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

Optimal. Leaf size=265 \[ \frac{3 b f^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^4}+\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac{3 f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{2 a f x}{g^3}-\frac{2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac{b e f^3 n \log (d+e x)}{g^4 (e f-d g)}+\frac{b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac{b d n x}{2 e g^2}+\frac{2 b f n x}{g^3}-\frac{b n x^2}{4 g^2} \]

[Out]

(-2*a*f*x)/g^3 + (2*b*f*n*x)/g^3 + (b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2
) - (b*e*f^3*n*Log[d + e*x])/(g^4*(e*f - d*g)) - (2*b*f*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) + (x^2*(a + b*Lo
g[c*(d + e*x)^n]))/(2*g^2) + (f^3*(a + b*Log[c*(d + e*x)^n]))/(g^4*(f + g*x)) + (b*e*f^3*n*Log[f + g*x])/(g^4*
(e*f - d*g)) + (3*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 + (3*b*f^2*n*PolyLog[2, -
((g*(d + e*x))/(e*f - d*g))])/g^4

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Rubi [A]  time = 0.260384, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {43, 2416, 2389, 2295, 2395, 36, 31, 2394, 2393, 2391} \[ \frac{3 b f^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^4}+\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac{3 f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{2 a f x}{g^3}-\frac{2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac{b e f^3 n \log (d+e x)}{g^4 (e f-d g)}+\frac{b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac{b d n x}{2 e g^2}+\frac{2 b f n x}{g^3}-\frac{b n x^2}{4 g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*f*x)/g^3 + (2*b*f*n*x)/g^3 + (b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2
) - (b*e*f^3*n*Log[d + e*x])/(g^4*(e*f - d*g)) - (2*b*f*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) + (x^2*(a + b*Lo
g[c*(d + e*x)^n]))/(2*g^2) + (f^3*(a + b*Log[c*(d + e*x)^n]))/(g^4*(f + g*x)) + (b*e*f^3*n*Log[f + g*x])/(g^4*
(e*f - d*g)) + (3*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 + (3*b*f^2*n*PolyLog[2, -
((g*(d + e*x))/(e*f - d*g))])/g^4

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 2389

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx &=\int \left (-\frac{2 f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}-\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)^2}+\frac{3 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}\right ) \, dx\\ &=-\frac{(2 f) \int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^3}+\frac{\left (3 f^2\right ) \int \frac{a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g^3}-\frac{f^3 \int \frac{a+b \log \left (c (d+e x)^n\right )}{(f+g x)^2} \, dx}{g^3}+\frac{\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^2}\\ &=-\frac{2 a f x}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac{3 f^2 \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^4}-\frac{(2 b f) \int \log \left (c (d+e x)^n\right ) \, dx}{g^3}-\frac{\left (3 b e f^2 n\right ) \int \frac{\log \left (\frac{e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{g^4}-\frac{\left (b e f^3 n\right ) \int \frac{1}{(d+e x) (f+g x)} \, dx}{g^4}-\frac{(b e n) \int \frac{x^2}{d+e x} \, dx}{2 g^2}\\ &=-\frac{2 a f x}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac{3 f^2 \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^4}-\frac{(2 b f) \operatorname{Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g^3}-\frac{\left (3 b f^2 n\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^4}-\frac{(b e n) \int \left (-\frac{d}{e^2}+\frac{x}{e}+\frac{d^2}{e^2 (d+e x)}\right ) \, dx}{2 g^2}-\frac{\left (b e^2 f^3 n\right ) \int \frac{1}{d+e x} \, dx}{g^4 (e f-d g)}+\frac{\left (b e f^3 n\right ) \int \frac{1}{f+g x} \, dx}{g^3 (e f-d g)}\\ &=-\frac{2 a f x}{g^3}+\frac{2 b f n x}{g^3}+\frac{b d n x}{2 e g^2}-\frac{b n x^2}{4 g^2}-\frac{b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac{b e f^3 n \log (d+e x)}{g^4 (e f-d g)}-\frac{2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac{f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac{b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac{3 f^2 \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^4}+\frac{3 b f^2 n \text{Li}_2\left (-\frac{g (d+e x)}{e f-d g}\right )}{g^4}\\ \end{align*}

