3.4.38 \(\int \frac {x^5 \log (c (d+e x^2)^p)}{f+g x^2} \, dx\) [338]

Optimal. Leaf size=188 \[ \frac {f p x^2}{2 g^2}+\frac {d p x^2}{4 e g}-\frac {p x^4}{8 g}-\frac {d^2 p \log \left (d+e x^2\right )}{4 e^2 g}+\frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {f \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}+\frac {f^2 \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^3}+\frac {f^2 p \text {Li}_2\left (-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^3} \]

[Out]

1/2*f*p*x^2/g^2+1/4*d*p*x^2/e/g-1/8*p*x^4/g-1/4*d^2*p*ln(e*x^2+d)/e^2/g+1/4*x^4*ln(c*(e*x^2+d)^p)/g-1/2*f*(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)*ln(e*(g*x^2+f)/(-d*g+e*f))/g^3+1/2*f^2*p*polylog(2,-g*
(e*x^2+d)/(-d*g+e*f))/g^3

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Rubi [A]
time = 0.19, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2525, 45, 2463, 2436, 2332, 2442, 2441, 2440, 2438} \begin {gather*} \frac {f^2 p \text {PolyLog}\left (2,-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^3}+\frac {f^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 )}{2 g^3}-\frac {f \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e g^2}+\frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {d^2 p \log \left (d+e x^2\right )}{4 e^2 g}+\frac {d p x^2}{4 e g}+\frac {f p x^2}{2 g^2}-\frac {p x^4}{8 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*p*x^2)/(2*g^2) + (d*p*x^2)/(4*e*g) - (p*x^4)/(8*g) - (d^2*p*Log[d + e*x^2])/(4*e^2*g) + (x^4*Log[c*(d + e*x
^2)^p])/(4*g) - (f*(d + e*x^2)*Log[c*(d + e*x^2)^p])/(2*e*g^2) + (f^2*Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))
/(e*f - d*g)])/(2*g^3) + (f^2*p*PolyLog[2, -((g*(d + e*x^2))/(e*f - d*g))])/(2*g^3)

Rule 45

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 2332

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

Rule 2436

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 2438

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 2440

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 2441

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 2442

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 2463

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 2525

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 )}{f+g x^2} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2 \log \left (c (d+e x)^p\right )}{f+g x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (-\frac {f \log \left (c (d+e x)^p\right )}{g^2}+\frac {x \log \left (c (d+e x)^p\right )}{g}+\frac {f^2 \log \left (c (d+e x)^p\right )}{g^2 (f+g x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {f \text {Subst}\left (\int \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )}{2 g^2}+\frac {f^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{f+g x} \, dx,x,x^2\right )}{2 g^2}+\frac {\text {Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^2\right )}{2 g}\\ &=\frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}+\frac {f^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 )}{2 g^3}-\frac {f \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^2\right )}{2 e g^2}-\frac {\left (e f^2 p\right ) \text {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^3}-\frac {(e p) \text {Subst}\left (\int \frac {x^2}{d+e x} \, dx,x,x^2\right )}{4 g}\\ &=\frac {f p x^2}{2 g^2}+\frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {f \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 ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^3}-\frac {\left (f^2 p\right ) \text {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^3}-\frac {(e p) \text {Subst}\left (\int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx,x,x^2\right )}{4 g}\\ &=\frac {f p x^2}{2 g^2}+\frac {d p x^2}{4 e g}-\frac {p x^4}{8 g}-\frac {d^2 p \log \left (d+e x^2\right )}{4 e^2 g}+\frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{4 g}-\frac {f \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 ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{2 g^3}+\frac {f^2 p \text {Li}_2\left (-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{2 g^3}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 143, normalized size = 0.76 \begin {gather*} \frac {e g p x^2 \left (4 e f+2 d g-e g x^2\right )-2 d^2 g^2 p \log \left (d+e x^2\right )+e \log \left (c \left (d+e x^2\right )^p\right ) \left (2 g \left (-2 d f-2 e f x^2+e g x^4\right )+4 e f^2 \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )\right )+4 e^2 f^2 p \text {Li}_2\left (\frac {g \left (d+e x^2\right )}{-e f+d g}\right )}{8 e^2 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*g*p*x^2*(4*e*f + 2*d*g - e*g*x^2) - 2*d^2*g^2*p*Log[d + e*x^2] + e*Log[c*(d + e*x^2)^p]*(2*g*(-2*d*f - 2*e*
f*x^2 + e*g*x^4) + 4*e*f^2*Log[(e*(f + g*x^2))/(e*f - d*g)]) + 4*e^2*f^2*p*PolyLog[2, (g*(d + e*x^2))/(-(e*f)
+ d*g)])/(8*e^2*g^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.80, size = 902, normalized size = 4.80

