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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.44, size = 119, normalized size = 2.38 \[ \left [\frac {p x \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right ) - p \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, p - \log \relax (c)}{x}, \frac {2 \, p x \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right ) - p \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, p - \log \relax (c)}{x}\right ] \]

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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giac [A]  time = 0.17, size = 54, normalized size = 1.08 \[ \frac {2 \, a p \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b}} - \frac {p \log \left (a x^{2} + b\right )}{x} + \frac {p \log \left (x^{2}\right )}{x} + \frac {2 \, p - \log \relax (c)}{x} \]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{x^{2}}\, dx \]

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 [A]  time = 1.41, size = 49, normalized size = 0.98 \[ 2 \, b p {\left (\frac {a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {1}{b x}\right )} - \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{x} \]

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]

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

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mupad [B]  time = 0.24, size = 42, normalized size = 0.84 \[ \frac {2\,p}{x}-\frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{x}+\frac {2\,\sqrt {a}\,p\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x^2)^p)/x^2,x)

[Out]

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

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sympy [A]  time = 25.45, size = 129, normalized size = 2.58 \[ \begin {cases} - \frac {\log {\left (0^{p} c \right )}}{x} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {\log {\left (a^{p} c \right )}}{x} & \text {for}\: b = 0 \\- \frac {p \log {\relax (b )}}{x} + \frac {2 p \log {\relax (x )}}{x} + \frac {2 p}{x} - \frac {\log {\relax (c )}}{x} & \text {for}\: a = 0 \\- \frac {p \log {\left (a + \frac {b}{x^{2}} \right )}}{x} + \frac {2 p}{x} - \frac {\log {\relax (c )}}{x} - \frac {i p \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{\sqrt {b} \sqrt {\frac {1}{a}}} + \frac {i p \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{\sqrt {b} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

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)), (-log(a**p*c)/x, Eq(b, 0)), (-p*log(b)/x + 2*p*log(x)/x + 2*p
/x - log(c)/x, Eq(a, 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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