3.343 \(\int \frac{a+b \log (c x)}{(d+\frac{e}{x}) x^3} \, dx\)

Optimal. Leaf size=84 \[ \frac{b d \text{PolyLog}\left (2,-\frac{d x}{e}\right )}{e^2}-\frac{d (a+b \log (c x))^2}{2 b e^2}+\frac{d \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))}{e^2}-\frac{a+b \log (c x)}{e x}-\frac{b}{e x} \]

[Out]

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

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Rubi [A]  time = 0.131292, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {263, 44, 2351, 2304, 2301, 2317, 2391} \[ \frac{b d \text{PolyLog}\left (2,-\frac{d x}{e}\right )}{e^2}-\frac{d (a+b \log (c x))^2}{2 b e^2}+\frac{d \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))}{e^2}-\frac{a+b \log (c x)}{e x}-\frac{b}{e x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x])/((d + e/x)*x^3),x]

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, 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 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^3} \, dx &=\int \left (\frac{a+b \log (c x)}{e x^2}-\frac{d (a+b \log (c x))}{e^2 x}+\frac{d^2 (a+b \log (c x))}{e^2 (e+d x)}\right ) \, dx\\ &=-\frac{d \int \frac{a+b \log (c x)}{x} \, dx}{e^2}+\frac{d^2 \int \frac{a+b \log (c x)}{e+d x} \, dx}{e^2}+\frac{\int \frac{a+b \log (c x)}{x^2} \, dx}{e}\\ &=-\frac{b}{e x}-\frac{a+b \log (c x)}{e x}-\frac{d (a+b \log (c x))^2}{2 b e^2}+\frac{d (a+b \log (c x)) \log \left (1+\frac{d x}{e}\right )}{e^2}-\frac{(b d) \int \frac{\log \left (1+\frac{d x}{e}\right )}{x} \, dx}{e^2}\\ &=-\frac{b}{e x}-\frac{a+b \log (c x)}{e x}-\frac{d (a+b \log (c x))^2}{2 b e^2}+\frac{d (a+b \log (c x)) \log \left (1+\frac{d x}{e}\right )}{e^2}+\frac{b d \text{Li}_2\left (-\frac{d x}{e}\right )}{e^2}\\ \end{align*}

Mathematica [A]  time = 0.0883366, size = 77, normalized size = 0.92 \[ -\frac{-2 b d \text{PolyLog}\left (2,-\frac{d x}{e}\right )-2 d \log \left (\frac{d x}{e}+1\right ) (a+b \log (c x))+\frac{d (a+b \log (c x))^2}{b}+\frac{2 e (a+b \log (c x))}{x}+\frac{2 b e}{x}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x])/((d + e/x)*x^3),x]

[Out]

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

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Maple [A]  time = 0.076, size = 120, normalized size = 1.4 \begin{align*}{\frac{ad\ln \left ( cdx+ce \right ) }{{e}^{2}}}-{\frac{a}{ex}}-{\frac{ad\ln \left ( cx \right ) }{{e}^{2}}}+{\frac{bd}{{e}^{2}}{\it dilog} \left ({\frac{cdx+ce}{ce}} \right ) }+{\frac{bd\ln \left ( cx \right ) }{{e}^{2}}\ln \left ({\frac{cdx+ce}{ce}} \right ) }-{\frac{bd \left ( \ln \left ( cx \right ) \right ) ^{2}}{2\,{e}^{2}}}-{\frac{b\ln \left ( cx \right ) }{ex}}-{\frac{b}{ex}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x))/(d+e/x)/x^3,x)

[Out]

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

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Maxima [A]  time = 1.36649, size = 130, normalized size = 1.55 \begin{align*} \frac{{\left (\log \left (\frac{d x}{e} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{d x}{e}\right )\right )} b d}{e^{2}} + \frac{{\left (b d \log \left (c\right ) + a d\right )} \log \left (d x + e\right )}{e^{2}} - \frac{b d x \log \left (x\right )^{2} + 2 \,{\left (e \log \left (c\right ) + e\right )} b + 2 \, a e + 2 \,{\left (b e +{\left (b d \log \left (c\right ) + a d\right )} x\right )} \log \left (x\right )}{2 \, e^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left (c x\right ) + a}{d x^{3} + e x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x) + a)/(d*x^3 + e*x^2), x)

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Sympy [A]  time = 52.3139, size = 187, normalized size = 2.23 \begin{align*} \frac{a d^{2} \left (\begin{cases} \frac{x}{e} & \text{for}\: d = 0 \\\frac{\log{\left (d x + e \right )}}{d} & \text{otherwise} \end{cases}\right )}{e^{2}} - \frac{a d \log{\left (x \right )}}{e^{2}} - \frac{a}{e x} - \frac{b d^{2} \left (\begin{cases} \frac{x}{e} & \text{for}\: d = 0 \\\frac{\begin{cases} \log{\left (e \right )} \log{\left (x \right )} - \operatorname{Li}_{2}\left (\frac{d x e^{i \pi }}{e}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (e \right )} \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{\left (e \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (e \right )} - \operatorname{Li}_{2}\left (\frac{d x e^{i \pi }}{e}\right ) & \text{otherwise} \end{cases}}{d} & \text{otherwise} \end{cases}\right )}{e^{2}} + \frac{b d^{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 \right )}}{e^{2}} + \frac{b d \log{\left (x \right )}^{2}}{2 e^{2}} - \frac{b d \log{\left (x \right )} \log{\left (c x \right )}}{e^{2}} - \frac{b \log{\left (c x \right )}}{e x} - \frac{b}{e x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x))/(d+e/x)/x**3,x)

[Out]

a*d**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/e**2 - a*d*log(x)/e**2 - a/(e*x) - b*d**2*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))/e**2 + b*d**2*Pie
cewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/e**2 + b*d*log(x)**2/(2*e**2) - b*d*log(x)*log(c*x)/e
**2 - b*log(c*x)/(e*x) - b/(e*x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x\right ) + a}{{\left (d + \frac{e}{x}\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x))/(d+e/x)/x^3,x, algorithm="giac")

[Out]

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