\(\int \frac {1}{(a+\frac {b}{x^4})^{3/2} x^4} \, dx\) [2097]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 241 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=-\frac {1}{2 a \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\sqrt {a+\frac {b}{x^4}}}{2 a \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {342, 296, 311, 226, 1210} \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {a+\frac {b}{x^4}}}{2 a \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )} \]

[In]

Int[1/((a + b/x^4)^(3/2)*x^4),x]

[Out]

-1/2*1/(a*Sqrt[a + b/x^4]*x^3) + Sqrt[a + b/x^4]/(2*a*Sqrt[b]*(Sqrt[a] + Sqrt[b]/x^2)*x) - (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])/(2*a^(3/4)*b^
(3/4)*Sqrt[a + b/x^4]) + (Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcC
ot[(a^(1/4)*x)/b^(1/4)], 1/2])/(4*a^(3/4)*b^(3/4)*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 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 342

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

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 {x^2}{\left (a+b x^4\right )^{3/2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {1}{2 a \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{2 a} \\ & = -\frac {1}{2 a \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{2 \sqrt {a} \sqrt {b}}-\frac {\text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{2 \sqrt {a} \sqrt {b}} \\ & = -\frac {1}{2 a \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\sqrt {a+\frac {b}{x^4}}}{2 a \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {\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 )}{2 a^{3/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {\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 )}{4 a^{3/4} b^{3/4} \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 = 7.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\frac {x \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{3 b \sqrt {a+\frac {b}{x^4}}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.75

method result size
default \(\frac {\left (a \,x^{4}+b \right ) \left (x^{3} \sqrt {b}\, \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a}-i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+i \sqrt {\frac {-i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, b E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right )}{2 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {a}}\) \(181\)

[In]

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

[Out]

1/2*(a*x^4+b)*(x^3*b^(1/2)*(I*a^(1/2)/b^(1/2))^(1/2)*a^(1/2)-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)*b*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)+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)*b*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I))/((a*x^4+b)/x^4)^
(3/2)/x^6/b^(3/2)/(I*a^(1/2)/b^(1/2))^(1/2)/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\frac {a x^{5} \sqrt {\frac {a x^{4} + b}{x^{4}}} + {\left (a x^{4} + b\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (a x^{4} + b\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1)}{2 \, {\left (a^{2} b x^{4} + a b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.74 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

-gamma(3/4)*hyper((3/4, 3/2), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(3/2)*x**3*gamma(7/4))

Maxima [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} x^{4}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^4} \, dx=\int \frac {1}{x^4\,{\left (a+\frac {b}{x^4}\right )}^{3/2}} \,d x \]

[In]

int(1/(x^4*(a + b/x^4)^(3/2)),x)

[Out]

int(1/(x^4*(a + b/x^4)^(3/2)), x)