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

Optimal. Leaf size=107 \[ \frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {a d x}{e^2}-\frac {b d x \log \left (c x^n\right )}{e^2}+\frac {b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )}{e^3}+\frac {b d n x}{e^2}-\frac {b n x^2}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {b d^2 n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3}+\frac {d^2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {a d x}{e^2}-\frac {b d x \log \left (c x^n\right )}{e^2}+\frac {b d n x}{e^2}-\frac {b n x^2}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 105, normalized size = 0.98 \[ \frac {4 a d^2 \log \left (\frac {e x}{d}+1\right )-4 a d e x+2 a e^2 x^2+2 b \log \left (c x^n\right ) \left (2 d^2 \log \left (\frac {e x}{d}+1\right )+e x (e x-2 d)\right )+4 b d^2 n \text {Li}_2\left (-\frac {e x}{d}\right )+4 b d e n x-b e^2 n x^2}{4 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \log \left (c x^{n}\right ) + a x^{2}}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.21, size = 521, normalized size = 4.87 \[ -\frac {b d x \ln \left (x^{n}\right )}{e^{2}}+\frac {b \,d^{2} \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{e^{3}}-\frac {b \,d^{2} n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{e^{3}}+\frac {b \,d^{2} \ln \relax (c ) \ln \left (e x +d \right )}{e^{3}}-\frac {b d x \ln \relax (c )}{e^{2}}-\frac {b \,d^{2} n \dilog \left (-\frac {e x}{d}\right )}{e^{3}}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi b d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 e^{2}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 e}+\frac {a \,x^{2}}{2 e}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 e^{3}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 e}-\frac {i \pi b d x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{2}}-\frac {i \pi b d x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e^{2}}+\frac {b \,x^{2} \ln \left (x^{n}\right )}{2 e}+\frac {5 b \,d^{2} n}{4 e^{3}}+\frac {a \,d^{2} \ln \left (e x +d \right )}{e^{3}}+\frac {b \,x^{2} \ln \relax (c )}{2 e}-\frac {b n \,x^{2}}{4 e}-\frac {i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 e^{3}}+\frac {i \pi b d x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 e^{2}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}+\frac {b d n x}{e^{2}}-\frac {a d x}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*ln(x^n)/e^2*x*d+b*ln(x^n)*d^2/e^3*ln(e*x+d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d^2/e^3*ln(e*x+d
)-b*n*d^2/e^3*ln(e*x+d)*ln(-1/d*e*x)+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e*x^2+1/2*I*b*Pi*csgn(I*c*x^n)^3/e
^2*x*d+b*ln(c)*d^2/e^3*ln(e*x+d)-b*n*d^2/e^3*dilog(-1/d*e*x)-b*ln(c)/e^2*x*d-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x
^n)*csgn(I*c)/e*x^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d^2/e^3*ln(e*x+d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x
^n)^2/e^2*x*d+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d^2/e^3*ln(e*x+d)+1/2*a/e*x^2-1/2*I*b*Pi*csgn(I*c*x^n)^2*cs
gn(I*c)/e^2*x*d-1/2*I*b*Pi*csgn(I*c*x^n)^3*d^2/e^3*ln(e*x+d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^
2*x*d+1/2*b*ln(x^n)/e*x^2-1/4*I*b*Pi*csgn(I*c*x^n)^3/e*x^2+5/4*b*n*d^2/e^3+a*d^2/e^3*ln(e*x+d)+1/2*b*ln(c)/e*x
^2+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e*x^2-1/4*b*n*x^2/e+b*d*n*x/e^2-a*d*x/e^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} + b \int \frac {x^{2} \log \relax (c) + x^{2} \log \left (x^{n}\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + b*integrate((x^2*log(c) + x^2*log(x^n))/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 32.81, size = 199, normalized size = 1.86 \[ \frac {a d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {a d x}{e^{2}} + \frac {a x^{2}}{2 e} - \frac {b d^{2} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} + \frac {b d n x}{e^{2}} - \frac {b d x \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n x^{2}}{4 e} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**2 - a*d*x/e**2 + a*x**2/(2*e) - b*d**2*n*Piecewis
e((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) + meije
rg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e**2 + b*d**2*P
iecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**2 + b*d*n*x/e**2 - b*d*x*log(c*x**n)/e**2 - b
*n*x**2/(4*e) + b*x**2*log(c*x**n)/(2*e)

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