\(\int \frac {1}{(a+\frac {b}{x^4})^{5/2}} \, dx\) [2103]

Optimal result

Integrand size = 11, antiderivative size = 277 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}-\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}} \]

[Out]

-1/6*x/a/(a+b/x^4)^(3/2)-7/12*x/a^2/(a+b/x^4)^(1/2)+7/4*x*(a+b/x^4)^(1/2)/a^3-7/4*b^(1/2)*(a+b/x^4)^(1/2)/a^3/
x/(a^(1/2)+b^(1/2)/x^2)+7/4*b^(1/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4))
)*EllipticE(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x^
2)^2)^(1/2)/a^(11/4)/(a+b/x^4)^(1/2)-7/8*b^(1/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/
4)*x/b^(1/4)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/
2)+b^(1/2)/x^2)^2)^(1/2)/a^(11/4)/(a+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {248, 296, 331, 311, 226, 1210} \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=-\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}+\frac {7 x \sqrt {a+\frac {b}{x^4}}}{4 a^3}-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}} \]

[In]

Int[(a + b/x^4)^(-5/2),x]

[Out]

(-7*Sqrt[b]*Sqrt[a + b/x^4])/(4*a^3*(Sqrt[a] + Sqrt[b]/x^2)*x) - x/(6*a*(a + b/x^4)^(3/2)) - (7*x)/(12*a^2*Sqr
t[a + b/x^4]) + (7*Sqrt[a + b/x^4]*x)/(4*a^3) + (7*b^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a
] + Sqrt[b]/x^2)*EllipticE[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(4*a^(11/4)*Sqrt[a + b/x^4]) - (7*b^(1/4)*Sqrt
[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])
/(8*a^(11/4)*Sqrt[a + b/x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

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 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^4\right )^{5/2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^4\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{6 a} \\ & = -\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}-\frac {7 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^2} \\ & = -\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}-\frac {(7 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^3} \\ & = -\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}-\frac {\left (7 \sqrt {b}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^{5/2}}+\frac {\left (7 \sqrt {b}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^{5/2}} \\ & = -\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}-\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {7 b x+3 a x^5-7 x \left (b+a x^4\right ) \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{3 a^2 \sqrt {a+\frac {b}{x^4}} \left (b+a x^4\right )} \]

[In]

Integrate[(a + b/x^4)^(-5/2),x]

[Out]

(7*b*x + 3*a*x^5 - 7*x*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[3/4, 5/2, 7/4, -((a*x^4)/b)])/(3*a^2*
Sqrt[a + b/x^4]*(b + a*x^4))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 485, normalized size of antiderivative = 1.75

method	result	size
default	\(\frac {-9 a^{\frac {9}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x^{11}+21 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) \sqrt {b}\, a^{4} x^{8}-21 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) \sqrt {b}\, a^{4} x^{8}-16 a^{\frac {7}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b \,x^{7}+42 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) b^{\frac {3}{2}} a^{3} x^{4}-42 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) b^{\frac {3}{2}} a^{3} x^{4}+21 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) b^{\frac {5}{2}} a^{2}-21 i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) b^{\frac {5}{2}} a^{2}-7 a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2} x^{3}}{12 a^{\frac {9}{2}} \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{10} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(485\)

[In]

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

[Out]

1/12*(-9*a^(9/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^11+21*I*((-I*x^2*a^(1/2)+b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+
b^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^(1/2)*a^4*x^8-21*I*((-I*x^2*a^(1/2)+b^(1/2)
)/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^(1/2)*a^4*
x^8-16*a^(7/2)*(I*a^(1/2)/b^(1/2))^(1/2)*b*x^7+42*I*((-I*x^2*a^(1/2)+b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b
^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^(3/2)*a^3*x^4-42*I*((-I*x^2*a^(1/2)+b^(1/2))
/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^(3/2)*a^3*x
^4+21*I*((-I*x^2*a^(1/2)+b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2
)/b^(1/2))^(1/2),I)*b^(5/2)*a^2-21*I*((-I*x^2*a^(1/2)+b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2)
)^(1/2)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*b^(5/2)*a^2-7*a^(5/2)*(I*a^(1/2)/b^(1/2))^(1/2)*b^2*x^3)/a^(9
/2)/((a*x^4+b)/x^4)^(5/2)/x^10/(I*a^(1/2)/b^(1/2))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {21 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 21 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (12 \, a^{2} x^{9} + 35 \, a b x^{5} + 21 \, b^{2} x\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, {\left (a^{5} x^{8} + 2 \, a^{4} b x^{4} + a^{3} b^{2}\right )}} \]

[In]

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

[Out]

1/12*(21*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin((-b/a)^(1/4)/x), -1) - 21*(a^2*x^8
+ 2*a*b*x^4 + b^2)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin((-b/a)^(1/4)/x), -1) + (12*a^2*x^9 + 35*a*b*x^5 + 2
1*b^2*x)*sqrt((a*x^4 + b)/x^4))/(a^5*x^8 + 2*a^4*b*x^4 + a^3*b^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

-x*gamma(-1/4)*hyper((-1/4, 5/2), (3/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(5/2)*gamma(3/4))

Maxima [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate((a + b/x^4)^(-5/2), x)

Giac [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate((a + b/x^4)^(-5/2), x)

Mupad [B] (verification not implemented)

Time = 6.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx=\frac {x\,{\left (\frac {a\,x^4}{b}+1\right )}^{5/2}\,\sqrt {x^{20}}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {11}{4};\ \frac {15}{4};\ -\frac {a\,x^4}{b}\right )}{11\,{\left (a\,x^4+b\right )}^{5/2}} \]

[In]

int(1/(a + b/x^4)^(5/2),x)

[Out]

(x*((a*x^4)/b + 1)^(5/2)*(x^20)^(1/2)*hypergeom([5/2, 11/4], 15/4, -(a*x^4)/b))/(11*(b + a*x^4)^(5/2))