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

Optimal. Leaf size=138 \[ \frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {b n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{g^2}-\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {43, 2416, 2395, 36, 31, 2394, 2393, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}-\frac {b e f n \log (d+e x)}{g^2 (e f-d g)}+\frac {b e f n \log (f+g x)}{g^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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]

Rule 2393

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 2394

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 2395

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 2416

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

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Mathematica [A]  time = 0.11, size = 114, normalized size = 0.83 \[ \frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+b n \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \log \left ({\left (e x + d\right )}^{n} c\right ) + a x}{g^{2} x^{2} + 2 \, f g x + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.23, size = 519, normalized size = 3.76 \[ \frac {b e f n \ln \left (d g -e f +\left (g x +f \right ) e \right )}{\left (d g -e f \right ) g^{2}}-\frac {b e f n \ln \left (g x +f \right )}{\left (d g -e f \right ) g^{2}}-\frac {i \pi b f \,\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 \left (g x +f \right ) g^{2}}+\frac {i \pi b f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 \left (g x +f \right ) g^{2}}+\frac {i \pi b f \,\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 \left (g x +f \right ) g^{2}}-\frac {i \pi b f \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 \left (g x +f \right ) g^{2}}-\frac {i \pi b \,\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 ) \ln \left (g x +f \right )}{2 g^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g x +f \right )}{2 g^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g x +f \right )}{2 g^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g x +f \right )}{2 g^{2}}-\frac {b n \ln \left (\frac {d g -e f +\left (g x +f \right ) e}{d g -e f}\right ) \ln \left (g x +f \right )}{g^{2}}+\frac {b f \ln \relax (c )}{\left (g x +f \right ) g^{2}}+\frac {b f \ln \left (\left (e x +d \right )^{n}\right )}{\left (g x +f \right ) g^{2}}-\frac {b n \dilog \left (\frac {d g -e f +\left (g x +f \right ) e}{d g -e f}\right )}{g^{2}}+\frac {b \ln \relax (c ) \ln \left (g x +f \right )}{g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g^{2}}+\frac {a f}{\left (g x +f \right ) g^{2}}+\frac {a \ln \left (g x +f \right )}{g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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