3.348 \(\int \frac {x^5 \log (c (d+e x^2)^p)}{(f+g x^2)^2} \, dx\)

Optimal. Leaf size=199 \[ -\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{g^3}+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}+\frac {e f^2 p \log \left (d+e x^2\right )}{2 g^3 (e f-d g)}-\frac {e f^2 p \log \left (f+g x^2\right )}{2 g^3 (e f-d g)}-\frac {f p \text {Li}_2\left (-\frac {g \left (e x^2+d\right )}{e f-d g}\right )}{g^3}-\frac {p x^2}{2 g^2} \]

[Out]

-1/2*p*x^2/g^2+1/2*e*f^2*p*ln(e*x^2+d)/g^3/(-d*g+e*f)+1/2*(e*x^2+d)*ln(c*(e*x^2+d)^p)/e/g^2-1/2*f^2*ln(c*(e*x^
2+d)^p)/g^3/(g*x^2+f)-1/2*e*f^2*p*ln(g*x^2+f)/g^3/(-d*g+e*f)-f*ln(c*(e*x^2+d)^p)*ln(e*(g*x^2+f)/(-d*g+e*f))/g^
3-f*p*polylog(2,-g*(e*x^2+d)/(-d*g+e*f))/g^3

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Rubi [A]  time = 0.28, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2475, 43, 2416, 2389, 2295, 2395, 36, 31, 2394, 2393, 2391} \[ -\frac {f p \text {PolyLog}\left (2,-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{g^3}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{g^3}+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}+\frac {e f^2 p \log \left (d+e x^2\right )}{2 g^3 (e f-d g)}-\frac {e f^2 p \log \left (f+g x^2\right )}{2 g^3 (e f-d g)}-\frac {p x^2}{2 g^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]

[Out]

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

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 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^5 \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^2 \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 {\log \left (c (d+e x)^p\right )}{g^2}+\frac {f^2 \log \left (c (d+e x)^p\right )}{g^2 (f+g x)^2}-\frac {2 f \log \left (c (d+e x)^p\right )}{g^2 (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^2}-\frac {f \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{f+g x} \, dx,x,x^2\right )}{g^2}+\frac {f^2 \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{(f+g x)^2} \, dx,x,x^2\right )}{2 g^2}\\ &=-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 g^3 \left (f+g x^2\right )}-\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 )}{g^3}+\frac {\operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^2\right )}{2 e g^2}+\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 )}{g^3}+\frac {\left (e f^2 p\right ) \operatorname {Subst}\left (\int \frac {1}{(d+e x) (f+g x)} \, dx,x,x^2\right )}{2 g^3}\\ &=-\frac {p x^2}{2 g^2}+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 g^3 \left (f+g x^2\right )}-\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 )}{g^3}+\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 )}{g^3}+\frac {\left (e^2 f^2 p\right ) \operatorname {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^2\right )}{2 g^3 (e f-d g)}-\frac {\left (e f^2 p\right ) \operatorname {Subst}\left (\int \frac {1}{f+g x} \, dx,x,x^2\right )}{2 g^2 (e f-d g)}\\ &=-\frac {p x^2}{2 g^2}+\frac {e f^2 p \log \left (d+e x^2\right )}{2 g^3 (e f-d g)}+\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}-\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 g^3 \left (f+g x^2\right )}-\frac {e f^2 p \log \left (f+g x^2\right )}{2 g^3 (e f-d 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 )}{g^3}-\frac {f p \text {Li}_2\left (-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{g^3}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 166, normalized size = 0.83 \[ -\frac {\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right )}{g \left (f+g x^2\right )}+\frac {2 f \left (\log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )+p \text {Li}_2\left (\frac {g \left (e x^2+d\right )}{d g-e f}\right )\right )}{g}-\frac {\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac {e f^2 p \left (\log \left (d+e x^2\right )-\log \left (f+g x^2\right )\right )}{g (d g-e f)}+p x^2}{2 g^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]

[Out]

-1/2*(p*x^2 - ((d + e*x^2)*Log[c*(d + e*x^2)^p])/e + (f^2*Log[c*(d + e*x^2)^p])/(g*(f + g*x^2)) + (e*f^2*p*(Lo
g[d + e*x^2] - Log[f + g*x^2]))/(g*(-(e*f) + d*g)) + (2*f*(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)]))/g)/g^2

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{5} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral(x^5*log((e*x^2 + d)^p*c)/(g^2*x^4 + 2*f*g*x^2 + f^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="giac")

