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

Optimal. Leaf size=93 \[ \frac{1}{2} g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}+\frac{1}{2} g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )-\frac{e f p \log \left (d+e x^2\right )}{2 d}+\frac{e f p \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.126835, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2475, 43, 2416, 2395, 36, 29, 31, 2394, 2315} \[ \frac{1}{2} g p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}+\frac{1}{2} g \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )-\frac{e f p \log \left (d+e x^2\right )}{2 d}+\frac{e f p \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0276093, size = 92, normalized size = 0.99 \[ \frac{1}{2} g \left (p \text{PolyLog}\left (2,\frac{d+e x^2}{d}\right )+\log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )-\frac{f \log \left (c \left (d+e x^2\right )^p\right )}{2 x^2}-\frac{e f p \log \left (d+e x^2\right )}{2 d}+\frac{e f p \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.499, size = 421, normalized size = 4.5 \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)*ln(c*(e*x^2+d)^p)/x^3,x)

[Out]

ln((e*x^2+d)^p)*g*ln(x)-1/2*ln((e*x^2+d)^p)*f/x^2+e*f*p*ln(x)/d-1/2*e*f*p*ln(e*x^2+d)/d-p*g*ln(x)*ln((-e*x+(-d
*e)^(1/2))/(-d*e)^(1/2))-p*g*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-p*g*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1
/2))-p*g*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*csgn(I*c)*g
*ln(x)+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*g*ln(x)+1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2
*g*ln(x)-1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^2*csgn(I*c)*f/x^2+1/4*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*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)*f/x^2-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3*g*ln(x)-1/4*I*Pi*csgn(I
*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)^2*f/x^2+ln(c)*g*ln(x)-1/2*ln(c)*f/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 )} \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)*log(c*(e*x^2+d)^p)/x^3,x, algorithm="maxima")

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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