method	result	size
risch	\(\text {Expression too large to display}\)	\(902\)

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),x,method=_RETURNVERBOSE)

[Out]

1/4*ln((e*x^2+d)^p)/g*x^4-1/2*ln((e*x^2+d)^p)/g^2*f*x^2+1/2*ln((e*x^2+d)^p)*f^2/g^3*ln(g*x^2+f)-1/2*p*f^2/g^3*
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)/Root
Of(_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)/R
ootOf(_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+_a
lpha)/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))-1/8*p*x^4/g+1/4*d*p*x^2
/e/g+1/2*f*p*x^2/g^2-1/4*d^2*p*ln(e*x^2+d)/e^2/g-1/2*p/e/g^2*d*ln(e*x^2+d)*f-1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^3/
g*x^4-1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)/g*x^4+1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn
(I*c)/g*x^4+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f^2/g^3*ln(g*x^2+f)+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3/g^
2*f*x^2+1/4*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*f^2/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)*csgn(I*c)*f^2/g^3*ln(g*x^2+f)+1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2/g*x^4
-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)/g^2*f*x^2-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*f^2/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*f*x^2+1/4*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)
*csgn(I*c)/g^2*f*x^2+1/4*ln(c)/g*x^4-1/2*ln(c)/g^2*f*x^2+1/2*ln(c)*f^2/g^3*ln(g*x^2+f)

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Maxima [A]
time = 0.58, size = 181, normalized size = 0.96 \begin {gather*} \frac {{\left (\log \left (x^{2} e + d\right ) \log \left (-\frac {g x^{2} e + d g}{d g - f e} + 1\right ) + {\rm Li}_2\left (\frac {g x^{2} e + d g}{d g - f e}\right )\right )} f^{2} p}{2 \, g^{3}} + \frac {f^{2} \log \left (g x^{2} + f\right ) \log \left (c\right )}{2 \, g^{3}} - \frac {{\left ({\left (g p - 2 \, g \log \left (c\right )\right )} x^{4} e^{2} - 2 \, {\left (d g p e + 2 \, {\left (f p - f \log \left (c\right )\right )} e^{2}\right )} x^{2} - 2 \, {\left (g p x^{4} e^{2} - 2 \, f p x^{2} e^{2} - d^{2} g p - 2 \, d f p e\right )} \log \left (x^{2} e + d\right )\right )} e^{\left (-2\right )}}{8 \, g^{2}} \end {gather*}

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),x, algorithm="maxima")

[Out]

1/2*(log(x^2*e + d)*log(-(g*x^2*e + d*g)/(d*g - f*e) + 1) + dilog((g*x^2*e + d*g)/(d*g - f*e)))*f^2*p/g^3 + 1/
2*f^2*log(g*x^2 + f)*log(c)/g^3 - 1/8*((g*p - 2*g*log(c))*x^4*e^2 - 2*(d*g*p*e + 2*(f*p - f*log(c))*e^2)*x^2 -
2*(g*p*x^4*e^2 - 2*f*p*x^2*e^2 - d^2*g*p - 2*d*f*p*e)*log(x^2*e + d))*e^(-2)/g^2

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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),x, algorithm="fricas")

[Out]

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

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

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),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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),x, algorithm="giac")

[Out]

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

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

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),x)

[Out]

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

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