Integrand size = 13, antiderivative size = 893 \[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=-\frac {2 \sqrt {2+\sqrt {2}} \sqrt {a} \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} E\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{b x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}+\frac {2 \sqrt {2+\sqrt {2}} \sqrt {a} \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} E\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right )|-2 \left (1-\sqrt {2}\right )\right )}{b x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )}+\frac {2 \sqrt {a} \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{\sqrt {2+\sqrt {2}} b x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}-\frac {2 \sqrt {a} \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{\sqrt {2+\sqrt {2}} b x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )} \] Output:
-2*(2+2^(1/2))^(1/2)*a^(1/2)*(-b*x^2/a^(1/2)/(b*x^2-a)^(1/2))^(1/2)*(b*x^2 -a)^(3/8)*((a^(1/4)+(b*x^2-a)^(1/4))^2/a^(1/4)/(b*x^2-a)^(1/4))^(1/2)*Elli pticE(1/2*(-a^(1/4)*(2^(1/2)-2*(b*x^2-a)^(1/4)/a^(1/4)+2^(1/2)*(b*x^2-a)^( 1/2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2))^(1/2))/b/x/(a^(1/4)+(b *x^2-a)^(1/4))+2*(2+2^(1/2))^(1/2)*a^(1/2)*(-b*x^2/a^(1/2)/(b*x^2-a)^(1/2) )^(1/2)*(b*x^2-a)^(3/8)*(-(a^(1/4)-(b*x^2-a)^(1/4))^2/a^(1/4)/(b*x^2-a)^(1 /4))^(1/2)*EllipticE(1/2*(a^(1/4)*(2^(1/2)+2*(b*x^2-a)^(1/4)/a^(1/4)+2^(1/ 2)*(b*x^2-a)^(1/2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2))^(1/2))/b /x/(a^(1/4)-(b*x^2-a)^(1/4))+2*a^(1/2)*(-b*x^2/a^(1/2)/(b*x^2-a)^(1/2))^(1 /2)*(b*x^2-a)^(3/8)*((a^(1/4)+(b*x^2-a)^(1/4))^2/a^(1/4)/(b*x^2-a)^(1/4))^ (1/2)*EllipticF(1/2*(-a^(1/4)*(2^(1/2)-2*(b*x^2-a)^(1/4)/a^(1/4)+2^(1/2)*( b*x^2-a)^(1/2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^ (1/2))^(1/2)/b/x/(a^(1/4)+(b*x^2-a)^(1/4))-2*a^(1/2)*(-b*x^2/a^(1/2)/(b*x^ 2-a)^(1/2))^(1/2)*(b*x^2-a)^(3/8)*(-(a^(1/4)-(b*x^2-a)^(1/4))^2/a^(1/4)/(b *x^2-a)^(1/4))^(1/2)*EllipticF(1/2*(a^(1/4)*(2^(1/2)+2*(b*x^2-a)^(1/4)/a^( 1/4)+2^(1/2)*(b*x^2-a)^(1/2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2) )^(1/2))/(2+2^(1/2))^(1/2)/b/x/(a^(1/4)-(b*x^2-a)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.05 \[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\frac {x \left (1-\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )}{\left (-a+b x^2\right )^{3/8}} \] Input:
Integrate[(-a + b*x^2)^(-3/8),x]
Output:
(x*(1 - (b*x^2)/a)^(3/8)*Hypergeometric2F1[3/8, 1/2, 3/2, (b*x^2)/a])/(-a + b*x^2)^(3/8)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {238, 237}
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.
\(\displaystyle \int \frac {1}{\left (b x^2-a\right )^{3/8}} \, dx\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {\left (1-\frac {b x^2}{a}\right )^{3/8} \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/8}}dx}{\left (b x^2-a\right )^{3/8}}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {x \left (1-\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )}{\left (b x^2-a\right )^{3/8}}\) |
Input:
Int[(-a + b*x^2)^(-3/8),x]
Output:
(x*(1 - (b*x^2)/a)^(3/8)*Hypergeometric2F1[3/8, 1/2, 3/2, (b*x^2)/a])/(-a + b*x^2)^(3/8)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
\[\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {3}{8}}}d x\]
Input:
int(1/(b*x^2-a)^(3/8),x)
Output:
int(1/(b*x^2-a)^(3/8),x)
\[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}}} \,d x } \] Input:
integrate(1/(b*x^2-a)^(3/8),x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(-3/8), x)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.03 \[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\frac {x e^{- \frac {3 i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{a^{\frac {3}{8}}} \] Input:
integrate(1/(b*x**2-a)**(3/8),x)
Output:
x*exp(-3*I*pi/8)*hyper((3/8, 1/2), (3/2,), b*x**2/a)/a**(3/8)
\[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}}} \,d x } \] Input:
integrate(1/(b*x^2-a)^(3/8),x, algorithm="maxima")
Output:
integrate((b*x^2 - a)^(-3/8), x)
\[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}}} \,d x } \] Input:
integrate(1/(b*x^2-a)^(3/8),x, algorithm="giac")
Output:
integrate((b*x^2 - a)^(-3/8), x)
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.04 \[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\frac {x\,{\left (1-\frac {b\,x^2}{a}\right )}^{3/8}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{8},\frac {1}{2};\ \frac {3}{2};\ \frac {b\,x^2}{a}\right )}{{\left (b\,x^2-a\right )}^{3/8}} \] Input:
int(1/(b*x^2 - a)^(3/8),x)
Output:
(x*(1 - (b*x^2)/a)^(3/8)*hypergeom([3/8, 1/2], 3/2, (b*x^2)/a))/(b*x^2 - a )^(3/8)
\[ \int \frac {1}{\left (-a+b x^2\right )^{3/8}} \, dx=\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {3}{8}}}d x \] Input:
int(1/(b*x^2-a)^(3/8),x)
Output:
int(1/( - a + b*x**2)**(3/8),x)