3.1.39 \(\int x \log (c (a+\frac {b}{x^2})^p) \, dx\) [39]

Optimal. Leaf size=37 \[ \frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {b p \log \left (b+a x^2\right )}{2 a} \]

[Out]

1/2*x^2*ln(c*(a+b/x^2)^p)+1/2*b*p*ln(a*x^2+b)/a

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2505, 269, 266} \begin {gather*} \frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {b p \log \left (a x^2+b\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b/x^2)^p],x]

[Out]

(x^2*Log[c*(a + b/x^2)^p])/2 + (b*p*Log[b + a*x^2])/(2*a)

Rule 266

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

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx &=\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+(b p) \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x} \, dx\\ &=\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+(b p) \int \frac {x}{b+a x^2} \, dx\\ &=\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {b p \log \left (b+a x^2\right )}{2 a}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 45, normalized size = 1.22 \begin {gather*} \frac {b p \log \left (a+\frac {b}{x^2}\right )}{2 a}+\frac {1}{2} x^2 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {b p \log (x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b/x^2)^p],x]

[Out]

(b*p*Log[a + b/x^2])/(2*a) + (x^2*Log[c*(a + b/x^2)^p])/2 + (b*p*Log[x])/a

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int x \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(c*(a+b/x^2)^p),x)

[Out]

int(x*ln(c*(a+b/x^2)^p),x)

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Maxima [A]
time = 0.28, size = 33, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x^{2} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right ) + \frac {b p \log \left (a x^{2} + b\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*(a+b/x^2)^p),x, algorithm="maxima")

[Out]

1/2*x^2*log((a + b/x^2)^p*c) + 1/2*b*p*log(a*x^2 + b)/a

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Fricas [A]
time = 0.37, size = 42, normalized size = 1.14 \begin {gather*} \frac {a p x^{2} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + a x^{2} \log \left (c\right ) + b p \log \left (a x^{2} + b\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*(a+b/x^2)^p),x, algorithm="fricas")

[Out]

1/2*(a*p*x^2*log((a*x^2 + b)/x^2) + a*x^2*log(c) + b*p*log(a*x^2 + b))/a

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Sympy [A]
time = 0.68, size = 53, normalized size = 1.43 \begin {gather*} \begin {cases} \frac {x^{2} \log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{2} + \frac {b p \log {\left (a x^{2} + b \right )}}{2 a} & \text {for}\: a \neq 0 \\\frac {p x^{2}}{2} + \frac {x^{2} \log {\left (c \left (\frac {b}{x^{2}}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(c*(a+b/x**2)**p),x)

[Out]

Piecewise((x**2*log(c*(a + b/x**2)**p)/2 + b*p*log(a*x**2 + b)/(2*a), Ne(a, 0)), (p*x**2/2 + x**2*log(c*(b/x**
2)**p)/2, True))

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Giac [A]
time = 4.50, size = 47, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, p x^{2} \log \left (a x^{2} + b\right ) - \frac {1}{2} \, p x^{2} \log \left (x^{2}\right ) + \frac {1}{2} \, x^{2} \log \left (c\right ) + \frac {b p \log \left (a x^{2} + b\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*(a+b/x^2)^p),x, algorithm="giac")

[Out]

1/2*p*x^2*log(a*x^2 + b) - 1/2*p*x^2*log(x^2) + 1/2*x^2*log(c) + 1/2*b*p*log(a*x^2 + b)/a

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Mupad [B]
time = 0.21, size = 33, normalized size = 0.89 \begin {gather*} \frac {x^2\,\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{2}+\frac {b\,p\,\ln \left (a\,x^2+b\right )}{2\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(c*(a + b/x^2)^p),x)

[Out]

(x^2*log(c*(a + b/x^2)^p))/2 + (b*p*log(b + a*x^2))/(2*a)

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