3.42 \(\int \frac{\log (c (a+\frac{b}{x^2})^p)}{x^2} \, dx\)

Optimal. Leaf size=50 \[ -\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}+\frac{2 \sqrt{a} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{\sqrt{b}}+\frac{2 p}{x} \]

[Out]

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

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Rubi [A]  time = 0.0271548, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 263, 325, 205} \[ -\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}+\frac{2 \sqrt{a} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{\sqrt{b}}+\frac{2 p}{x} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x^2)^p]/x^2,x]

[Out]

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

Rule 2455

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]

Rule 263

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x^2} \, dx &=-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}-(2 b p) \int \frac{1}{\left (a+\frac{b}{x^2}\right ) x^4} \, dx\\ &=-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}-(2 b p) \int \frac{1}{x^2 \left (b+a x^2\right )} \, dx\\ &=\frac{2 p}{x}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}+(2 a p) \int \frac{1}{b+a x^2} \, dx\\ &=\frac{2 p}{x}+\frac{2 \sqrt{a} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{\sqrt{b}}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.0139083, size = 52, normalized size = 1.04 \[ -\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{x}-\frac{2 \sqrt{a} p \tan ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} x}\right )}{\sqrt{b}}+\frac{2 p}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x^2)^p]/x^2,x]

[Out]

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

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Maple [F]  time = 0.235, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\ln \left ( c \left ( a+{\frac{b}{{x}^{2}}} \right ) ^{p} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27257, size = 261, normalized size = 5.22 \begin{align*} \left [\frac{p x \sqrt{-\frac{a}{b}} \log \left (\frac{a x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - b}{a x^{2} + b}\right ) - p \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + 2 \, p - \log \left (c\right )}{x}, \frac{2 \, p x \sqrt{\frac{a}{b}} \arctan \left (x \sqrt{\frac{a}{b}}\right ) - p \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + 2 \, p - \log \left (c\right )}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(p*x*sqrt(-a/b)*log((a*x^2 + 2*b*x*sqrt(-a/b) - b)/(a*x^2 + b)) - p*log((a*x^2 + b)/x^2) + 2*p - log(c))/x, (
2*p*x*sqrt(a/b)*arctan(x*sqrt(a/b)) - p*log((a*x^2 + b)/x^2) + 2*p - log(c))/x]

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Sympy [A]  time = 59.9988, size = 129, normalized size = 2.58 \begin{align*} \begin{cases} - \frac{\log{\left (0^{p} c \right )}}{x} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{p \log{\left (b \right )}}{x} + \frac{2 p \log{\left (x \right )}}{x} + \frac{2 p}{x} - \frac{\log{\left (c \right )}}{x} & \text{for}\: a = 0 \\- \frac{\log{\left (a^{p} c \right )}}{x} & \text{for}\: b = 0 \\- \frac{p \log{\left (a + \frac{b}{x^{2}} \right )}}{x} + \frac{2 p}{x} - \frac{\log{\left (c \right )}}{x} - \frac{i p \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{\sqrt{b} \sqrt{\frac{1}{a}}} + \frac{i p \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{\sqrt{b} \sqrt{\frac{1}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-log(0**p*c)/x, Eq(a, 0) & Eq(b, 0)), (-p*log(b)/x + 2*p*log(x)/x + 2*p/x - log(c)/x, Eq(a, 0)), (-
log(a**p*c)/x, Eq(b, 0)), (-p*log(a + b/x**2)/x + 2*p/x - log(c)/x - I*p*log(-I*sqrt(b)*sqrt(1/a) + x)/(sqrt(b
)*sqrt(1/a)) + I*p*log(I*sqrt(b)*sqrt(1/a) + x)/(sqrt(b)*sqrt(1/a)), True))

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Giac [A]  time = 1.26026, size = 73, normalized size = 1.46 \begin{align*} \frac{2 \, a p \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{\sqrt{a b}} - \frac{p \log \left (a x^{2} + b\right )}{x} + \frac{p \log \left (x^{2}\right )}{x} + \frac{2 \, p - \log \left (c\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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