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]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2455, 263, 325, 205} \[ -\frac {2 a^{3/2} p \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{3 b^{3/2}}-\frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{3 x^3}-\frac {2 a p}{3 b x}+\frac {2 p}{9 x^3} \]

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 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^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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.47, size = 154, normalized size = 2.26 \[ \left [\frac {3 \, a p x^{3} \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right ) - 6 \, a p x^{2} - 3 \, b p \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, b p - 3 \, b \log \relax (c)}{9 \, b x^{3}}, -\frac {6 \, a p x^{3} \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right ) + 6 \, a p x^{2} + 3 \, b p \log \left (\frac {a x^{2} + b}{x^{2}}\right ) - 2 \, b p + 3 \, b \log \relax (c)}{9 \, b x^{3}}\right ] \]

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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giac [A]  time = 0.19, size = 73, normalized size = 1.07 \[ -\frac {2 \, a^{2} p \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} b} - \frac {p \log \left (a x^{2} + b\right )}{3 \, x^{3}} + \frac {p \log \left (x^{2}\right )}{3 \, x^{3}} - \frac {6 \, a p x^{2} - 2 \, b p + 3 \, b \log \relax (c)}{9 \, b x^{3}} \]

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)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{x^{4}}\, dx \]

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 [A]  time = 1.58, size = 62, normalized size = 0.91 \[ -\frac {2}{9} \, b p {\left (\frac {3 \, a^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {3 \, a x^{2} - b}{b^{2} x^{3}}\right )} - \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{3 \, x^{3}} \]

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]

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

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mupad [B]  time = 0.25, size = 55, normalized size = 0.81 \[ \frac {\frac {2\,p}{3}-\frac {2\,a\,p\,x^2}{b}}{3\,x^3}-\frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{3\,x^3}-\frac {2\,a^{3/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{3\,b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x^2)^p)/x^4,x)

[Out]

((2*p)/3 - (2*a*p*x^2)/b)/(3*x^3) - log(c*(a + b/x^2)^p)/(3*x^3) - (2*a^(3/2)*p*atan((a^(1/2)*x)/b^(1/2)))/(3*
b^(3/2))

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sympy [A]  time = 72.06, size = 177, normalized size = 2.60 \[ \begin {cases} - \frac {\log {\left (0^{p} c \right )}}{3 x^{3}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {\log {\left (a^{p} c \right )}}{3 x^{3}} & \text {for}\: b = 0 \\- \frac {p \log {\relax (b )}}{3 x^{3}} + \frac {2 p \log {\relax (x )}}{3 x^{3}} + \frac {2 p}{9 x^{3}} - \frac {\log {\relax (c )}}{3 x^{3}} & \text {for}\: a = 0 \\- \frac {2 a p}{3 b x} + \frac {i a p \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{3 b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {i a p \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{3 b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {p \log {\left (a + \frac {b}{x^{2}} \right )}}{3 x^{3}} + \frac {2 p}{9 x^{3}} - \frac {\log {\relax (c )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

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)), (-log(a**p*c)/(3*x**3), 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)), (-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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