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

Optimal. Leaf size=155 \[ \frac {f \log \left (c \left (d+e x^2\right )^p\right )}{2 g^2 \left (f+g x^2\right )}+\frac {\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^2}+\frac {p \text {Li}_2\left (-\frac {g \left (e x^2+d\right )}{e f-d g}\right )}{2 g^2}-\frac {e f p \log \left (d+e x^2\right )}{2 g^2 (e f-d g)}+\frac {e f p \log \left (f+g x^2\right )}{2 g^2 (e f-d g)} \]

[Out]

-1/2*e*f*p*ln(e*x^2+d)/g^2/(-d*g+e*f)+1/2*f*ln(c*(e*x^2+d)^p)/g^2/(g*x^2+f)+1/2*e*f*p*ln(g*x^2+f)/g^2/(-d*g+e*
f)+1/2*ln(c*(e*x^2+d)^p)*ln(e*(g*x^2+f)/(-d*g+e*f))/g^2+1/2*p*polylog(2,-g*(e*x^2+d)/(-d*g+e*f))/g^2

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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]

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 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 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 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 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])

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 = 0.09, size = 131, normalized size = 0.85 \[ \frac {\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )+p \text {Li}_2\left (\frac {g \left (e x^2+d\right )}{d g-e f}\right )+\frac {e f p \log \left (d+e x^2\right )}{d g-e f}+\frac {e f p \log \left (f+g x^2\right )}{e f-d g}}{2 g^2} \]

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)

________________________________________________________________________________________

fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}}\,{d x} \]

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)

________________________________________________________________________________________

maple [C]  time = 0.76, size = 732, normalized size = 4.72 \[ \text {result too large to display} \]

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

________________________________________________________________________________________

maxima [A]  time = 1.29, size = 181, normalized size = 1.17 \[ \frac {{\left (e f p + {\left (e f - d g\right )} \log \relax (c)\right )} \log \left (g x^{2} + f\right )}{2 \, {\left (e f g^{2} - d g^{3}\right )}} - \frac {{\left (e f g p x^{2} + d f g p\right )} \log \left (e x^{2} + d\right ) - {\left (e f^{2} - d f g\right )} \log \relax (c)}{2 \, {\left (e f^{2} g^{2} - d f g^{3} + {\left (e f g^{3} - d g^{4}\right )} x^{2}\right )}} + \frac {{\left (\log \left (e x^{2} + d\right ) \log \left (\frac {e g x^{2} + d g}{e f - d g} + 1\right ) + {\rm Li}_2\left (-\frac {e g x^{2} + d g}{e f - d g}\right )\right )} p}{2 \, g^{2}} \]

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]

1/2*(e*f*p + (e*f - d*g)*log(c))*log(g*x^2 + f)/(e*f*g^2 - d*g^3) - 1/2*((e*f*g*p*x^2 + d*f*g*p)*log(e*x^2 + d
) - (e*f^2 - d*f*g)*log(c))/(e*f^2*g^2 - d*f*g^3 + (e*f*g^3 - d*g^4)*x^2) + 1/2*(log(e*x^2 + d)*log((e*g*x^2 +
d*g)/(e*f - d*g) + 1) + dilog(-(e*g*x^2 + d*g)/(e*f - d*g)))*p/g^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{{\left (g\,x^2+f\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________