3.50 \(\int \frac{(a+b \log (c (d+e x)^n))^2}{(f+g x)^3} \, dx\)

Optimal. Leaf size=202 \[ \frac{b^2 e^2 n^2 \text{PolyLog}\left (2,-\frac{e f-d g}{g (d+e x)}\right )}{g (e f-d g)^2}-\frac{b e^2 n \log \left (\frac{e f-d g}{g (d+e x)}+1\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)^2}-\frac{b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x) (e f-d g)^2}-\frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (f+g x)^2}+\frac{b^2 e^2 n^2 \log (f+g x)}{g (e f-d g)^2} \]

[Out]

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

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Rubi [A]  time = 0.38548, antiderivative size = 233, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2398, 2411, 2347, 2344, 2301, 2317, 2391, 2314, 31} \[ -\frac{b^2 e^2 n^2 \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)^2}+\frac{e^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (e f-d g)^2}-\frac{b e^2 n \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)^2}-\frac{b e n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x) (e f-d g)^2}-\frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (f+g x)^2}+\frac{b^2 e^2 n^2 \log (f+g x)}{g (e f-d g)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^3,x]

[Out]

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

Rule 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 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 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 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 2314

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

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.216715, size = 204, normalized size = 1.01 \[ \frac{\frac{e (f+g x) \left (-2 b^2 e n^2 (f+g x) \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+2 b n (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 b e n (f+g x) \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+e (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2-2 b^2 e n^2 (f+g x) (\log (d+e x)-\log (f+g x))\right )}{(e f-d g)^2}-\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{2 g (f+g x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^3,x]

[Out]

