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

Optimal. Leaf size=181 \[ \frac{b f^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^3}+\frac{f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{a f x}{g^2}-\frac{b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}-\frac{b d^2 n \log (d+e x)}{2 e^2 g}+\frac{b d n x}{2 e g}+\frac{b f n x}{g^2}-\frac{b n x^2}{4 g} \]

[Out]

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

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Rubi [A]  time = 0.19333, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {43, 2416, 2389, 2295, 2395, 2394, 2393, 2391} \[ \frac{b f^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^3}+\frac{f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{a f x}{g^2}-\frac{b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}-\frac{b d^2 n \log (d+e x)}{2 e^2 g}+\frac{b d n x}{2 e g}+\frac{b f n x}{g^2}-\frac{b n x^2}{4 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.13688, size = 170, normalized size = 0.94 \[ \frac{b f^2 n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^3}+\frac{f^2 \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{a f x}{g^2}-\frac{b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac{b n \left (-\frac{2 d^2 \log (d+e x)}{e^2}+\frac{2 d x}{e}-x^2\right )}{4 g}+\frac{b f n x}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.561, size = 724, 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^2*(a+b*ln(c*(e*x+d)^n))/(g*x+f),x)

[Out]

1/2*b*ln((e*x+d)^n)/g*x^2+a*f^2/g^3*ln(g*x+f)+1/2*b*ln(c)/g*x^2+5/4*b*n/g^3*f^2-b*n/g^3*f^2*ln(g*x+f)*ln(((g*x
+f)*e+d*g-f*e)/(d*g-e*f))-1/2*b/e^2*n/g*d^2*ln((g*x+f)*e+d*g-f*e)+1/2*b/e*n/g^2*d*f+1/2*I*b*Pi*csgn(I*c*(e*x+d
)^n)^3/g^2*f*x+1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/g*x^2+b*ln(c)*f^2/g^3*ln(g*x+f)-b*ln(c)/g^2*
f*x-b*n/g^3*f^2*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))+1/2*a/g*x^2+1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I
*c*(e*x+d)^n)/g^2*f*x-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/g*x^2+b*ln((e*x+d)^n)*f^2/g^3*ln(g*x+f)+1/4*I*b*Pi*csgn
(I*c)*csgn(I*c*(e*x+d)^n)^2/g*x^2-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*f^2/g^3*ln(g*x+f)
-b/e*n/g^2*d*ln((g*x+f)*e+d*g-f*e)*f-b*ln((e*x+d)^n)/g^2*f*x-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3*f^2/g^3*ln(g*x+f
)-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/g*x^2+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+
d)^n)^2*f^2/g^3*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2*f^2/g^3*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*f*x-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/g^2*f*x-1/4*b*n*x^2/g+1/2*b*d*n
*x/e/g-a*f*x/g^2+b*f*n*x/g^2

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

[Out]

1/2*a*(2*f^2*log(g*x + f)/g^3 + (g*x^2 - 2*f*x)/g^2) + b*integrate((x^2*log((e*x + d)^n) + x^2*log(c))/(g*x +
f), 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 x + f}, 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),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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}}{g x + f}\,{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),x, algorithm="giac")

[Out]

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