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

Optimal. Leaf size=278 \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}-\frac{b f n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 g^2}-\frac{f \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{b d^2 n \log (d+e x)}{2 e^2 g}+\frac{b d n x}{2 e g}-\frac{b n x^2}{4 g} \]

[Out]

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

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Rubi [A]  time = 0.325094, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {266, 43, 2416, 2395, 260, 2394, 2393, 2391} \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}-\frac{b f n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 g^2}-\frac{f \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac{f \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{b d^2 n \log (d+e x)}{2 e^2 g}+\frac{b d n x}{2 e g}-\frac{b n x^2}{4 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx &=\int \left (\frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac{f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \left (f+g x^2\right )}\right ) \, dx\\ &=\frac{\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}-\frac{f \int \frac{x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx}{g}\\ &=\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{f \int \left (-\frac{a+b \log \left (c (d+e x)^n\right )}{2 \sqrt{g} \left (\sqrt{-f}-\sqrt{g} x\right )}+\frac{a+b \log \left (c (d+e x)^n\right )}{2 \sqrt{g} \left (\sqrt{-f}+\sqrt{g} x\right )}\right ) \, dx}{g}-\frac{(b e n) \int \frac{x^2}{d+e x} \, dx}{2 g}\\ &=\frac{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac{f \int \frac{a+b \log \left (c (d+e x)^n\right )}{\sqrt{-f}-\sqrt{g} x} \, dx}{2 g^{3/2}}-\frac{f \int \frac{a+b \log \left (c (d+e x)^n\right )}{\sqrt{-f}+\sqrt{g} x} \, dx}{2 g^{3/2}}-\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{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{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{e \sqrt{-f}+d \sqrt{g}}\right )}{2 g^2}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}+\frac{(b e f n) \int \frac{\log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{e \sqrt{-f}+d \sqrt{g}}\right )}{d+e x} \, dx}{2 g^2}+\frac{(b e f n) \int \frac{\log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right )}{d+e x} \, dx}{2 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{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{e \sqrt{-f}+d \sqrt{g}}\right )}{2 g^2}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}+\frac{(b f n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\sqrt{g} x}{e \sqrt{-f}-d \sqrt{g}}\right )}{x} \, dx,x,d+e x\right )}{2 g^2}+\frac{(b f n) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{\sqrt{g} x}{e \sqrt{-f}+d \sqrt{g}}\right )}{x} \, dx,x,d+e x\right )}{2 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{x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{e \sqrt{-f}+d \sqrt{g}}\right )}{2 g^2}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}-\frac{b f n \text{Li}_2\left (-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 g^2}-\frac{b f n \text{Li}_2\left (\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}+d \sqrt{g}}\right )}{2 g^2}\\ \end{align*}

Mathematica [A]  time = 0.172062, size = 243, normalized size = 0.87 \[ -\frac{2 b f n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )+2 b f n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )+2 f \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+2 f \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 g x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b g n \left (2 d^2 \log (d+e x)+e x (e x-2 d)\right )}{e^2}}{4 g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.456, size = 631, normalized size = 2.3 \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^2+f),x)

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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