Integrand size = 11, antiderivative size = 258 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=-\frac {3 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{2 a^2 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}+\frac {3 \sqrt {a+\frac {b}{x^4}} x}{2 a^2}+\frac {3 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{7/4} \sqrt {a+\frac {b}{x^4}}}-\frac {3 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{4 a^{7/4} \sqrt {a+\frac {b}{x^4}}} \] Output:
-3/2*b^(1/2)*(a+b/x^4)^(1/2)/a^2/(a^(1/2)+b^(1/2)/x^2)/x-1/2*x/a/(a+b/x^4) ^(1/2)+3/2*(a+b/x^4)^(1/2)*x/a^2+3/2*b^(1/4)*((a+b/x^4)/(a^(1/2)+b^(1/2)/x ^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*EllipticE(sin(2*arccot(a^(1/4)*x/b^(1/4 ))),1/2*2^(1/2))/a^(7/4)/(a+b/x^4)^(1/2)-3/4*b^(1/4)*((a+b/x^4)/(a^(1/2)+b ^(1/2)/x^2)^2)^(1/2)*(a^(1/2)+b^(1/2)/x^2)*InverseJacobiAM(2*arccot(a^(1/4 )*x/b^(1/4)),1/2*2^(1/2))/a^(7/4)/(a+b/x^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.78 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {x-x \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{a \sqrt {a+\frac {b}{x^4}}} \] Input:
Integrate[(a + b/x^4)^(-3/2),x]
Output:
(x - x*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[3/4, 3/2, 7/4, -((a*x^4)/b)]) /(a*Sqrt[a + b/x^4])
Time = 0.60 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {773, 819, 847, 834, 27, 761, 1510}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 773 |
| \(\displaystyle -\int \frac {x^2}{\left (a+\frac {b}{x^4}\right )^{3/2}}d\frac {1}{x}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle -\frac {3 \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle -\frac {3 \left (\frac {b \int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle -\frac {3 \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 \left (\frac {b \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle -\frac {3 \left (\frac {b \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle -\frac {3 \left (\frac {b \left (\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}\right )}{a}-\frac {x \sqrt {a+\frac {b}{x^4}}}{a}\right )}{2 a}-\frac {x}{2 a \sqrt {a+\frac {b}{x^4}}}\) |
Input:
Int[(a + b/x^4)^(-3/2),x]
Output:
-1/2*x/(a*Sqrt[a + b/x^4]) - (3*(-((Sqrt[a + b/x^4]*x)/a) + (b*(-((-(Sqrt[ a + b/x^4]/((Sqrt[a] + Sqrt[b]/x^2)*x)) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[ a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*ArcTan[b^(1/4)/(a ^(1/4)*x)], 1/2])/(b^(1/4)*Sqrt[a + b/x^4]))/Sqrt[b]) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*Arc Tan[b^(1/4)/(a^(1/4)*x)], 1/2])/(2*b^(3/4)*Sqrt[a + b/x^4])))/a))/(2*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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] && !IntegerQ[p]
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] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.73
| method | result | size |
| default | \(\frac {\left (a \,x^{4}+b \right ) \left (-x^{3} a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}+3 i \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )-3 i \sqrt {b}\, \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )\right )}{2 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} x^{6} a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(188\) |
Input:
int(1/(a+b/x^4)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2*(a*x^4+b)*(-x^3*a^(3/2)*(I*a^(1/2)/b^(1/2))^(1/2)+3*I*b^(1/2)*(-(I*a^( 1/2)*x^2-b^(1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*a *EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)-3*I*b^(1/2)*(-(I*a^(1/2)*x^2-b^( 1/2))/b^(1/2))^(1/2)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*a*EllipticE(x *(I*a^(1/2)/b^(1/2))^(1/2),I))/((a*x^4+b)/x^4)^(3/2)/x^6/a^(5/2)/(I*a^(1/2 )/b^(1/2))^(1/2)
Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {3 \, {\left (a x^{4} + b\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 3 \, {\left (a x^{4} + b\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, a x^{5} + 3 \, b x\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{2 \, {\left (a^{3} x^{4} + a^{2} b\right )}} \] Input:
integrate(1/(a+b/x^4)^(3/2),x, algorithm="fricas")
Output:
1/2*(3*(a*x^4 + b)*sqrt(a)*(-b/a)^(3/4)*elliptic_e(arcsin((-b/a)^(1/4)/x), -1) - 3*(a*x^4 + b)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin((-b/a)^(1/4)/x ), -1) + (2*a*x^5 + 3*b*x)*sqrt((a*x^4 + b)/x^4))/(a^3*x^4 + a^2*b)
Result contains complex when optimal does not.
Time = 0.77 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.16 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=- \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate(1/(a+b/x**4)**(3/2),x)
Output:
-x*gamma(-1/4)*hyper((-1/4, 3/2), (3/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a **(3/2)*gamma(3/4))
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b/x^4)^(3/2),x, algorithm="maxima")
Output:
integrate((a + b/x^4)^(-3/2), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b/x^4)^(3/2),x, algorithm="giac")
Output:
integrate((a + b/x^4)^(-3/2), x)
Time = 0.52 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.17 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {x\,{\left (\frac {a\,x^4}{b}+1\right )}^{3/2}\,\sqrt {x^{12}}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{4};\ \frac {11}{4};\ -\frac {a\,x^4}{b}\right )}{7\,{\left (a\,x^4+b\right )}^{3/2}} \] Input:
int(1/(a + b/x^4)^(3/2),x)
Output:
(x*((a*x^4)/b + 1)^(3/2)*(x^12)^(1/2)*hypergeom([3/2, 7/4], 11/4, -(a*x^4) /b))/(7*(b + a*x^4)^(3/2))
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2}} \, dx=\frac {\sqrt {a \,x^{4}+b}\, x^{3}-3 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{2} x^{8}+2 a b \,x^{4}+b^{2}}d x \right ) a b \,x^{4}-3 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{2} x^{8}+2 a b \,x^{4}+b^{2}}d x \right ) b^{2}}{a \left (a \,x^{4}+b \right )} \] Input:
int(1/(a+b/x^4)^(3/2),x)
Output:
(sqrt(a*x**4 + b)*x**3 - 3*int((sqrt(a*x**4 + b)*x**2)/(a**2*x**8 + 2*a*b* x**4 + b**2),x)*a*b*x**4 - 3*int((sqrt(a*x**4 + b)*x**2)/(a**2*x**8 + 2*a* b*x**4 + b**2),x)*b**2)/(a*(a*x**4 + b))