3.3.47 \(\int \frac {a+b \log (c (d+e x)^n)}{x^2 (f+g x)} \, dx\) [247]

Optimal. Leaf size=162 \[ \frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}-\frac {g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^2}+\frac {b g n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^2}-\frac {b g n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {46, 2463, 2442, 36, 29, 31, 2441, 2352, 2440, 2438} \begin {gather*} \frac {b g n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^2}-\frac {b g n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}-\frac {g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {g \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}+\frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

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]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 141, normalized size = 0.87 \begin {gather*} \frac {\frac {b e f n (\log (x)-\log (d+e x))}{d}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+b g n \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )-b g n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.39, size = 669, normalized size = 4.13

method	result	size
risch	\(\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} g \ln \left (g x +f \right )}{2 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} g \ln \left (x \right )}{2 f^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2 f x}+\frac {b \ln \left (c \right ) g \ln \left (g x +f \right )}{f^{2}}-\frac {b \ln \left (c \right ) g \ln \left (x \right )}{f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} g \ln \left (x \right )}{2 f^{2}}-\frac {b n g \dilog \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}-\frac {a}{f x}-\frac {b \ln \left (c \right )}{f x}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{f^{2}}-\frac {b n g \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{2}}+\frac {b n g \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} g \ln \left (g x +f \right )}{2 f^{2}}+\frac {a g \ln \left (g x +f \right )}{f^{2}}-\frac {a g \ln \left (x \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (g x +f \right )}{f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 f x}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} g \ln \left (g x +f \right )}{2 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 f x}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} g \ln \left (x \right )}{2 f^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 f x}+\frac {b n g \dilog \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{2 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) g \ln \left (g x +f \right )}{2 f^{2}}+\frac {b e n \ln \left (x \right )}{d f}-\frac {b e n \ln \left (e x +d \right )}{d f}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f x}\)	\(669\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/x^2/(g*x+f),x,method=_RETURNVERBOSE)

[Out]

1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(g*x+f)-b*n/f^2*g*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+b*n/
f^2*g*dilog((e*x+d)/d)-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(x)+b*ln(c)/f^2*g*ln(g*x+f)-b*ln(c)/
f^2*g*ln(x)+1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f^2*g*ln(x)-1/2*I*b*Pi*csgn(I*(e*x+d)^n
)*csgn(I*c*(e*x+d)^n)^2/f/x-a/f/x+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(g*x+f)-1/2*I*b*P
i*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(x)-b*ln(c)/f/x-b*ln((e*x+d)^n)/f^2*g*ln(x)-b*n/f^2*g*ln(g*x
+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+b*n/f^2*g*ln(x)*ln((e*x+d)/d)+a/f^2*g*ln(g*x+f)-a/f^2*g*ln(x)+b*ln((e*x+
d)^n)/f^2*g*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f/x-1/2*I*b*Pi*csgn(I*c)*csgn
(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f^2*g*ln(g*x+f)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f^2*g*ln(g*x+f)-1/2*I*b*Pi*
csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f/x+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f^2*g*ln(x)+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n
)^3/f/x+b*e*n*ln(x)/d/f-b*e*n*ln(e*x+d)/d/f-b*ln((e*x+d)^n)/f/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*(e*x+d)^n))/x^2/(g*x+f),x, algorithm="maxima")

[Out]

a*(g*log(g*x + f)/f^2 - g*log(x)/f^2 - 1/(f*x)) + b*integrate((log((x*e + d)^n) + log(c))/(g*x^3 + f*x^2), 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*(e*x+d)^n))/x^2/(g*x+f),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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*(e*x+d)^n))/x^2/(g*x+f),x, algorithm="giac")

[Out]

integrate((b*log((x*e + d)^n*c) + a)/((g*x + f)*x^2), 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\,{\left (d+e\,x\right )}^n\right )}{x^2\,\left (f+g\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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