3.340 \(\int \frac{x \log (c (d+e x^2)^p)}{f+g x^2} \, dx\)

Optimal. Leaf size=70 \[ \frac{p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g}+\frac{\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g} \]

[Out]

(Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*g)])/(2*g) + (p*PolyLog[2, -((g*(d + e*x^2))/(e*f - d*g))])
/(2*g)

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Rubi [A]  time = 0.09544, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2475, 2394, 2393, 2391} \[ \frac{p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g}+\frac{\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*g)])/(2*g) + (p*PolyLog[2, -((g*(d + e*x^2))/(e*f - d*g))])
/(2*g)

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

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

Mathematica [A]  time = 0.0069087, size = 64, normalized size = 0.91 \[ \frac{p \text{PolyLog}\left (2,\frac{g \left (d+e x^2\right )}{d g-e f}\right )+\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*g)] + p*PolyLog[2, (g*(d + e*x^2))/(-(e*f) + d*g)])/(2*g)

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Maple [C]  time = 0.649, size = 472, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/g*ln(g*x^2+f)*ln((e*x^2+d)^p)-1/2/g*p*sum(ln(x-_alpha)*ln(g*x^2+f)-ln(x-_alpha)*(ln((RootOf(_Z^2*e*g+2*_Z*
_alpha*e*g-d*g+e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1))+ln((RootOf(_Z^2*e*g+2*
_Z*_alpha*e*g-d*g+e*f,index=2)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=2)))-dilog((RootOf(_Z^2
*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1))-dilog((RootO
f(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=2)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=2)),_alpha
=RootOf(_Z^2*e+d))+1/4*I/g*ln(g*x^2+f)*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2-1/4*I/g*ln(g*x^2+f)*Pi*c
sgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)-1/4*I/g*ln(g*x^2+f)*Pi*csgn(I*c*(e*x^2+d)^p)^3+1/4*I/g*ln(g
*x^2+f)*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)+1/2/g*ln(g*x^2+f)*ln(c)

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Maxima [B]  time = 1.01482, size = 186, normalized size = 2.66 \begin{align*} \frac{e p{\left (\frac{\log \left (e x^{2} + d\right ) \log \left (g x^{2} + f\right )}{e} - \frac{\log \left (g x^{2} + f\right ) \log \left (-\frac{e g x^{2} + e f}{e f - d g} + 1\right ) +{\rm Li}_2\left (\frac{e g x^{2} + e f}{e f - d g}\right )}{e}\right )}}{2 \, g} - \frac{p \log \left (e x^{2} + d\right ) \log \left (g x^{2} + f\right )}{2 \, g} + \frac{\log \left (g x^{2} + f\right ) \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{2 \, g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e*p*(log(e*x^2 + d)*log(g*x^2 + f)/e - (log(g*x^2 + f)*log(-(e*g*x^2 + e*f)/(e*f - d*g) + 1) + dilog((e*g*
x^2 + e*f)/(e*f - d*g)))/e)/g - 1/2*p*log(e*x^2 + d)*log(g*x^2 + f)/g + 1/2*log(g*x^2 + f)*log((e*x^2 + d)^p*c
)/g

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g x^{2} + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*(e*x^2+d)^p)/(g*x^2+f),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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