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

Optimal. Leaf size=44 \[ \frac {b n \text {Li}_2\left (-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2337, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

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

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Mathematica [A]  time = 0.03, size = 63, normalized size = 1.43 \[ \frac {\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-2 b n \log \left (\frac {d x}{e}+1\right )\right )}{2 b e n}-\frac {b n \text {Li}_2\left (-\frac {d x}{e}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{d x^{2} + e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.18, size = 336, normalized size = 7.64 \[ -\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2 e}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (d x +e \right )}{2 e}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 e}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 e}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 e}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (d x +e \right )}{2 e}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 e}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (d x +e \right )}{2 e}-\frac {b n \ln \relax (x )^{2}}{2 e}+\frac {b n \ln \left (-\frac {d x}{e}\right ) \ln \left (d x +e \right )}{e}+\frac {b n \dilog \left (-\frac {d x}{e}\right )}{e}+\frac {b \ln \relax (c ) \ln \relax (x )}{e}-\frac {b \ln \relax (c ) \ln \left (d x +e \right )}{e}+\frac {b \ln \relax (x ) \ln \left (x^{n}\right )}{e}-\frac {b \ln \left (x^{n}\right ) \ln \left (d x +e \right )}{e}+\frac {a \ln \relax (x )}{e}-\frac {a \ln \left (d x +e \right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,\left (d+\frac {e}{x}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.69, size = 156, normalized size = 3.55 \[ \frac {2 a d \left (\begin {cases} - \frac {x}{e} - \frac {1}{2 d} & \text {for}\: d = 0 \\\frac {\log {\left (2 d x \right )}}{2 d} & \text {otherwise} \end {cases}\right )}{e} - \frac {2 a d \left (\begin {cases} \frac {x}{e} + \frac {1}{2 d} & \text {for}\: d = 0 \\\frac {\log {\left (2 d x + 2 e \right )}}{2 d} & \text {otherwise} \end {cases}\right )}{e} + b n \left (\begin {cases} - \frac {1}{d x} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} \frac {1}{d x} & \text {for}\: e = 0 \\\frac {\log {\left (d + \frac {e}{x} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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