\(\int (a+\frac {b}{x^4})^{3/2} \, dx\) [2071]

Optimal result

Integrand size = 11, antiderivative size = 250 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {12 a \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{5 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\left (a+\frac {b}{x^4}\right )^{3/2} x+\frac {12 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}}-\frac {6 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}} \]

[Out]

(a+b/x^4)^(3/2)*x-6/5*b*(a+b/x^4)^(1/2)/x^3-12/5*a*b^(1/2)*(a+b/x^4)^(1/2)/x/(a^(1/2)+b^(1/2)/x^2)+12/5*a^(5/4
)*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+b/x^4)^(1/2
)-6/5*a^(5/4)*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)))*EllipticF(si
n(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
+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {248, 283, 285, 311, 226, 1210} \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {6 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}}+\frac {12 a^{5/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+\frac {b}{x^4}}}+x \left (a+\frac {b}{x^4}\right )^{3/2}-\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {12 a \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{5 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )} \]

[In]

Int[(a + b/x^4)^(3/2),x]

[Out]

(-6*b*Sqrt[a + b/x^4])/(5*x^3) - (12*a*Sqrt[b]*Sqrt[a + b/x^4])/(5*(Sqrt[a] + Sqrt[b]/x^2)*x) + (a + b/x^4)^(3
/2)*x + (12*a^(5/4)*b^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(5*Sqrt[a + b/x^4]) - (6*a^(5/4)*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])/(5*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 283

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

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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 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 {\left (a+b x^4\right )^{3/2}}{x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \left (a+\frac {b}{x^4}\right )^{3/2} x-(6 b) \text {Subst}\left (\int x^2 \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}+\left (a+\frac {b}{x^4}\right )^{3/2} x-\frac {1}{5} (12 a b) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}+\left (a+\frac {b}{x^4}\right )^{3/2} x-\frac {1}{5} \left (12 a^{3/2} \sqrt {b}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )+\frac {1}{5} \left (12 a^{3/2} \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 ) \\ & = -\frac {6 b \sqrt {a+\frac {b}{x^4}}}{5 x^3}-\frac {12 a \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{5 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\left (a+\frac {b}{x^4}\right )^{3/2} x+\frac {12 a^{5/4} \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 )}{5 \sqrt {a+\frac {b}{x^4}}}-\frac {6 a^{5/4} \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 )}{5 \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 = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.21 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {b \sqrt {a+\frac {b}{x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {a x^4}{b}\right )}{5 x^3 \sqrt {1+\frac {a x^4}{b}}} \]

[In]

Integrate[(a + b/x^4)^(3/2),x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.56

method	result	size
risch	\(-\frac {\left (7 a \,x^{4}+b \right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}{5 x^{3}}+\frac {12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right ) \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}{5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \left (a \,x^{4}+b \right )}\)	\(140\)
default	\(\frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x \left (12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-12 i a^{\frac {3}{2}} \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, x^{5} E\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}-8 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a b \,x^{4}-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2}\right )}{5 \left (a \,x^{4}+b \right )^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(229\)

[In]

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

[Out]

-1/5*(7*a*x^4+b)/x^3*((a*x^4+b)/x^4)^(1/2)+12/5*I*a^(3/2)*b^(1/2)/(I*a^(1/2)/b^(1/2))^(1/2)*(1-I*a^(1/2)/b^(1/
2)*x^2)^(1/2)*(1+I*a^(1/2)/b^(1/2)*x^2)^(1/2)/(a*x^4+b)*(EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)-EllipticE(x*
(I*a^(1/2)/b^(1/2))^(1/2),I))*((a*x^4+b)/x^4)^(1/2)*x^2

Fricas [F]

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

[In]

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

[Out]

integral((a*x^4 + b)*sqrt((a*x^4 + b)/x^4)/x^4, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.17 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=- \frac {a^{\frac {3}{2}} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((a + b/x^4)^(3/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((a + b/x^4)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 6.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.15 \[ \int \left (a+\frac {b}{x^4}\right )^{3/2} \, dx=-\frac {x\,{\left (a+\frac {b}{x^4}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {5}{4};\ -\frac {1}{4};\ -\frac {a\,x^4}{b}\right )}{5\,{\left (\frac {a\,x^4}{b}+1\right )}^{3/2}} \]

[In]

int((a + b/x^4)^(3/2),x)

[Out]

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