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

Optimal. Leaf size=41 \[ \frac {b \text {Li}_2\left (-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2337, 2391} \[ \frac {b \text {PolyLog}\left (2,-\frac {e}{d x}\right )}{e}-\frac {\log \left (\frac {e}{d x}+1\right ) (a+b \log (c x))}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 54, normalized size = 1.32 \[ \frac {(a+b \log (c x)) \left (a+b \log (c x)-2 b \log \left (\frac {d x}{e}+1\right )\right )-2 b^2 \text {Li}_2\left (-\frac {d x}{e}\right )}{2 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 86, normalized size = 2.10 \[ \frac {b \ln \left (c x \right )^{2}}{2 e}-\frac {b \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{c e}\right )}{e}+\frac {a \ln \left (c x \right )}{e}-\frac {a \ln \left (c d x +c e \right )}{e}-\frac {b \dilog \left (\frac {c d x +c e}{c e}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.85, size = 67, normalized size = 1.63 \[ \frac {b \log \relax (x)^{2}}{2 \, e} - \frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b}{e} - \frac {{\left (b \log \relax (c) + a\right )} \log \left (d x + e\right )}{e} + \frac {{\left (b \log \relax (c) + a\right )} \log \relax (x)}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*log(x)^2/e - (log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b/e - (b*log(c) + a)*log(d*x + e)/e + (b*log(c) + a
)*log(x)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.26, size = 153, normalized size = 3.73 \[ \frac {2 a d \left (\begin {cases} - \frac {x}{e} - \frac {1}{2 d} & \text {for}\: d = 0 \\\frac {\log {\left (2 d x \right )}}{2 d} & \text {otherwise} \end {cases}\right )}{e} - \frac {2 a d \left (\begin {cases} \frac {x}{e} + \frac {1}{2 d} & \text {for}\: d = 0 \\\frac {\log {\left (2 d x + 2 e \right )}}{2 d} & \text {otherwise} \end {cases}\right )}{e} + b \left (\begin {cases} - \frac {1}{d x} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} + \operatorname {Li}_{2}\left (\frac {e e^{i \pi }}{d x}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} \frac {1}{d x} & \text {for}\: e = 0 \\\frac {\log {\left (d + \frac {e}{x} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x))/(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*Piecewise((-1/(d*x), Eq(e, 0)), (Piecewise((log(d)*log(x) + polyl
og(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) + polyl
og(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))*lo
g(c*x)

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