3.4.35 \(\int \frac {a+b \log (c x^n)}{(d+\frac {e}{x}) x^3} \, dx\) [335]

Optimal. Leaf size=95 \[ -\frac {b n}{e x}-\frac {a+b \log \left (c x^n\right )}{e x}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^2 n}+\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^2}+\frac {b d n \text {Li}_2\left (-\frac {d x}{e}\right )}{e^2} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {269, 46, 2393, 2341, 2338, 2354, 2438} \begin {gather*} \frac {b d n \text {PolyLog}\left (2,-\frac {d x}{e}\right )}{e^2}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^2 n}+\frac {d \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {a+b \log \left (c x^n\right )}{e x}-\frac {b n}{e x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 269

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 2338

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 2341

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 2354

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 2393

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 2438

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^3} \, dx &=\int \left (\frac {a+b \log \left (c x^n\right )}{e x^2}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 x}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 (e+d x)}\right ) \, dx\\ &=-\frac {d \int \frac {a+b \log \left (c x^n\right )}{x} \, dx}{e^2}+\frac {d^2 \int \frac {a+b \log \left (c x^n\right )}{e+d x} \, dx}{e^2}+\frac {\int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx}{e}\\ &=-\frac {b n}{e x}-\frac {a+b \log \left (c x^n\right )}{e x}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^2 n}+\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^2}-\frac {(b d n) \int \frac {\log \left (1+\frac {d x}{e}\right )}{x} \, dx}{e^2}\\ &=-\frac {b n}{e x}-\frac {a+b \log \left (c x^n\right )}{e x}-\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b e^2 n}+\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{e^2}+\frac {b d n \text {Li}_2\left (-\frac {d x}{e}\right )}{e^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 88, normalized size = 0.93 \begin {gather*} -\frac {\frac {2 b e n}{x}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )-2 b d n \text {Li}_2\left (-\frac {d x}{e}\right )}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.06, size = 504, normalized size = 5.31

method	result	size
risch	\(-\frac {b \ln \left (x^{n}\right ) d \ln \left (x \right )}{e^{2}}+\frac {b n d \ln \left (x \right )^{2}}{2 e^{2}}-\frac {b n d \dilog \left (-\frac {d x}{e}\right )}{e^{2}}-\frac {b \ln \left (c \right )}{e x}-\frac {b n d \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{e^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d \ln \left (d x +e \right )}{2 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e x}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 e x}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d \ln \left (x \right )}{2 e^{2}}-\frac {a}{e x}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 e x}-\frac {b \ln \left (c \right ) d \ln \left (x \right )}{e^{2}}+\frac {b \ln \left (c \right ) d \ln \left (d x +e \right )}{e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (d x +e \right )}{2 e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (d x +e \right )}{2 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (x \right )}{2 e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 e x}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (x \right )}{2 e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d \ln \left (x \right )}{2 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d \ln \left (d x +e \right )}{2 e^{2}}-\frac {a d \ln \left (x \right )}{e^{2}}+\frac {a d \ln \left (d x +e \right )}{e^{2}}+\frac {b \ln \left (x^{n}\right ) d \ln \left (d x +e \right )}{e^{2}}-\frac {b \ln \left (x^{n}\right )}{e x}-\frac {b n}{e x}\)	\(504\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*ln(x^n)*d/e^2*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*d/e^2*ln(x)-1/2*I*b*Pi*csgn(I*c*x^n)^3*d
/e^2*ln(d*x+e)-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/e/x-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e/x-b*n*d/e^2*d
ilog(-d*x/e)+1/2*b*n*d/e^2*ln(x)^2+1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/e/x-b*ln(c)/e/x-b*n*d/e^2*ln
(d*x+e)*ln(-d*x/e)+1/2*I*b*Pi*csgn(I*c*x^n)^3/e/x-a/e/x+1/2*I*b*Pi*csgn(I*c*x^n)^3*d/e^2*ln(x)-b*ln(c)*d/e^2*l
n(x)+b*ln(c)*d/e^2*ln(d*x+e)-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*d/e^2*ln(x)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*
x^n)^2*d/e^2*ln(d*x+e)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*d/e^2*ln(d*x+e)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^
n)^2*d/e^2*ln(x)-a*d/e^2*ln(x)+a*d/e^2*ln(d*x+e)+b*ln(x^n)*d/e^2*ln(d*x+e)-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*cs
gn(I*c*x^n)*d/e^2*ln(d*x+e)-b*ln(x^n)/e/x-b*n/e/x

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*e^(-2)*log(d*x + e) - d*e^(-2)*log(x) - e^(-1)/x)*a + b*integrate((log(c) + log(x^n))/(d*x^3 + x^2*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 59.41, size = 216, normalized size = 2.27 \begin {gather*} \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} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\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^{n} \right )}}{e^{2}} + \frac {b d n \log {\left (x \right )}^{2}}{2 e^{2}} - \frac {b d \log {\left (x \right )} \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n}{e x} - \frac {b \log {\left (c x^{n} \right )}}{e x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/(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*n*Piecewis
e((x/e, Eq(d, 0)), (Piecewise((-polylog(2, d*x*exp_polar(I*pi)/e), (Abs(x) < 1) & (1/Abs(x) < 1)), (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*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d
, True))*log(c*x**n)/e**2 + b*d*n*log(x)**2/(2*e**2) - b*d*log(x)*log(c*x**n)/e**2 - b*n/(e*x) - b*log(c*x**n)
/(e*x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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