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

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

[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

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Rubi [A]  time = 0.152176, antiderivative size = 138, normalized size of antiderivative = 1., 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 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 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]

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 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 31

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

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 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 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{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*}

Mathematica [A]  time = 0.109138, size = 114, normalized size = 0.83 \[ \frac{b n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+\log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-\frac{b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}}{g^2} \]

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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Maple [C]  time = 0.504, size = 519, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 \left (c\right )}{g^{2} x^{2} + 2 \, f g x + f^{2}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)