3.44 \(\int \frac{\log (c (a+\frac{b}{x^2})^p)}{x^4} \, dx\)

Optimal. Leaf size=68 \[ -\frac{2 a^{3/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{3 b^{3/2}}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{2 a p}{3 b x}+\frac{2 p}{9 x^3} \]

[Out]

(2*p)/(9*x^3) - (2*a*p)/(3*b*x) - (2*a^(3/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(3*b^(3/2)) - Log[c*(a + b/x^2)^p]
/(3*x^3)

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Rubi [A]  time = 0.0355895, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2455, 263, 325, 205} \[ -\frac{2 a^{3/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{3 b^{3/2}}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{2 a p}{3 b x}+\frac{2 p}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x^2)^p]/x^4,x]

[Out]

(2*p)/(9*x^3) - (2*a*p)/(3*b*x) - (2*a^(3/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(3*b^(3/2)) - Log[c*(a + b/x^2)^p]
/(3*x^3)

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^4} \, dx &=-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{1}{3} (2 b p) \int \frac{1}{\left (a+\frac{b}{x^2}\right ) x^6} \, dx\\ &=-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{1}{3} (2 b p) \int \frac{1}{x^4 \left (b+a x^2\right )} \, dx\\ &=\frac{2 p}{9 x^3}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}+\frac{1}{3} (2 a p) \int \frac{1}{x^2 \left (b+a x^2\right )} \, dx\\ &=\frac{2 p}{9 x^3}-\frac{2 a p}{3 b x}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{\left (2 a^2 p\right ) \int \frac{1}{b+a x^2} \, dx}{3 b}\\ &=\frac{2 p}{9 x^3}-\frac{2 a p}{3 b x}-\frac{2 a^{3/2} p \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{3 b^{3/2}}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0212296, size = 70, normalized size = 1.03 \[ \frac{2 a^{3/2} p \tan ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a} x}\right )}{3 b^{3/2}}-\frac{\log \left (c \left (a+\frac{b}{x^2}\right )^p\right )}{3 x^3}-\frac{2 a p}{3 b x}+\frac{2 p}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x^2)^p]/x^4,x]

[Out]

(2*p)/(9*x^3) - (2*a*p)/(3*b*x) + (2*a^(3/2)*p*ArcTan[Sqrt[b]/(Sqrt[a]*x)])/(3*b^(3/2)) - Log[c*(a + b/x^2)^p]
/(3*x^3)

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Maple [F]  time = 0.273, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\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^4,x)

[Out]

int(ln(c*(a+b/x^2)^p)/x^4,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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.31024, size = 362, normalized size = 5.32 \begin{align*} \left [\frac{3 \, a p x^{3} \sqrt{-\frac{a}{b}} \log \left (\frac{a x^{2} - 2 \, b x \sqrt{-\frac{a}{b}} - b}{a x^{2} + b}\right ) - 6 \, a p x^{2} - 3 \, b p \log \left (\frac{a x^{2} + b}{x^{2}}\right ) + 2 \, b p - 3 \, b \log \left (c\right )}{9 \, b x^{3}}, -\frac{6 \, a p x^{3} \sqrt{\frac{a}{b}} \arctan \left (x \sqrt{\frac{a}{b}}\right ) + 6 \, a p x^{2} + 3 \, b p \log \left (\frac{a x^{2} + b}{x^{2}}\right ) - 2 \, b p + 3 \, b \log \left (c\right )}{9 \, b x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x^2)^p)/x^4,x, algorithm="fricas")

[Out]

[1/9*(3*a*p*x^3*sqrt(-a/b)*log((a*x^2 - 2*b*x*sqrt(-a/b) - b)/(a*x^2 + b)) - 6*a*p*x^2 - 3*b*p*log((a*x^2 + b)
/x^2) + 2*b*p - 3*b*log(c))/(b*x^3), -1/9*(6*a*p*x^3*sqrt(a/b)*arctan(x*sqrt(a/b)) + 6*a*p*x^2 + 3*b*p*log((a*
x^2 + b)/x^2) - 2*b*p + 3*b*log(c))/(b*x^3)]

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Sympy [A]  time = 126.087, size = 177, normalized size = 2.6 \begin{align*} \begin{cases} - \frac{\log{\left (0^{p} c \right )}}{3 x^{3}} & \text{for}\: a = 0 \wedge b = 0 \\- \frac{p \log{\left (b \right )}}{3 x^{3}} + \frac{2 p \log{\left (x \right )}}{3 x^{3}} + \frac{2 p}{9 x^{3}} - \frac{\log{\left (c \right )}}{3 x^{3}} & \text{for}\: a = 0 \\- \frac{\log{\left (a^{p} c \right )}}{3 x^{3}} & \text{for}\: b = 0 \\- \frac{2 a p}{3 b x} + \frac{i a p \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{3 b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} - \frac{i a p \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + x \right )}}{3 b^{\frac{3}{2}} \sqrt{\frac{1}{a}}} - \frac{p \log{\left (a + \frac{b}{x^{2}} \right )}}{3 x^{3}} + \frac{2 p}{9 x^{3}} - \frac{\log{\left (c \right )}}{3 x^{3}} & \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**4,x)

[Out]

Piecewise((-log(0**p*c)/(3*x**3), Eq(a, 0) & Eq(b, 0)), (-p*log(b)/(3*x**3) + 2*p*log(x)/(3*x**3) + 2*p/(9*x**
3) - log(c)/(3*x**3), Eq(a, 0)), (-log(a**p*c)/(3*x**3), Eq(b, 0)), (-2*a*p/(3*b*x) + I*a*p*log(-I*sqrt(b)*sqr
t(1/a) + x)/(3*b**(3/2)*sqrt(1/a)) - I*a*p*log(I*sqrt(b)*sqrt(1/a) + x)/(3*b**(3/2)*sqrt(1/a)) - p*log(a + b/x
**2)/(3*x**3) + 2*p/(9*x**3) - log(c)/(3*x**3), True))

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Giac [A]  time = 1.28948, size = 99, normalized size = 1.46 \begin{align*} -\frac{2 \, a^{2} p \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} b} - \frac{p \log \left (a x^{2} + b\right )}{3 \, x^{3}} + \frac{p \log \left (x^{2}\right )}{3 \, x^{3}} - \frac{6 \, a p x^{2} - 2 \, b p + 3 \, b \log \left (c\right )}{9 \, b x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x^2)^p)/x^4,x, algorithm="giac")

[Out]

-2/3*a^2*p*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b) - 1/3*p*log(a*x^2 + b)/x^3 + 1/3*p*log(x^2)/x^3 - 1/9*(6*a*p*x^
2 - 2*b*p + 3*b*log(c))/(b*x^3)