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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {b e^2 n \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3}+\frac {e^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d}-\frac {a e x}{d^2}-\frac {b e x \log \left (c x^n\right )}{d^2}+\frac {b e n x}{d^2}-\frac {b n x^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \log \left (c x^{n}\right ) + a x^{2}}{d x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 521, normalized size = 4.87 \[ -\frac {b \,e^{2} n \dilog \left (-\frac {d x}{e}\right )}{d^{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 d}-\frac {i \pi b e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2}}-\frac {b \,e^{2} n \ln \left (-\frac {d x}{e}\right ) \ln \left (d x +e \right )}{d^{3}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 d}-\frac {b e x \ln \relax (c )}{d^{2}}+\frac {b \,e^{2} \ln \relax (c ) \ln \left (d x +e \right )}{d^{3}}+\frac {b \,e^{2} \ln \left (x^{n}\right ) \ln \left (d x +e \right )}{d^{3}}-\frac {b e x \ln \left (x^{n}\right )}{d^{2}}+\frac {a \,x^{2}}{2 d}+\frac {i \pi b e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 d^{2}}-\frac {i \pi b \,e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (d x +e \right )}{2 d^{3}}+\frac {b \,x^{2} \ln \left (x^{n}\right )}{2 d}+\frac {a \,e^{2} \ln \left (d x +e \right )}{d^{3}}+\frac {b \,x^{2} \ln \relax (c )}{2 d}+\frac {5 b \,e^{2} n}{4 d^{3}}-\frac {i \pi b e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d^{2}}+\frac {i \pi b \,e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 d^{3}}+\frac {i \pi b \,e^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 d^{3}}-\frac {a e x}{d^{2}}+\frac {b e n x}{d^{2}}+\frac {i \pi b e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 d^{2}}-\frac {i \pi b \,e^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (d x +e \right )}{2 d^{3}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 d}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 d}-\frac {b n \,x^{2}}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 114.15, size = 199, normalized size = 1.86 \[ \frac {a x^{2}}{2 d} + \frac {a e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e x}{d^{2}} - \frac {b n x^{2}}{4 d} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 d} - \frac {b e^{2} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b e n x}{d^{2}} - \frac {b e x \log {\left (c x^{n} \right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**2/(2*d) + a*e**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d**2 - a*e*x/d**2 - b*n*x**2/(4*d) +
b*x**2*log(c*x**n)/(2*d) - b*e**2*n*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) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_pola
r(I*pi)/e), True))/d, True))/d**2 + b*e**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x**n)/d**2
+ b*e*n*x/d**2 - b*e*x*log(c*x**n)/d**2

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