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

Optimal. Leaf size=74 \[ \frac {e \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {a+b \log \left (c x^n\right )}{d x}-\frac {b e n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^2}-\frac {b n}{d x} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 95, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {44, 2351, 2304, 2301, 2317, 2391} \[ \frac {b e n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2}-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^2 n}+\frac {e \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {a+b \log \left (c x^n\right )}{d x}-\frac {b n}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.09, size = 88, normalized size = 1.19 \[ -\frac {-2 e \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b e n \text {Li}_2\left (-\frac {e x}{d}\right )+\frac {2 b d n}{x}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e x^{3} + d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 504, normalized size = 6.81 \[ -\frac {b e n \dilog \left (-\frac {e x}{d}\right )}{d^{2}}+\frac {b e \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {b e \ln \relax (x ) \ln \left (x^{n}\right )}{d^{2}}+\frac {b e n \ln \relax (x )^{2}}{2 d^{2}}-\frac {b e \ln \relax (c ) \ln \relax (x )}{d^{2}}+\frac {b e \ln \relax (c ) \ln \left (e x +d \right )}{d^{2}}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 d x}-\frac {b e n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {a}{d x}-\frac {b \ln \left (x^{n}\right )}{d x}-\frac {i \pi b e \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}+\frac {i \pi b e \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{2}}-\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}+\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{2}}-\frac {a e \ln \relax (x )}{d^{2}}+\frac {a e \ln \left (e x +d \right )}{d^{2}}-\frac {b \ln \relax (c )}{d x}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 d x}+\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2 d^{2}}-\frac {b n}{d x}+\frac {i \pi b e \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 d^{2}}-\frac {i \pi b e \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d x}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 d x}-\frac {i \pi b e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +d \right )}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \relax (x)}{d^{2}} - \frac {1}{d x}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e x^{3} + d x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,\left (d+e\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 61.19, size = 197, normalized size = 2.66 \[ - \frac {a}{d x} + \frac {a e^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e \log {\relax (x )}}{d^{2}} - \frac {b n}{d x} - \frac {b \log {\left (c x^{n} \right )}}{d x} - \frac {b e^{2} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} \log {\relax (d )} \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (d )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (d )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (d )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b e n \log {\relax (x )}^{2}}{2 d^{2}} - \frac {b e \log {\relax (x )} \log {\left (c x^{n} \right )}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**2/(e*x+d),x)

[Out]

-a/(d*x) + a*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**2 - a*e*log(x)/d**2 - b*n/(d*x) - b*lo
g(c*x**n)/(d*x) - b*e**2*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((log(d)*log(x) - polylog(2, e*x*exp_polar(I*p
i)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1
)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d
), True))/e, True))/d**2 + b*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**2 + b*e*n*
log(x)**2/(2*d**2) - b*e*log(x)*log(c*x**n)/d**2

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