Mathematica [A]  time = 0.324783, size = 220, normalized size = 0.83 \[ \frac{12 b f^2 n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )+\frac{4 f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+12 f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+2 g^2 x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )-8 a f g x-\frac{8 b f g (d+e x) \log \left (c (d+e x)^n\right )}{e}-\frac{b g^2 n \left (2 d^2 \log (d+e x)+e x (e x-2 d)\right )}{e^2}-\frac{4 b e f^3 n (\log (d+e x)-\log (f+g x))}{e f-d g}+8 b f g n x}{4 g^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*f*g*x + 8*b*f*g*n*x - (b*g^2*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x]))/e^2 - (8*b*f*g*(d + e*x)*Log[c*(
d + e*x)^n])/e + 2*g^2*x^2*(a + b*Log[c*(d + e*x)^n]) + (4*f^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x) - (4*b*e*
f^3*n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g) + 12*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f -
d*g)] + 12*b*f^2*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)])/(4*g^4)

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Maple [C]  time = 0.565, size = 1063, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a/g^4*f^2*ln(g*x+f)+a*f^3/g^4/(g*x+f)+1/2*b*ln(c)/g^2*x^2-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g^2*x^2+1/2*b/e*n
/g^3*d*f-3*b*n/g^4*f^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))+9/4*b*n/g^4*f^2-3/2*I*b*Pi*csgn(I*c*(e*x+d)
^n)^3/g^4*f^2*ln(g*x+f)-2*b*ln((e*x+d)^n)/g^3*f*x+3*b*ln((e*x+d)^n)/g^4*f^2*ln(g*x+f)+b*ln((e*x+d)^n)*f^3/g^4/
(g*x+f)+2*b*n/g^3/(d*g-e*f)*ln((g*x+f)*e+d*g-f*e)*d*f^2+b*ln(c)*f^3/g^4/(g*x+f)-2*b*ln(c)/g^3*f*x+3*b*ln(c)/g^
4*f^2*ln(g*x+f)-3*b*n/g^4*f^2*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))+1/2*b*ln((e*x+d)^n)/g^2*x^2+1/2*a/g^2*x^2+b
*e*n/g^4/(d*g-e*f)*ln((g*x+f)*e+d*g-f*e)*f^3-b*e*n/g^4*f^3/(d*g-e*f)*ln(g*x+f)-1/2*b/e^2*n/g/(d*g-e*f)*ln((g*x
+f)*e+d*g-f*e)*d^3-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^2*x^2+1/2*I*b*Pi*csgn(I*(e*x+d
)^n)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4/(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*f^3/g^4/(g*x+f)+3/2*I*b*
Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^4*f^2*ln(g*x+f)-I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^3*f*x+3/2*I*b*Pi
*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^4*f^2*ln(g*x+f)-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g^3*
f*x-3/2*b/e*n/g^2/(d*g-e*f)*ln((g*x+f)*e+d*g-f*e)*d^2*f-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3*f^3/g^4/(g*x+f)+I*b*P
i*csgn(I*c*(e*x+d)^n)^3/g^3*f*x+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^2*x^2+1/4*I*b*Pi*csgn(I*(e*x+d)^n
)*csgn(I*c*(e*x+d)^n)^2/g^2*x^2-2*a*f*x/g^3-1/4*b*n*x^2/g^2+2*b*f*n*x/g^3-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^
n)*csgn(I*c*(e*x+d)^n)*f^3/g^4/(g*x+f)+1/2*b*d*n*x/e/g^2+I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n
)/g^3*f*x-3/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g^4*f^2*ln(g*x+f)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (\frac{2 \, f^{3}}{g^{5} x + f g^{4}} + \frac{6 \, f^{2} \log \left (g x + f\right )}{g^{4}} + \frac{g x^{2} - 4 \, f x}{g^{3}}\right )} a + b \int \frac{x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{3} \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^3*(a+b*log(c*(e*x+d)^n))/(g*x+f)^2,x, algorithm="maxima")

[Out]

1/2*(2*f^3/(g^5*x + f*g^4) + 6*f^2*log(g*x + f)/g^4 + (g*x^2 - 4*f*x)/g^3)*a + b*integrate((x^3*log((e*x + d)^
n) + x^3*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^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{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(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f)^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3/(g*x + f)^2, x)