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

Optimal. Leaf size=153 \[ \frac{1}{2} f^2 p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\frac{1}{2} f^2 \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac{1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac{d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac{d g^2 p x^2}{4 e}-f g p x^2-\frac{1}{8} g^2 p x^4 \]

[Out]

-(f*g*p*x^2) + (d*g^2*p*x^2)/(4*e) - (g^2*p*x^4)/8 - (d^2*g^2*p*Log[d + e*x^2])/(4*e^2) + (g^2*x^4*Log[c*(d +
e*x^2)^p])/4 + (f*g*(d + e*x^2)*Log[c*(d + e*x^2)^p])/e + (f^2*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/2 + (f^
2*p*PolyLog[2, 1 + (e*x^2)/d])/2

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Rubi [A]  time = 0.202965, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2475, 43, 2416, 2389, 2295, 2394, 2315, 2395} \[ \frac{1}{2} f^2 p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\frac{1}{2} f^2 \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{f g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}+\frac{1}{4} g^2 x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac{d^2 g^2 p \log \left (d+e x^2\right )}{4 e^2}+\frac{d g^2 p x^2}{4 e}-f g p x^2-\frac{1}{8} g^2 p x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(f*g*p*x^2) + (d*g^2*p*x^2)/(4*e) - (g^2*p*x^4)/8 - (d^2*g^2*p*Log[d + e*x^2])/(4*e^2) + (g^2*x^4*Log[c*(d +
e*x^2)^p])/4 + (f*g*(d + e*x^2)*Log[c*(d + e*x^2)^p])/e + (f^2*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/2 + (f^
2*p*PolyLog[2, 1 + (e*x^2)/d])/2

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 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]

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]

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

Mathematica [A]  time = 0.0815272, size = 121, normalized size = 0.79 \[ \frac{4 e^2 f^2 p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+2 e \log \left (c \left (d+e x^2\right )^p\right ) \left (2 e f^2 \log \left (-\frac{e x^2}{d}\right )+g \left (4 d f+4 e f x^2+e g x^4\right )\right )-2 d^2 g^2 p \log \left (d+e x^2\right )-e g p x^2 \left (-2 d g+8 e f+e g x^2\right )}{8 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(e*g*p*x^2*(8*e*f - 2*d*g + e*g*x^2)) - 2*d^2*g^2*p*Log[d + e*x^2] + 2*e*(g*(4*d*f + 4*e*f*x^2 + e*g*x^4) +
2*e*f^2*Log[-((e*x^2)/d)])*Log[c*(d + e*x^2)^p] + 4*e^2*f^2*p*PolyLog[2, 1 + (e*x^2)/d])/(8*e^2)

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Maple [C]  time = 0.569, size = 652, normalized size = 4.3 \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,x)

[Out]

1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*x^4*g^2+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p
)^2*f^2*ln(x)-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*x^2*f*g+1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*x^4*g^2-1/2*
I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*x^2*f*g+ln((e*x^2+d)^p)*x^2*f*g+ln(c)*x^2*f*g-p*f^2*l
n(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*f^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+p/e*g*d*ln(e*x^2+d)*
f+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f^2*ln(x)-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn
(I*c)*f^2*ln(x)+ln(c)*f^2*ln(x)-p*f^2*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*f^2*dilog((e*x+(-d*e)^(1/2))/(
-d*e)^(1/2))-1/4*d^2*g^2*p*ln(e*x^2+d)/e^2-f*g*p*x^2-1/8*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*x^4*g^2-1/2*I*Pi*csgn(I*
c*(e*x^2+d)^p)^3*f^2*ln(x)-1/8*g^2*p*x^4+1/4*ln(c)*x^4*g^2+1/4*ln((e*x^2+d)^p)*x^4*g^2+ln((e*x^2+d)^p)*f^2*ln(
x)+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*x^2*f*g+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*x^2
*f*g-1/8*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*x^4*g^2+1/4*d*g^2*p*x^2/e

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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}\,{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,x, algorithm="maxima")

[Out]

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

[Out]

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

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

[Out]

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

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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}\,{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,x, algorithm="giac")

[Out]

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