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

Optimal. Leaf size=135 \[ f g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}+f g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{g^2 \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}-\frac{e f^2 p \log \left (d+e x^2\right )}{2 d}+\frac{e f^2 p \log (x)}{d}-\frac{1}{2} g^2 p x^2 \]

[Out]

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

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Rubi [A]  time = 0.193142, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {2475, 43, 2416, 2389, 2295, 2395, 36, 29, 31, 2394, 2315} \[ f g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}+f g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{g^2 \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}-\frac{e f^2 p \log \left (d+e x^2\right )}{2 d}+\frac{e f^2 p \log (x)}{d}-\frac{1}{2} g^2 p x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0776067, size = 126, normalized size = 0.93 \[ \frac{1}{2} \left (2 f g \left (p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )-\frac{f^2 \log \left (c \left (d+e x^2\right )^p\right )}{x^2}+\frac{g^2 \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac{e f^2 p \left (2 \log (x)-\log \left (d+e x^2\right )\right )}{d}-g^2 p x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.583, size = 642, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x^{2} + f\right )}^{2} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (g^{2} x^{4} + 2 \, f g x^{2} + f^{2}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x^{2} + f\right )}^{2} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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