\(\int \frac {x \log (c (d+e x^2)^p)}{(f+g x^2)^2} \, dx\) [350]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {e p \log \left (d+e x^2\right )}{2 g (e f-d g)}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 g \left (f+g x^2\right )}-\frac {e p \log \left (f+g x^2\right )}{2 g (e f-d g)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2525, 2442, 36, 31} \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 g \left (f+g x^2\right )}+\frac {e p \log \left (d+e x^2\right )}{2 g (e f-d g)}-\frac {e p \log \left (f+g x^2\right )}{2 g (e f-d g)} \]

[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 31

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

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 2442

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 2525

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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {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) \text {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) \text {Subst}\left (\int \frac {1}{f+g x} \, dx,x,x^2\right )}{2 (e f-d g)}+\frac {\left (e^2 p\right ) \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {\frac {e p \log \left (d+e x^2\right )}{e f-d g}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\frac {e p \log \left (f+g x^2\right )}{-e f+d g}}{2 g} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90

method	result	size
parts	\(-\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2 g \left (g \,x^{2}+f \right )}+\frac {p e \left (-\frac {\ln \left (e \,x^{2}+d \right )}{2 \left (d g -e f \right )}+\frac {\ln \left (g \,x^{2}+f \right )}{2 d g -2 e f}\right )}{g}\)	\(75\)
parallelrisch	\(-\frac {\ln \left (e \,x^{2}+d \right ) x^{2} e^{2} g p -\ln \left (g \,x^{2}+f \right ) x^{2} e^{2} g p +\ln \left (e \,x^{2}+d \right ) e^{2} f p -\ln \left (g \,x^{2}+f \right ) e^{2} f p +\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d e g -\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{2} f}{2 \left (d g -e f \right ) \left (g \,x^{2}+f \right ) e g}\)	\(127\)
risch	\(-\frac {\ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{2 g \left (g \,x^{2}+f \right )}-\frac {i \pi d g \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}-i \pi d g \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3} d g +i \pi d g {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-i \pi e f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+i \pi e f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+i \pi e f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-i \pi e f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 \ln \left (g \,x^{2}+f \right ) e g p \,x^{2}+2 \ln \left (-e \,x^{2}-d \right ) e g p \,x^{2}-2 \ln \left (g \,x^{2}+f \right ) e f p +2 \ln \left (-e \,x^{2}-d \right ) e f p +2 \ln \left (c \right ) d g -2 \ln \left (c \right ) e f}{4 g \left (g \,x^{2}+f \right ) \left (d g -e f \right )}\)	\(371\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {{\left (e g p x^{2} + d g p\right )} \log \left (e x^{2} + d\right ) - {\left (e g p x^{2} + e f p\right )} \log \left (g x^{2} + f\right ) - {\left (e f - d g\right )} \log \left (c\right )}{2 \, {\left (e f^{2} g - d f g^{2} + {\left (e f g^{2} - d g^{3}\right )} x^{2}\right )}} \]

[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)]

Timed out. \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {e p {\left (\frac {\log \left (e x^{2} + d\right )}{e f - d g} - \frac {\log \left (g x^{2} + f\right )}{e f - d g}\right )}}{2 \, g} - \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{2 \, {\left (g x^{2} + f\right )} g} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {e p \log \left (e x^{2} + d\right )}{2 \, {\left (e f g + {\left (e x^{2} + d\right )} g^{2} - d g^{2}\right )}} + \frac {e p \log \left (e x^{2} + d\right )}{2 \, {\left (e f g - d g^{2}\right )}} - \frac {e p \log \left (e f + {\left (e x^{2} + d\right )} g - d g\right )}{2 \, {\left (e f g - d g^{2}\right )}} - \frac {e \log \left (c\right )}{2 \, {\left (e f g + {\left (e x^{2} + d\right )} g^{2} - d g^{2}\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.63 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{2\,g\,\left (g\,x^2+f\right )}-\frac {e\,p\,\mathrm {atan}\left (\frac {x^2\,\left (d\,g\,1{}\mathrm {i}-e\,f\,1{}\mathrm {i}\right )}{2\,d\,f+d\,g\,x^2+e\,f\,x^2}\right )\,1{}\mathrm {i}}{d\,g^2-e\,f\,g} \]

[In]

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

[Out]

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