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

Optimal. Leaf size=83 \[ -\frac{\log \left (c \left (d+e x^2\right )^p\right )}{2 g \left (f+g x^2\right )}+\frac{e p \log \left (d+e x^2\right )}{2 g (e f-d g)}-\frac{e p \log \left (f+g x^2\right )}{2 g (e f-d g)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.074348, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2475, 2395, 36, 31} \[ -\frac{\log \left (c \left (d+e x^2\right )^p\right )}{2 g \left (f+g x^2\right )}+\frac{e p \log \left (d+e x^2\right )}{2 g (e f-d g)}-\frac{e p \log \left (f+g x^2\right )}{2 g (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 31

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

Rubi steps

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

Mathematica [A]  time = 0.051894, size = 63, normalized size = 0.76 \[ \frac{\frac{e p \left (\log \left (d+e x^2\right )-\log \left (f+g x^2\right )\right )}{e f-d g}-\frac{\log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}}{2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.382, size = 371, normalized size = 4.5 \begin{align*} -{\frac{\ln \left ( \left ( e{x}^{2}+d \right ) ^{p} \right ) }{2\,g \left ( g{x}^{2}+f \right ) }}-{\frac{i\pi \,dg{\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}-i\pi \,dg{\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 ) -i\pi \,dg \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}+i\pi \,dg \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \,ef{\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}+i\pi \,ef{\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 ) +i\pi \,ef \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{3}-i\pi \,ef \left ({\it csgn} \left ( ic \left ( e{x}^{2}+d \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +2\,\ln \left ( -e{x}^{2}-d \right ) egp{x}^{2}-2\,\ln \left ( g{x}^{2}+f \right ) egp{x}^{2}+2\,\ln \left ( -e{x}^{2}-d \right ) efp-2\,\ln \left ( g{x}^{2}+f \right ) efp+2\,\ln \left ( c \right ) dg-2\,\ln \left ( c \right ) ef}{4\,g \left ( g{x}^{2}+f \right ) \left ( dg-fe \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.0145, size = 100, normalized size = 1.2 \begin{align*} \frac{e p{\left (\frac{\log \left (e x^{2} + d\right )}{e f - d g} - \frac{\log \left (g x^{2} + f\right )}{e f - d g}\right )}}{2 \, g} - \frac{\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{2 \,{\left (g x^{2} + f\right )} g} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e*p*(log(e*x^2 + d)/(e*f - d*g) - log(g*x^2 + f)/(e*f - d*g))/g - 1/2*log((e*x^2 + d)^p*c)/((g*x^2 + f)*g)

________________________________________________________________________________________

Fricas [A]  time = 2.06877, size = 194, normalized size = 2.34 \begin{align*} \frac{{\left (e g p x^{2} + d g p\right )} \log \left (e x^{2} + d\right ) -{\left (e g p x^{2} + e f p\right )} \log \left (g x^{2} + f\right ) -{\left (e f - d g\right )} \log \left (c\right )}{2 \,{\left (e f^{2} g - d f g^{2} +{\left (e f g^{2} - d g^{3}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((e*g*p*x^2 + d*g*p)*log(e*x^2 + d) - (e*g*p*x^2 + e*f*p)*log(g*x^2 + f) - (e*f - d*g)*log(c))/(e*f^2*g -
d*f*g^2 + (e*f*g^2 - d*g^3)*x^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.16787, size = 246, normalized size = 2.96 \begin{align*} -\frac{{\left (x^{2} e + d\right )} g p e \log \left (x^{2} e + d\right ) -{\left (x^{2} e + d\right )} g p e \log \left ({\left (x^{2} e + d\right )} g - d g + f e\right ) + d g p e \log \left ({\left (x^{2} e + d\right )} g - d g + f e\right ) - f p e^{2} \log \left ({\left (x^{2} e + d\right )} g - d g + f e\right ) + d g e \log \left (c\right ) - f e^{2} \log \left (c\right )}{2 \,{\left ({\left (x^{2} e + d\right )} d g^{3} - d^{2} g^{3} -{\left (x^{2} e + d\right )} f g^{2} e + 2 \, d f g^{2} e - f^{2} g e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((x^2*e + d)*g*p*e*log(x^2*e + d) - (x^2*e + d)*g*p*e*log((x^2*e + d)*g - d*g + f*e) + d*g*p*e*log((x^2*e
+ d)*g - d*g + f*e) - f*p*e^2*log((x^2*e + d)*g - d*g + f*e) + d*g*e*log(c) - f*e^2*log(c))/((x^2*e + d)*d*g^
3 - d^2*g^3 - (x^2*e + d)*f*g^2*e + 2*d*f*g^2*e - f^2*g*e^2)