\(\int \frac {1}{(a+\frac {b}{x^4})^{5/2} x^3} \, dx\) [2101]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {b}{3 a^2 \left (a+\frac {b}{x^4}\right )^{3/2} x^6}-\frac {1}{2 a \left (a+\frac {b}{x^4}\right )^{3/2} x^2} \]

[Out]

-1/3*b/a^2/(a+b/x^4)^(3/2)/x^6-1/2/a/(a+b/x^4)^(3/2)/x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {277, 270} \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {b}{3 a^2 x^6 \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {1}{2 a x^2 \left (a+\frac {b}{x^4}\right )^{3/2}} \]

[In]

Int[1/((a + b/x^4)^(5/2)*x^3),x]

[Out]

-1/3*b/(a^2*(a + b/x^4)^(3/2)*x^6) - 1/(2*a*(a + b/x^4)^(3/2)*x^2)

Rule 270

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 a \left (a+\frac {b}{x^4}\right )^{3/2} x^2}+\frac {(2 b) \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^7} \, dx}{a} \\ & = -\frac {b}{3 a^2 \left (a+\frac {b}{x^4}\right )^{3/2} x^6}-\frac {1}{2 a \left (a+\frac {b}{x^4}\right )^{3/2} x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=\frac {\left (-2 b-3 a x^4\right ) \left (b+a x^4\right )}{6 a^2 \left (a+\frac {b}{x^4}\right )^{5/2} x^{10}} \]

[In]

Integrate[1/((a + b/x^4)^(5/2)*x^3),x]

[Out]

((-2*b - 3*a*x^4)*(b + a*x^4))/(6*a^2*(a + b/x^4)^(5/2)*x^10)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89

method	result	size
gosper	\(-\frac {\left (a \,x^{4}+b \right ) \left (3 a \,x^{4}+2 b \right )}{6 a^{2} x^{10} \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}}}\)	\(39\)
default	\(-\frac {\left (a \,x^{4}+b \right ) \left (3 a \,x^{4}+2 b \right )}{6 a^{2} x^{10} \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}}}\)	\(39\)
trager	\(-\frac {x^{2} \left (3 a \,x^{4}+2 b \right ) \sqrt {-\frac {-a \,x^{4}-b}{x^{4}}}}{6 a^{2} \left (a \,x^{4}+b \right )^{2}}\)	\(45\)

[In]

int(1/(a+b/x^4)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/6*(a*x^4+b)*(3*a*x^4+2*b)/a^2/x^10/((a*x^4+b)/x^4)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {{\left (3 \, a x^{6} + 2 \, b x^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{6 \, {\left (a^{4} x^{8} + 2 \, a^{3} b x^{4} + a^{2} b^{2}\right )}} \]

[In]

integrate(1/(a+b/x^4)^(5/2)/x^3,x, algorithm="fricas")

[Out]

-1/6*(3*a*x^6 + 2*b*x^2)*sqrt((a*x^4 + b)/x^4)/(a^4*x^8 + 2*a^3*b*x^4 + a^2*b^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (39) = 78\).

Time = 0.83 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.39 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=- \frac {3 a x^{4}}{6 a^{3} \sqrt {b} x^{4} \sqrt {\frac {a x^{4}}{b} + 1} + 6 a^{2} b^{\frac {3}{2}} \sqrt {\frac {a x^{4}}{b} + 1}} - \frac {2 b}{6 a^{3} \sqrt {b} x^{4} \sqrt {\frac {a x^{4}}{b} + 1} + 6 a^{2} b^{\frac {3}{2}} \sqrt {\frac {a x^{4}}{b} + 1}} \]

[In]

integrate(1/(a+b/x**4)**(5/2)/x**3,x)

[Out]

-3*a*x**4/(6*a**3*sqrt(b)*x**4*sqrt(a*x**4/b + 1) + 6*a**2*b**(3/2)*sqrt(a*x**4/b + 1)) - 2*b/(6*a**3*sqrt(b)*
x**4*sqrt(a*x**4/b + 1) + 6*a**2*b**(3/2)*sqrt(a*x**4/b + 1))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {3 \, {\left (a + \frac {b}{x^{4}}\right )} x^{4} - b}{6 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} x^{6}} \]

[In]

integrate(1/(a+b/x^4)^(5/2)/x^3,x, algorithm="maxima")

[Out]

-1/6*(3*(a + b/x^4)*x^4 - b)/((a + b/x^4)^(3/2)*a^2*x^6)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {3 \, a x^{4} + 2 \, b}{6 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{2}} \]

[In]

integrate(1/(a+b/x^4)^(5/2)/x^3,x, algorithm="giac")

[Out]

-1/6*(3*a*x^4 + 2*b)/((a*x^4 + b)^(3/2)*a^2)

Mupad [B] (verification not implemented)

Time = 6.00 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^3} \, dx=-\frac {x^2\,\sqrt {a+\frac {b}{x^4}}\,\left (3\,a\,x^4+2\,b\right )}{6\,a^2\,{\left (a\,x^4+b\right )}^2} \]

[In]

int(1/(x^3*(a + b/x^4)^(5/2)),x)

[Out]

-(x^2*(a + b/x^4)^(1/2)*(2*b + 3*a*x^4))/(6*a^2*(b + a*x^4)^2)