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

Optimal. Leaf size=186 \[ -\frac{2 b f n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^3}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}-\frac{2 f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{a x}{g^2}+\frac{b (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac{b e f^2 n \log (d+e x)}{g^3 (e f-d g)}-\frac{b e f^2 n \log (f+g x)}{g^3 (e f-d g)}-\frac{b n x}{g^2} \]

[Out]

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

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Rubi [A]  time = 0.203305, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ -\frac{2 b f n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^3}-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}-\frac{2 f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{a x}{g^2}+\frac{b (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac{b e f^2 n \log (d+e x)}{g^3 (e f-d g)}-\frac{b e f^2 n \log (f+g x)}{g^3 (e f-d g)}-\frac{b n x}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.153424, size = 153, normalized size = 0.82 \[ \frac{-2 b f n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )-\frac{f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-2 f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+a g x+\frac{b g (d+e x) \log \left (c (d+e x)^n\right )}{e}+\frac{b e f^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}-b g n x}{g^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.566, size = 791, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^2/(g*x + f)^2, x)