Integrand size = 15, antiderivative size = 262 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=-\frac {1}{6 a \left (a+\frac {b}{x^4}\right )^{3/2} x^3}-\frac {1}{4 a^2 \sqrt {a+\frac {b}{x^4}} x^3}+\frac {\sqrt {a+\frac {b}{x^4}}}{4 a^2 \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{8 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}} \] Output:
-1/6/a/(a+b/x^4)^(3/2)/x^3-1/4/a^2/(a+b/x^4)^(1/2)/x^3+1/4*(a+b/x^4)^(1/2) /a^2/b^(1/2)/(a^(1/2)+b^(1/2)/x^2)/x-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)/b^(3/4)/(a+b/x^4)^(1/2)+1/8*((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)/b^(3/4)/(a+b/x^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\frac {-b x+x \left (b+a x^4\right ) \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {a x^4}{b}\right )}{3 a b \sqrt {a+\frac {b}{x^4}} \left (b+a x^4\right )} \] Input:
Integrate[1/((a + b/x^4)^(5/2)*x^4),x]
Output:
(-(b*x) + x*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[3/4, 5/2, 7/ 4, -((a*x^4)/b)])/(3*a*b*Sqrt[a + b/x^4]*(b + a*x^4))
Time = 0.61 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {858, 819, 819, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a+\frac {b}{x^4}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 819 |
\(\displaystyle -\frac {\int \frac {1}{\left (a+\frac {b}{x^4}\right )^{3/2} x^2}d\frac {1}{x}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
\(\Big \downarrow \) 819 |
\(\displaystyle -\frac {\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {1}{\sqrt {a+\frac {b}{x^4}} x^2}d\frac {1}{x}}{2 a}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle -\frac {\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a} \sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}}{2 a}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt {a} \int \frac {1}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}}{2 a}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\int \frac {\sqrt {a}-\frac {\sqrt {b}}{x^2}}{\sqrt {a+\frac {b}{x^4}}}d\frac {1}{x}}{\sqrt {b}}}{2 a}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle -\frac {\frac {1}{2 a x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\frac {\sqrt [4]{a} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{b}}{\sqrt [4]{a} x}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {a+\frac {b}{x^4}}}{x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}}{\sqrt {b}}}{2 a}}{2 a}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}}\) |
Input:
Int[1/((a + b/x^4)^(5/2)*x^4),x]
Output:
-1/6*1/(a*(a + b/x^4)^(3/2)*x^3) - (1/(2*a*Sqrt[a + b/x^4]*x^3) - (-((-(Sq rt[a + b/x^4]/((Sqrt[a] + Sqrt[b]/x^2)*x)) + (a^(1/4)*Sqrt[(a + b/x^4)/(Sq rt[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* ArcTan[b^(1/4)/(a^(1/4)*x)], 1/2])/(2*b^(3/4)*Sqrt[a + b/x^4]))/(2*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[((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[(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] && Int egerQ[m]
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 = 0.43 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.92
method | result | size |
default | \(-\frac {-3 a^{\frac {7}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {b}\, x^{11}+3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{3} b \,x^{8}-3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{3} b \,x^{8}-4 a^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {3}{2}} x^{7}+6 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{2} b^{2} x^{4}-6 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a^{2} b^{2} x^{4}-a^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {5}{2}} x^{3}+3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a \,b^{3}-3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, \operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right ) a \,b^{3}}{12 a^{\frac {5}{2}} \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} x^{10} b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\) | \(503\) |
Input:
int(1/(a+b/x^4)^(5/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/12*(-3*a^(7/2)*(I*a^(1/2)/b^(1/2))^(1/2)*b^(1/2)*x^11+3*I*(-(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)*Ellipt icF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^3*b*x^8-3*I*(-(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)*EllipticE(x*(I*a^( 1/2)/b^(1/2))^(1/2),I)*a^3*b*x^8-4*a^(5/2)*(I*a^(1/2)/b^(1/2))^(1/2)*b^(3/ 2)*x^7+6*I*(-(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)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^2*b^2*x^4-6*I *(-(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)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a^2*b^2*x^4-a^(3/2)*(I*a^ (1/2)/b^(1/2))^(1/2)*b^(5/2)*x^3+3*I*(-(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)*EllipticF(x*(I*a^(1/2)/b^(1/2) )^(1/2),I)*a*b^3-3*I*(-(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)*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*a*b^3 )/a^(5/2)/((a*x^4+b)/x^4)^(5/2)/x^10/b^(3/2)/(I*a^(1/2)/b^(1/2))^(1/2)
Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\frac {3 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) - 3 \, {\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (3 \, a^{2} x^{9} + a b x^{5}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, {\left (a^{4} b x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{2} b^{3}\right )}} \] Input:
integrate(1/(a+b/x^4)^(5/2)/x^4,x, algorithm="fricas")
Output:
1/12*(3*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin (x*(-a/b)^(1/4)), -1) - 3*(a^2*x^8 + 2*a*b*x^4 + b^2)*sqrt(b)*(-a/b)^(3/4) *elliptic_f(arcsin(x*(-a/b)^(1/4)), -1) + (3*a^2*x^9 + a*b*x^5)*sqrt((a*x^ 4 + b)/x^4))/(a^4*b*x^8 + 2*a^3*b^2*x^4 + a^2*b^3)
Result contains complex when optimal does not.
Time = 1.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {5}{2}} x^{3} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(1/(a+b/x**4)**(5/2)/x**4,x)
Output:
-gamma(3/4)*hyper((3/4, 5/2), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(5 /2)*x**3*gamma(7/4))
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:
integrate(1/(a+b/x^4)^(5/2)/x^4,x, algorithm="maxima")
Output:
integrate(1/((a + b/x^4)^(5/2)*x^4), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:
integrate(1/(a+b/x^4)^(5/2)/x^4,x, algorithm="giac")
Output:
integrate(1/((a + b/x^4)^(5/2)*x^4), x)
Timed out. \[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\int \frac {1}{x^4\,{\left (a+\frac {b}{x^4}\right )}^{5/2}} \,d x \] Input:
int(1/(x^4*(a + b/x^4)^(5/2)),x)
Output:
int(1/(x^4*(a + b/x^4)^(5/2)), x)
\[ \int \frac {1}{\left (a+\frac {b}{x^4}\right )^{5/2} x^4} \, dx=\frac {-\sqrt {a \,x^{4}+b}\, x^{3}+3 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) a^{2} b \,x^{8}+6 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) a \,b^{2} x^{4}+3 \left (\int \frac {\sqrt {a \,x^{4}+b}\, x^{2}}{a^{3} x^{12}+3 a^{2} b \,x^{8}+3 a \,b^{2} x^{4}+b^{3}}d x \right ) b^{3}}{3 a \left (a^{2} x^{8}+2 a b \,x^{4}+b^{2}\right )} \] Input:
int(1/(a+b/x^4)^(5/2)/x^4,x)
Output:
( - sqrt(a*x**4 + b)*x**3 + 3*int((sqrt(a*x**4 + b)*x**2)/(a**3*x**12 + 3* a**2*b*x**8 + 3*a*b**2*x**4 + b**3),x)*a**2*b*x**8 + 6*int((sqrt(a*x**4 + b)*x**2)/(a**3*x**12 + 3*a**2*b*x**8 + 3*a*b**2*x**4 + b**3),x)*a*b**2*x** 4 + 3*int((sqrt(a*x**4 + b)*x**2)/(a**3*x**12 + 3*a**2*b*x**8 + 3*a*b**2*x **4 + b**3),x)*b**3)/(3*a*(a**2*x**8 + 2*a*b*x**4 + b**2))