[Out]

integrate(x^5*log((e*x^2 + d)^p*c)/(g*x^2 + f)^2, x)

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maple [C]  time = 0.76, size = 985, normalized size = 4.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*ln(c*(e*x^2+d)^p)/(g*x^2+f)^2,x)

[Out]

1/2*ln((e*x^2+d)^p)/g^2*x^2-1/2*ln((e*x^2+d)^p)*f^2/g^3/(g*x^2+f)-ln((e*x^2+d)^p)*f/g^3*ln(g*x^2+f)-1/2*p*x^2/
g^2+1/2*p*e/g^3*f^2/(d*g-e*f)*ln(g*x^2+f)+1/2*p/e/g/(d*g-e*f)*ln(e*x^2+d)*d^2-1/2*p/g^2/(d*g-e*f)*ln(e*x^2+d)*
d*f-1/2*p*e/g^3/(d*g-e*f)*ln(e*x^2+d)*f^2+p*f/g^3*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((R
ootOf(_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))-di
log((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*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*f/g^3*ln(g*x^2+f)+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*c
sgn(I*c*(e*x^2+d)^p)*csgn(I*c)*f/g^3*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/g^2*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^2/g^3/(g*x^2+f)-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^
3/g^2*x^2-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f^2/g^3/(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^2/g^3/(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*x^2+1/4*I*Pi*
csgn(I*c*(e*x^2+d)^p)^3*f^2/g^3/(g*x^2+f)-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*f/g^3*ln(g*x^2+
f)-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f/g^3*ln(g*x^2+f)+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)/g^2
*x^2+1/2*ln(c)/g^2*x^2-1/2*ln(c)*f^2/g^3/(g*x^2+f)-ln(c)*f/g^3*ln(g*x^2+f)

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maxima [A]  time = 0.80, size = 337, normalized size = 1.69 \[ -\frac {{\left (e f^{2} p + 2 \, {\left (e f^{2} - d f g\right )} \log \relax (c)\right )} \log \left (g x^{2} + f\right )}{2 \, {\left (e f g^{3} - d g^{4}\right )}} - \frac {{\left (e^{2} f g^{2} p - d e g^{3} p - {\left (e^{2} f g^{2} - d e g^{3}\right )} \log \relax (c)\right )} x^{4} + {\left (e^{2} f^{2} g p - d e f g^{2} p - {\left (e^{2} f^{2} g - d e f g^{2}\right )} \log \relax (c)\right )} x^{2} - {\left (2 \, d e f^{2} g p - d^{2} f g^{2} p + {\left (e^{2} f g^{2} p - d e g^{3} p\right )} x^{4} + {\left (2 \, e^{2} f^{2} g p - d^{2} g^{3} p\right )} x^{2}\right )} \log \left (e x^{2} + d\right ) + {\left (e^{2} f^{3} - d e f^{2} g\right )} \log \relax (c)}{2 \, {\left (e^{2} f^{2} g^{3} - d e f g^{4} + {\left (e^{2} f g^{4} - d e g^{5}\right )} x^{2}\right )}} - \frac {{\left (\log \left (e x^{2} + d\right ) \log \left (\frac {e g x^{2} + d g}{e f - d g} + 1\right ) + {\rm Li}_2\left (-\frac {e g x^{2} + d g}{e f - d g}\right )\right )} f p}{g^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

-1/2*(e*f^2*p + 2*(e*f^2 - d*f*g)*log(c))*log(g*x^2 + f)/(e*f*g^3 - d*g^4) - 1/2*((e^2*f*g^2*p - d*e*g^3*p - (
e^2*f*g^2 - d*e*g^3)*log(c))*x^4 + (e^2*f^2*g*p - d*e*f*g^2*p - (e^2*f^2*g - d*e*f*g^2)*log(c))*x^2 - (2*d*e*f
^2*g*p - d^2*f*g^2*p + (e^2*f*g^2*p - d*e*g^3*p)*x^4 + (2*e^2*f^2*g*p - d^2*g^3*p)*x^2)*log(e*x^2 + d) + (e^2*
f^3 - d*e*f^2*g)*log(c))/(e^2*f^2*g^3 - d*e*f*g^4 + (e^2*f*g^4 - d*e*g^5)*x^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)))*f*p/g^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{{\left (g\,x^2+f\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*log(c*(d + e*x^2)^p))/(f + g*x^2)^2,x)

[Out]

int((x^5*log(c*(d + e*x^2)^p))/(f + g*x^2)^2, x)

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

[Out]

Timed out

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