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

Optimal. Leaf size=72 \[ \frac {2 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{5 a^{5/2}}-\frac {2 b^2 p x}{5 a^2}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {2 b p x^3}{15 a} \]

[Out]

-2/5*b^2*p*x/a^2+2/15*b*p*x^3/a+2/5*b^(5/2)*p*arctan(x*a^(1/2)/b^(1/2))/a^(5/2)+1/5*x^5*ln(c*(a+b/x^2)^p)

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Rubi [A]  time = 0.04, antiderivative size = 72, 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, 302, 205} \[ -\frac {2 b^2 p x}{5 a^2}+\frac {2 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{5 a^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {2 b p x^3}{15 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b/x^2)^p],x]

[Out]

(-2*b^2*p*x)/(5*a^2) + (2*b*p*x^3)/(15*a) + (2*b^(5/2)*p*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(5*a^(5/2)) + (x^5*Log[c
*(a + b/x^2)^p])/5

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {1}{5} (2 b p) \int \frac {x^2}{a+\frac {b}{x^2}} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {1}{5} (2 b p) \int \frac {x^4}{b+a x^2} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {1}{5} (2 b p) \int \left (-\frac {b}{a^2}+\frac {x^2}{a}+\frac {b^2}{a^2 \left (b+a x^2\right )}\right ) \, dx\\ &=-\frac {2 b^2 p x}{5 a^2}+\frac {2 b p x^3}{15 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {\left (2 b^3 p\right ) \int \frac {1}{b+a x^2} \, dx}{5 a^2}\\ &=-\frac {2 b^2 p x}{5 a^2}+\frac {2 b p x^3}{15 a}+\frac {2 b^{5/2} p \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{5 a^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.68 \[ \frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {2 b p x^3 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {b}{a x^2}\right )}{15 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b/x^2)^p],x]

[Out]

(2*b*p*x^3*Hypergeometric2F1[-3/2, 1, -1/2, -(b/(a*x^2))])/(15*a) + (x^5*Log[c*(a + b/x^2)^p])/5

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fricas [A]  time = 0.51, size = 178, normalized size = 2.47 \[ \left [\frac {3 \, a^{2} p x^{5} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \relax (c) + 2 \, a b p x^{3} + 3 \, b^{2} p \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}, \frac {3 \, a^{2} p x^{5} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \relax (c) + 2 \, a b p x^{3} + 6 \, b^{2} p \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(3*a^2*p*x^5*log((a*x^2 + b)/x^2) + 3*a^2*x^5*log(c) + 2*a*b*p*x^3 + 3*b^2*p*sqrt(-b/a)*log((a*x^2 + 2*a
*x*sqrt(-b/a) - b)/(a*x^2 + b)) - 6*b^2*p*x)/a^2, 1/15*(3*a^2*p*x^5*log((a*x^2 + b)/x^2) + 3*a^2*x^5*log(c) +
2*a*b*p*x^3 + 6*b^2*p*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b) - 6*b^2*p*x)/a^2]

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giac [A]  time = 0.21, size = 75, normalized size = 1.04 \[ \frac {1}{5} \, p x^{5} \log \left (a x^{2} + b\right ) - \frac {1}{5} \, p x^{5} \log \left (x^{2}\right ) + \frac {1}{5} \, x^{5} \log \relax (c) + \frac {2 \, b p x^{3}}{15 \, a} + \frac {2 \, b^{3} p \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} a^{2}} - \frac {2 \, b^{2} p x}{5 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*p*x^5*log(a*x^2 + b) - 1/5*p*x^5*log(x^2) + 1/5*x^5*log(c) + 2/15*b*p*x^3/a + 2/5*b^3*p*arctan(a*x/sqrt(a*
b))/(sqrt(a*b)*a^2) - 2/5*b^2*p*x/a^2

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int x^{4} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.47, size = 59, normalized size = 0.82 \[ \frac {1}{5} \, x^{5} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right ) + \frac {2}{15} \, b p {\left (\frac {3 \, b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {a x^{3} - 3 \, b x}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*log((a + b/x^2)^p*c) + 2/15*b*p*(3*b^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^2) + (a*x^3 - 3*b*x)/a^2)

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mupad [B]  time = 0.26, size = 56, normalized size = 0.78 \[ \frac {x^5\,\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{5}+\frac {2\,b^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{5\,a^{5/2}}+\frac {2\,b\,p\,x^3}{15\,a}-\frac {2\,b^2\,p\,x}{5\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(c*(a + b/x^2)^p),x)

[Out]

(x^5*log(c*(a + b/x^2)^p))/5 + (2*b^(5/2)*p*atan((a^(1/2)*x)/b^(1/2)))/(5*a^(5/2)) + (2*b*p*x^3)/(15*a) - (2*b
^2*p*x)/(5*a^2)

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sympy [A]  time = 92.91, size = 162, normalized size = 2.25 \[ \begin {cases} \frac {p x^{5} \log {\left (a + \frac {b}{x^{2}} \right )}}{5} + \frac {x^{5} \log {\relax (c )}}{5} + \frac {2 b p x^{3}}{15 a} - \frac {2 b^{2} p x}{5 a^{2}} - \frac {i b^{\frac {5}{2}} p \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{5 a^{3} \sqrt {\frac {1}{a}}} + \frac {i b^{\frac {5}{2}} p \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + x \right )}}{5 a^{3} \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\\frac {p x^{5} \log {\relax (b )}}{5} - \frac {2 p x^{5} \log {\relax (x )}}{5} + \frac {2 p x^{5}}{25} + \frac {x^{5} \log {\relax (c )}}{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*ln(c*(a+b/x**2)**p),x)

[Out]

Piecewise((p*x**5*log(a + b/x**2)/5 + x**5*log(c)/5 + 2*b*p*x**3/(15*a) - 2*b**2*p*x/(5*a**2) - I*b**(5/2)*p*l
og(-I*sqrt(b)*sqrt(1/a) + x)/(5*a**3*sqrt(1/a)) + I*b**(5/2)*p*log(I*sqrt(b)*sqrt(1/a) + x)/(5*a**3*sqrt(1/a))
, Ne(a, 0)), (p*x**5*log(b)/5 - 2*p*x**5*log(x)/5 + 2*p*x**5/25 + x**5*log(c)/5, True))

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