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

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Maple [C]  time = 0.759, size = 1473, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/(g*x+f)^2/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I/g*n*e^2/(d*g-e*f)^2*ln(g
*x+f)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/2*I/g*n*e^2/(d*g-e*f)^2*ln(g*x+f)*Pi*b^2*csgn(I*c*(e*x+
d)^n)^3+1/2*I/(g*x+f)^2/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-b/(g*x+f)^2/g*l
n((e*x+d)^n)*a-1/(g*x+f)^2/g*ln((e*x+d)^n)*b^2*ln(c)-1/2*I/g*n*e^2/(d*g-e*f)^2*ln(e*x+d)*Pi*b^2*csgn(I*c*(e*x+
d)^n)^3+1/2*I/g*n*e/(d*g-e*f)/(g*x+f)*Pi*b^2*csgn(I*c*(e*x+d)^n)^3-1/2*b^2/g*n^2*e^2/(d*g-e*f)^2*ln(e*x+d)^2-b
^2/g*n^2*e^2/(d*g-e*f)^2*ln(e*x+d)+b^2/g*n^2*e^2/(d*g-e*f)^2*ln(g*x+f)-1/8*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n
)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*
b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*a)^2/(g*x+f)^2/g+b^2/g*n^2*e^2/(d*g-e*f)^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-
f*e)/(d*g-e*f))-1/2*b^2/(g*x+f)^2/g*ln((e*x+d)^n)^2-1/2*I/(g*x+f)^2/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c)*csgn(I*c*
(e*x+d)^n)^2+b^2/g*n^2*e^2/(d*g-e*f)^2*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))-1/2*I/g*n*e/(d*g-e*f)/(g*x+f)*Pi*b
^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I/g*n*e/(d*g-e*f)/(g*x+f)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^
2+b^2/g*n*e^2*ln((e*x+d)^n)/(d*g-e*f)^2*ln(e*x+d)-1/2*I/g*n*e^2/(d*g-e*f)^2*ln(g*x+f)*Pi*b^2*csgn(I*c)*csgn(I*
c*(e*x+d)^n)^2-1/2*I/g*n*e^2/(d*g-e*f)^2*ln(e*x+d)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*
I/g*n*e/(d*g-e*f)/(g*x+f)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-b^2/g*n*e*ln((e*x+d)^n)/(d*g-
e*f)/(g*x+f)-b^2/g*n*e^2*ln((e*x+d)^n)/(d*g-e*f)^2*ln(g*x+f)+b/g*n*e^2/(d*g-e*f)^2*ln(e*x+d)*a-b/g*n*e/(d*g-e*
f)/(g*x+f)*a-b/g*n*e^2/(d*g-e*f)^2*ln(g*x+f)*a+1/g*n*e^2/(d*g-e*f)^2*ln(e*x+d)*b^2*ln(c)-1/g*n*e/(d*g-e*f)/(g*
x+f)*b^2*ln(c)-1/g*n*e^2/(d*g-e*f)^2*ln(g*x+f)*b^2*ln(c)+1/2*I/(g*x+f)^2/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c*(e*x+
d)^n)^3+1/2*I/g*n*e^2/(d*g-e*f)^2*ln(e*x+d)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/2*I/g*n*e^2/(d*g-
e*f)^2*ln(e*x+d)*Pi*b^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I/g*n*e^2/(d*g-e*f)^2*ln(g*x+f)*Pi*b^2*csgn(I*c)*c
sgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a b e n{\left (\frac{e \log \left (e x + d\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} - \frac{e \log \left (g x + f\right )}{e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}} + \frac{1}{e f^{2} g - d f g^{2} +{\left (e f g^{2} - d g^{3}\right )} x}\right )} - \frac{1}{2} \, b^{2}{\left (\frac{\log \left ({\left (e x + d\right )}^{n}\right )^{2}}{g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g} - 2 \, \int \frac{e g x \log \left (c\right )^{2} + d g \log \left (c\right )^{2} +{\left (e f n + 2 \, d g \log \left (c\right ) +{\left (e g n + 2 \, e g \log \left (c\right )\right )} x\right )} \log \left ({\left (e x + d\right )}^{n}\right )}{e g^{4} x^{4} + d f^{3} g +{\left (3 \, e f g^{3} + d g^{4}\right )} x^{3} + 3 \,{\left (e f^{2} g^{2} + d f g^{3}\right )} x^{2} +{\left (e f^{3} g + 3 \, d f^{2} g^{2}\right )} x}\,{d x}\right )} - \frac{a b \log \left ({\left (e x + d\right )}^{n} c\right )}{g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g} - \frac{a^{2}}{2 \,{\left (g^{3} x^{2} + 2 \, f g^{2} x + f^{2} g\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*e*n*(e*log(e*x + d)/(e^2*f^2*g - 2*d*e*f*g^2 + d^2*g^3) - e*log(g*x + f)/(e^2*f^2*g - 2*d*e*f*g^2 + d^2*g^
3) + 1/(e*f^2*g - d*f*g^2 + (e*f*g^2 - d*g^3)*x)) - 1/2*b^2*(log((e*x + d)^n)^2/(g^3*x^2 + 2*f*g^2*x + f^2*g)
- 2*integrate((e*g*x*log(c)^2 + d*g*log(c)^2 + (e*f*n + 2*d*g*log(c) + (e*g*n + 2*e*g*log(c))*x)*log((e*x + d)
^n))/(e*g^4*x^4 + d*f^3*g + (3*e*f*g^3 + d*g^4)*x^3 + 3*(e*f^2*g^2 + d*f*g^3)*x^2 + (e*f^3*g + 3*d*f^2*g^2)*x)
, x)) - a*b*log((e*x + d)^n*c)/(g^3*x^2 + 2*f*g^2*x + f^2*g) - 1/2*a^2/(g^3*x^2 + 2*f*g^2*x + f^2*g)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, a b \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{2}}{g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*log((e*x + d)^n*c)^2 + 2*a*b*log((e*x + d)^n*c) + a^2)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x + f^3)
, 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((a+b*ln(c*(e*x+d)**n))**2/(g*x+f)**3,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 )}^{2}}{{\left (g x + f\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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