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

Optimal. Leaf size=84 \[ -\frac {d (a+b \log (c x))^2}{2 b e^2}+\frac {d \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{e^2}-\frac {a+b \log (c x)}{e x}+\frac {b d \text {Li}_2\left (-\frac {d x}{e}\right )}{e^2}-\frac {b}{e x} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 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 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 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 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 {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*}

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Mathematica [A]  time = 0.09, size = 77, normalized size = 0.92 \[ -\frac {-2 d \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))+\frac {d (a+b \log (c x))^2}{b}+\frac {2 e (a+b \log (c x))}{x}-2 b d \text {Li}_2\left (-\frac {d x}{e}\right )+\frac {2 b e}{x}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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)])/e^2

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x\right ) + a}{d x^{3} + e x^{2}}, x\right ) \]

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

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)

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maple [A]  time = 0.05, size = 120, normalized size = 1.43 \[ -\frac {b d \ln \left (c x \right )^{2}}{2 e^{2}}+\frac {b d \ln \left (c x \right ) \ln \left (\frac {c d x +c e}{c e}\right )}{e^{2}}-\frac {a d \ln \left (c x \right )}{e^{2}}+\frac {a d \ln \left (c d x +c e \right )}{e^{2}}+\frac {b d \dilog \left (\frac {c d x +c e}{c e}\right )}{e^{2}}-\frac {b \ln \left (c x \right )}{e x}-\frac {a}{e x}-\frac {b}{e x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.93, size = 96, normalized size = 1.14 \[ \frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b d}{e^{2}} + \frac {{\left (b d \log \relax (c) + a d\right )} \log \left (d x + e\right )}{e^{2}} - \frac {b d x \log \relax (x)^{2} + 2 \, {\left (e \log \relax (c) + e\right )} b + 2 \, a e + 2 \, {\left (b e + {\left (b d \log \relax (c) + a d\right )} x\right )} \log \relax (x)}{2 \, e^{2} x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x\right )}{x^3\,\left (d+\frac {e}{x}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 62.91, size = 187, normalized size = 2.23 \[ \frac {a d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {a d \log {\relax (x )}}{e^{2}} - \frac {a}{e x} - \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} \log {\relax (e )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (e )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (e )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (e )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{e^{2}} + \frac {b d \log {\relax (x )}^{2}}{2 e^{2}} - \frac {b d \log {\relax (x )} \log {\left (c x \right )}}{e^{2}} - \frac {b \log {\left (c x \right )}}{e x} - \frac {b}{e x} \]

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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