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

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

[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)

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Rubi [A]
time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {2525, 2442, 36, 31} \begin {gather*} -\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)} \end {gather*}

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 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*} \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} \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*}

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Mathematica [A]
time = 0.04, size = 63, normalized size = 0.76 \begin {gather*} \frac {-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\frac {e p \left (\log \left (d+e x^2\right )-\log \left (f+g x^2\right )\right )}{e f-d g}}{2 g} \end {gather*}

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)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.51, size = 371, normalized size = 4.47

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

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 79, normalized size = 0.95 \begin {gather*} \frac {p {\left (\frac {\log \left (g x^{2} + f\right )}{d g - f e} - \frac {\log \left (x^{2} e + d\right )}{d g - f e}\right )} e}{2 \, g} - \frac {\log \left ({\left (x^{2} e + d\right )}^{p} c\right )}{2 \, {\left (g x^{2} + f\right )} g} \end {gather*}

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

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Fricas [A]
time = 0.35, size = 96, normalized size = 1.16 \begin {gather*} \frac {{\left (g p x^{2} + f p\right )} e \log \left (g x^{2} + f\right ) - {\left (g p x^{2} e + d g p\right )} \log \left (x^{2} e + d\right ) - {\left (d g - f e\right )} \log \left (c\right )}{2 \, {\left (d g^{3} x^{2} + d f g^{2} - {\left (f g^{2} x^{2} + f^{2} g\right )} e\right )}} \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (83) = 166\).
time = 4.44, size = 182, normalized size = 2.19 \begin {gather*} -\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 {gather*}

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)

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Mupad [B]
time = 1.46, size = 80, normalized size = 0.96 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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