\(\int x^4 \log (c (a+\frac {b}{x^2})^p) \, dx\) [36]

Optimal result

Integrand size = 16, antiderivative size = 72 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=-\frac {2 b^2 p x}{5 a^2}+\frac {2 b p x^3}{15 a}+\frac {2 b^{5/2} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{5 a^{5/2}}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2505, 269, 308, 211} \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\frac {2 b^{5/2} p \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{5 a^{5/2}}-\frac {2 b^2 p x}{5 a^2}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )+\frac {2 b p x^3}{15 a} \]

[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 211

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

Rule 269

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 308

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 2505

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*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.68 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\frac {2 b p x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {b}{a x^2}\right )}{15 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \]

[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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83

method	result	size
parts	\(\frac {x^{5} \ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{5}+\frac {2 p b \left (\frac {\frac {1}{3} x^{3} a -b x}{a^{2}}+\frac {b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\right )}{5}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.47 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\left [\frac {3 \, a^{2} p x^{5} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \left (c\right ) + 2 \, a b p x^{3} + 3 \, b^{2} p \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}, \frac {3 \, a^{2} p x^{5} \log \left (\frac {a x^{2} + b}{x^{2}}\right ) + 3 \, a^{2} x^{5} \log \left (c\right ) + 2 \, a b p x^{3} + 6 \, b^{2} p \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right ) - 6 \, b^{2} p x}{15 \, a^{2}}\right ] \]

[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]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (70) = 140\).

Time = 32.99 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.06 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\begin {cases} \frac {x^{5} \log {\left (0^{p} c \right )}}{5} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 p x^{5}}{25} + \frac {x^{5} \log {\left (c \left (\frac {b}{x^{2}}\right )^{p} \right )}}{5} & \text {for}\: a = 0 \\\frac {x^{5} \log {\left (a^{p} c \right )}}{5} & \text {for}\: b = 0 \\\frac {x^{5} \log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{5} + \frac {2 b p x^{3}}{15 a} - \frac {2 b^{2} p x}{5 a^{2}} + \frac {b^{3} p \log {\left (x - \sqrt {- \frac {b}{a}} \right )}}{5 a^{3} \sqrt {- \frac {b}{a}}} - \frac {b^{3} p \log {\left (x + \sqrt {- \frac {b}{a}} \right )}}{5 a^{3} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x**5*log(0**p*c)/5, Eq(a, 0) & Eq(b, 0)), (2*p*x**5/25 + x**5*log(c*(b/x**2)**p)/5, Eq(a, 0)), (x**
5*log(a**p*c)/5, Eq(b, 0)), (x**5*log(c*(a + b/x**2)**p)/5 + 2*b*p*x**3/(15*a) - 2*b**2*p*x/(5*a**2) + b**3*p*
log(x - sqrt(-b/a))/(5*a**3*sqrt(-b/a)) - b**3*p*log(x + sqrt(-b/a))/(5*a**3*sqrt(-b/a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right ) + \frac {2}{15} \, b p {\left (\frac {3 \, b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {a x^{3} - 3 \, b x}{a^{2}}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\frac {1}{5} \, p x^{5} \log \left (a x^{2} + b\right ) - \frac {1}{5} \, p x^{5} \log \left (x^{2}\right ) + \frac {1}{5} \, x^{5} \log \left (c\right ) + \frac {2 \, b p x^{3}}{15 \, a} + \frac {2 \, b^{3} p \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{5 \, \sqrt {a b} a^{2}} - \frac {2 \, b^{2} p x}{5 \, a^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int x^4 \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{5}+\frac {2\,b^{5/2}\,p\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{5\,a^{5/2}}+\frac {2\,b\,p\,x^3}{15\,a}-\frac {2\,b^2\,p\,x}{5\,a^2} \]

[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)