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

Optimal. Leaf size=190 \[ -\frac{6 b^2 e n^2 \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}+\frac{6 b^3 e n^3 \text{PolyLog}\left (3,-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)}-\frac{3 b e n \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g (e f-d g)}+\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(f+g x) (e f-d g)} \]

[Out]

((d + e*x)*(a + b*Log[c*(d + e*x)^n])^3)/((e*f - d*g)*(f + g*x)) - (3*b*e*n*(a + b*Log[c*(d + e*x)^n])^2*Log[(
e*(f + g*x))/(e*f - d*g)])/(g*(e*f - d*g)) - (6*b^2*e*n^2*(a + b*Log[c*(d + e*x)^n])*PolyLog[2, -((g*(d + e*x)
)/(e*f - d*g))])/(g*(e*f - d*g)) + (6*b^3*e*n^3*PolyLog[3, -((g*(d + e*x))/(e*f - d*g))])/(g*(e*f - d*g))

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Rubi [A]  time = 0.153738, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2397, 2396, 2433, 2374, 6589} \[ -\frac{6 b^2 e n^2 \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (e f-d g)}+\frac{6 b^3 e n^3 \text{PolyLog}\left (3,-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)}-\frac{3 b e n \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{g (e f-d g)}+\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)*(a + b*Log[c*(d + e*x)^n])^3)/((e*f - d*g)*(f + g*x)) - (3*b*e*n*(a + b*Log[c*(d + e*x)^n])^2*Log[(
e*(f + g*x))/(e*f - d*g)])/(g*(e*f - d*g)) - (6*b^2*e*n^2*(a + b*Log[c*(d + e*x)^n])*PolyLog[2, -((g*(d + e*x)
)/(e*f - d*g))])/(g*(e*f - d*g)) + (6*b^3*e*n^3*PolyLog[3, -((g*(d + e*x))/(e*f - d*g))])/(g*(e*f - d*g))

Rule 2397

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

Rule 2396

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

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(f+g x)^2} \, dx &=\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(e f-d g) (f+g x)}-\frac{(3 b e n) \int \frac{\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{f+g x} \, dx}{e f-d g}\\ &=\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(e f-d g) (f+g x)}-\frac{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g (e f-d g)}+\frac{\left (6 b^2 e^2 n^2\right ) \int \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 )}{d+e x} \, dx}{g (e f-d g)}\\ &=\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(e f-d g) (f+g x)}-\frac{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g (e f-d g)}+\frac{\left (6 b^2 e n^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (\frac{e \left (\frac{e f-d g}{e}+\frac{g x}{e}\right )}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)}\\ &=\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(e f-d g) (f+g x)}-\frac{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g (e f-d g)}-\frac{6 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text{Li}_2\left (-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)}+\frac{\left (6 b^3 e n^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g (e f-d g)}\\ &=\frac{(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{(e f-d g) (f+g x)}-\frac{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g (e f-d g)}-\frac{6 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \text{Li}_2\left (-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)}+\frac{6 b^3 e n^3 \text{Li}_3\left (-\frac{g (d+e x)}{e f-d g}\right )}{g (e f-d g)}\\ \end{align*}

Mathematica [B]  time = 0.400456, size = 410, normalized size = 2.16 \[ \frac{3 b^2 n^2 \left (\log (d+e x) \left (g (d+e x) \log (d+e x)-2 e (f+g x) \log \left (\frac{e (f+g x)}{e f-d g}\right )\right )-2 e (f+g x) \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+b^3 n^3 \left (6 e (f+g x) \text{PolyLog}\left (3,\frac{g (d+e x)}{d g-e f}\right )-6 e (f+g x) \log (d+e x) \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+\log ^2(d+e x) \left (g (d+e x) \log (d+e x)-3 e (f+g x) \log \left (\frac{e (f+g x)}{e f-d g}\right )\right )\right )-3 b n (e f-d g) \log (d+e x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2+3 b e n (f+g x) \log (d+e x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2-3 b e n (f+g x) \log (f+g x) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^2-(e f-d g) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )^3}{g (f+g x) (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(e*f - d*g)*n*Log[d + e*x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 + 3*b*e*n*(f + g*x)*Log[d + e
*x]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - (e*f - d*g)*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]
)^3 - 3*b*e*n*(f + g*x)*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2*Log[f + g*x] + 3*b^2*n^2*(a - b*n*Log[
d + e*x] + b*Log[c*(d + e*x)^n])*(Log[d + e*x]*(g*(d + e*x)*Log[d + e*x] - 2*e*(f + g*x)*Log[(e*(f + g*x))/(e*
f - d*g)]) - 2*e*(f + g*x)*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)]) + b^3*n^3*(Log[d + e*x]^2*(g*(d + e*x)*Lo
g[d + e*x] - 3*e*(f + g*x)*Log[(e*(f + g*x))/(e*f - d*g)]) - 6*e*(f + g*x)*Log[d + e*x]*PolyLog[2, (g*(d + e*x
))/(-(e*f) + d*g)] + 6*e*(f + g*x)*PolyLog[3, (g*(d + e*x))/(-(e*f) + d*g)]))/(g*(e*f - d*g)*(f + g*x))

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Maple [C]  time = 1.121, size = 5626, normalized size = 29.6 \begin{align*} \text{output 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))^3/(g*x+f)^2,x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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