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

Optimal. Leaf size=112 \[ -\frac{f p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^2}-\frac{f \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^2}+\frac{\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g}-\frac{p x^2}{2 g} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.187701, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2475, 43, 2416, 2389, 2295, 2394, 2393, 2391} \[ -\frac{f p \text{PolyLog}\left (2,-\frac{g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^2}-\frac{f \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^2}+\frac{\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g}-\frac{p x^2}{2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(p*x^2)/(2*g) + ((d + e*x^2)*Log[c*(d + e*x^2)^p])/(2*e*g) - (f*Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f
- d*g)])/(2*g^2) - (f*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 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 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 )}{f+g x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x \log \left (c (d+e x)^p\right )}{f+g x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{\log \left (c (d+e x)^p\right )}{g}-\frac{f \log \left (c (d+e x)^p\right )}{g (f+g x)}\right ) \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \log \left (c (d+e x)^p\right ) \, 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} \, dx,x,x^2\right )}{2 g}\\ &=-\frac{f \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{\operatorname{Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^2\right )}{2 e g}+\frac{(e f 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{p x^2}{2 g}+\frac{\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g}-\frac{f \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{(f 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{p x^2}{2 g}+\frac{\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g}-\frac{f \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{f 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 = 0.0411701, size = 91, normalized size = 0.81 \[ -\frac{e f p \text{PolyLog}\left (2,\frac{g \left (d+e x^2\right )}{d g-e f}\right )-\log \left (c \left (d+e x^2\right )^p\right ) \left (-e f \log \left (\frac{e \left (f+g x^2\right )}{e f-d g}\right )+d g+e g x^2\right )+e g p x^2}{2 e g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.68, size = 672, normalized size = 6. \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),x)

[Out]

1/2*ln((e*x^2+d)^p)/g*x^2-1/2*ln((e*x^2+d)^p)*f/g^2*ln(g*x^2+f)-1/2*p*x^2/g+1/2*p/e/g*d*ln(e*x^2+d)+1/2*p*f/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)/Ro
otOf(_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))+1/4*I*Pi*csgn(I*(e*x^
2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2/g*x^2-1/4*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*f/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)/g*x^2+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*ln(g*x^2+f)-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3/g*x^2+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*f/g
^2*ln(g*x^2+f)+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)/g*x^2-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f/g
^2*ln(g*x^2+f)+1/2*ln(c)/g*x^2-1/2*ln(c)*f/g^2*ln(g*x^2+f)

________________________________________________________________________________________

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 )}{g x^{2} + f}\,{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),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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 x^{2} + f}, 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),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

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 )}{g x^{2} + f}\,{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),x, algorithm="giac")

[Out]

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