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

Optimal. Leaf size=63 \[ -\frac {e \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^2}+\frac {a x}{d}+\frac {b x \log (c x)}{d}-\frac {b e \text {Li}_2\left (-\frac {d x}{e}\right )}{d^2}-\frac {b x}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {193, 43, 2330, 2295, 2317, 2391} \[ -\frac {b e \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^2}-\frac {e \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^2}+\frac {a x}{d}+\frac {b x \log (c x)}{d}-\frac {b x}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x])/(d + e/x),x]

[Out]

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

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 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

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 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 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && 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 {a+b \log (c x)}{d+\frac {e}{x}} \, dx &=\int \left (\frac {a+b \log (c x)}{d}-\frac {e (a+b \log (c x))}{d (e+d x)}\right ) \, dx\\ &=\frac {\int (a+b \log (c x)) \, dx}{d}-\frac {e \int \frac {a+b \log (c x)}{e+d x} \, dx}{d}\\ &=\frac {a x}{d}-\frac {e (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^2}+\frac {b \int \log (c x) \, dx}{d}+\frac {(b e) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{d^2}\\ &=\frac {a x}{d}-\frac {b x}{d}+\frac {b x \log (c x)}{d}-\frac {e (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^2}-\frac {b e \text {Li}_2\left (-\frac {d x}{e}\right )}{d^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 64, normalized size = 1.02 \[ -\frac {e \log \left (\frac {d x+e}{e}\right ) (a+b \log (c x))}{d^2}+\frac {a x}{d}+\frac {b x \log (c x)}{d}-\frac {b e \text {Li}_2\left (-\frac {d x}{e}\right )}{d^2}-\frac {b x}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 91, normalized size = 1.44 \[ \frac {b x \ln \left (c x \right )}{d}-\frac {b e \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{c e}\right )}{d^{2}}+\frac {a x}{d}-\frac {a e \ln \left (c d x +c e \right )}{d^{2}}-\frac {b x}{d}-\frac {b e \dilog \left (\frac {c d x +c e}{c e}\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.89, size = 69, normalized size = 1.10 \[ -\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b e}{d^{2}} + \frac {b x \log \relax (x) + {\left (b {\left (\log \relax (c) - 1\right )} + a\right )} x}{d} - \frac {{\left (b e \log \relax (c) + a e\right )} \log \left (d x + e\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*x))/(d + e/x),x)

[Out]

int((a + b*log(c*x))/(d + e/x), x)

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sympy [A]  time = 71.22, size = 138, normalized size = 2.19 \[ - \frac {a e \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d} + \frac {a x}{d} + \frac {b e \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d} - \frac {b e \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{d} + \frac {b x \log {\left (c x \right )}}{d} - \frac {b x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*e*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d + a*x/d + b*e*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_polar(I*pi)/e), True))/d, True))/d - b*e*Piecewise((x/e, Eq(d, 0)), (log(d*x +
e)/d, True))*log(c*x)/d + b*x*log(c*x)/d - b*x/d

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