3.4.49 \(\int \frac {x^3 \log (c (d+e x^2)^p)}{(f+g x^2)^2} \, dx\) [349]

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

[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

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Rubi [A]
time = 0.16, 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 = {2525, 45, 2463, 2442, 36, 31, 2441, 2440, 2438} \begin {gather*} \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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(e*f*p*Log[d + e*x^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 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 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^3 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx &=\frac {1}{2} \text {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} \text {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 {\text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{f+g x} \, dx,x,x^2\right )}{2 g}-\frac {f \text {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) \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^2}-\frac {(e f p) \text {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 \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^2}-\frac {\left (e^2 f p\right ) \text {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^2\right )}{2 g^2 (e f-d g)}+\frac {(e f p) \text {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*}

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Mathematica [A]
time = 0.07, size = 131, normalized size = 0.85 \begin {gather*} \frac {\frac {e f p \log \left (d+e x^2\right )}{-e f+d g}+\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\frac {e f p \log \left (f+g x^2\right )}{e f-d g}+\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 (d+e x^2\right )}{-e f+d g}\right )}{2 g^2} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.46, size = 732, normalized size = 4.72

method	result	size
risch	\(\text {Expression too large to display}\)	\(732\)

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,method=_RETURNVERBOSE)

[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 [A]
time = 0.54, size = 193, normalized size = 1.25 \begin {gather*} \frac {{\left (d g \log \left (c\right ) - {\left (f p + f \log \left (c\right )\right )} e\right )} \log \left (g x^{2} + f\right )}{2 \, {\left (d g^{3} - f g^{2} e\right )}} + \frac {d f g \log \left (c\right ) - f^{2} e \log \left (c\right ) + {\left (f g p x^{2} e + d f g p\right )} \log \left (x^{2} e + d\right )}{2 \, {\left (d f g^{3} - f^{2} g^{2} e + {\left (d g^{4} - f g^{3} e\right )} x^{2}\right )}} + \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 )} p}{2 \, g^{2}} \end {gather*}

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*(d*g*log(c) - (f*p + f*log(c))*e)*log(g*x^2 + f)/(d*g^3 - f*g^2*e) + 1/2*(d*f*g*log(c) - f^2*e*log(c) + (f
*g*p*x^2*e + d*f*g*p)*log(x^2*e + d))/(d*f*g^3 - f^2*g^2*e + (d*g^4 - f*g^3*e)*x^2) + 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)))*p/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^3*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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^3*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="giac")

[Out]

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

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

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)

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