3.349 \(\int \frac{x^3 \log (c (d+e x^2)^p)}{(f+g x^2)^2} \, dx\)

Optimal. Leaf size=155 \[ \frac{p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^2}+\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{2 g^2 \left (f+g x^2\right )}+\frac{\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^2}-\frac{e f p \log \left (d+e x^2\right )}{2 g^2 (e f-d g)}+\frac{e f p \log \left (f+g x^2\right )}{2 g^2 (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.221205, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {2475, 43, 2416, 2395, 36, 31, 2394, 2393, 2391} \[ \frac{p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^2}+\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{2 g^2 \left (f+g x^2\right )}+\frac{\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^2}-\frac{e f p \log \left (d+e x^2\right )}{2 g^2 (e f-d g)}+\frac{e f p \log \left (f+g x^2\right )}{2 g^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.103261, size = 131, normalized size = 0.85 \[ \frac{p \text{PolyLog}\left (2,\frac{g \left (d+e x^2\right )}{d g-e f}\right )+\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )+\frac{e f p \log \left (d+e x^2\right )}{d g-e f}+\frac{e f p \log \left (f+g x^2\right )}{e f-d g}}{2 g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.671, size = 732, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral(x^3*log((e*x^2 + d)^p*c)/(g^2*x^4 + 2*f*g*x^2 + 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*ln(c*(e*x**2+d)**p)/(g*x**2+f)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="giac")

[Out]

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