x))}{3 d}+\frac {a e^2 x}{d^3}+\frac {b e^2 x \log (c x)}{d^3}-\frac {b e^3 \text {Li}_2\left (-\frac {d x}{e}\right )}{d^4}-\frac {b e^2 x}{d^3}+\frac {b e x^2}{4 d^2}-\frac {b x^3}{9 d} \]

[Out]

a*e^2*x/d^3-b*e^2*x/d^3+1/4*b*e*x^2/d^2-1/9*b*x^3/d+b*e^2*x*ln(c*x)/d^3-1/2*e*x^2*(a+b*ln(c*x))/d^2+1/3*x^3*(a
+b*ln(c*x))/d-e^3*(a+b*ln(c*x))*ln(1+d*x/e)/d^4-b*e^3*polylog(2,-d*x/e)/d^4

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Rubi [A]  time = 0.15, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.07, size = 125, normalized size = 0.92 \[ \frac {12 d^3 x^3 (a+b \log (c x))-18 d^2 e x^2 (a+b \log (c x))-36 e^3 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))+36 a d e^2 x+36 b d e^2 x \log (c x)-4 b d^3 x^3+9 b d^2 e x^2-36 b e^3 \text {Li}_2\left (-\frac {d x}{e}\right )-36 b d e^2 x}{36 d^4} \]

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)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left (c x\right ) + a x^{3}}{d x + e}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x\right ) + a\right )} x^{2}}{d + \frac {e}{x}}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 171, normalized size = 1.26 \[ \frac {b \,x^{3} \ln \left (c x \right )}{3 d}+\frac {a \,x^{3}}{3 d}-\frac {b \,x^{3}}{9 d}-\frac {b e \,x^{2} \ln \left (c x \right )}{2 d^{2}}-\frac {a e \,x^{2}}{2 d^{2}}+\frac {b e \,x^{2}}{4 d^{2}}+\frac {b \,e^{2} x \ln \left (c x \right )}{d^{3}}-\frac {b \,e^{3} \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{c e}\right )}{d^{4}}+\frac {a \,e^{2} x}{d^{3}}-\frac {a \,e^{3} \ln \left (c d x +c e \right )}{d^{4}}-\frac {b \,e^{2} x}{d^{3}}-\frac {b \,e^{3} \dilog \left (\frac {c d x +c e}{c e}\right )}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a/d*x^3-1/2*a/d^2*e*x^2+a/d^3*e^2*x-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)

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maxima [A]  time = 0.94, size = 164, normalized size = 1.21 \[ -\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b e^{3}}{d^{4}} + \frac {4 \, {\left (3 \, a d^{2} + {\left (3 \, d^{2} \log \relax (c) - d^{2}\right )} b\right )} x^{3} - 9 \, {\left (2 \, a d e + {\left (2 \, d e \log \relax (c) - d e\right )} b\right )} x^{2} + 36 \, {\left (a e^{2} + {\left (e^{2} \log \relax (c) - e^{2}\right )} b\right )} x + 6 \, {\left (2 \, b d^{2} x^{3} - 3 \, b d e x^{2} + 6 \, b e^{2} x\right )} \log \relax (x)}{36 \, d^{3}} - \frac {{\left (b e^{3} \log \relax (c) + a e^{3}\right )} \log \left (d x + e\right )}{d^{4}} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\ln \left (c\,x\right )\right )}{d+\frac {e}{x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 136.43, size = 235, normalized size = 1.73 \[ \frac {a x^{3}}{3 d} - \frac {a e x^{2}}{2 d^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a e^{2} x}{d^{3}} + \frac {b x^{3} \log {\left (c x \right )}}{3 d} - \frac {b x^{3}}{9 d} - \frac {b e x^{2} \log {\left (c x \right )}}{2 d^{2}} + \frac {b e x^{2}}{4 d^{2}} + \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{d^{3}} + \frac {b e^{2} x \log {\left (c x \right )}}{d^{3}} - \frac {b e^{2} x}{d^{3}} \]

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

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