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

Optimal. Leaf size=82 \[ \frac{1}{2} f p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\frac{1}{2} f \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}-\frac{1}{2} g p x^2 \]

[Out]

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

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Rubi [A]  time = 0.111122, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2475, 43, 2416, 2389, 2295, 2394, 2315} \[ \frac{1}{2} f p \text{PolyLog}\left (2,\frac{e x^2}{d}+1\right )+\frac{1}{2} f \log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac{g \left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 e}-\frac{1}{2} g p x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0191056, size = 80, normalized size = 0.98 \[ \frac{1}{2} f \left (p \text{PolyLog}\left (2,\frac{d+e x^2}{d}\right )+\log \left (-\frac{e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )\right )+\frac{1}{2} g \left (\frac{\left (d+e x^2\right ) \log \left (c \left (d+e x^2\right )^p\right )}{e}-p x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.567, size = 419, normalized size = 5.1 \begin{align*}{\frac{\ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ){x}^{2}g}{2}}+\ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) f\ln \left ( x \right ) -{\frac{gp{x}^{2}}{2}}+{\frac{gpd\ln \left ( e{x}^{2}+d \right ) }{2\,e}}-pf\ln \left ( x \right ) \ln \left ({ \left ( -ex+\sqrt{-de} \right ){\frac{1}{\sqrt{-de}}}} \right ) -pf\ln \left ( x \right ) \ln \left ({ \left ( ex+\sqrt{-de} \right ){\frac{1}{\sqrt{-de}}}} \right ) -pf{\it dilog} \left ({ \left ( -ex+\sqrt{-de} \right ){\frac{1}{\sqrt{-de}}}} \right ) -pf{\it dilog} \left ({ \left ( ex+\sqrt{-de} \right ){\frac{1}{\sqrt{-de}}}} \right ) -{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ) f\ln \left ( x \right ) +{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}f\ln \left ( x \right ) -{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}f\ln \left ( x \right ) -{\frac{i}{4}}\pi \,{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ){\it csgn} \left ( ic \right ){x}^{2}g+{\frac{i}{4}}\pi \,{\it csgn} \left ( i \left ( e{x}^{2}+d \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{x}^{2}g+{\frac{i}{4}}\pi \, \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ){x}^{2}g+{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) f\ln \left ( x \right ) -{\frac{i}{4}}\pi \, \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}{x}^{2}g+{\frac{\ln \left ( c \right ){x}^{2}g}{2}}+\ln \left ( c \right ) f\ln \left ( x \right ) \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,x)

[Out]

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

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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}\,{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,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

[Out]

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