Integrand size = 25, antiderivative size = 94 \[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {\operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{2 b^2 f \sqrt {a \sin (e+f x)}} \] Output:
(a*sin(f*x+e))^(1/2)/a/b/f/(b*sec(f*x+e))^(1/2)+1/2*InverseJacobiAM(e-1/4* Pi+f*x,2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/b^2/f/(a*sin(f*x +e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {b \sec (e+f x)} \left (-1+\cos (2 (e+f x))-\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{2 b^2 f \sqrt {a \sin (e+f x)}} \] Input:
Integrate[1/((b*Sec[e + f*x])^(3/2)*Sqrt[a*Sin[e + f*x]]),x]
Output:
-1/2*(Cot[e + f*x]*Sqrt[b*Sec[e + f*x]]*(-1 + Cos[2*(e + f*x)] - Hypergeom etric2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(b^2*f*S qrt[a*Sin[e + f*x]])
Time = 0.85 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3062, 3042, 3065, 3042, 3053, 3042, 3120}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)} (b \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \sin (e+f x)} (b \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3062 |
\(\displaystyle \frac {\int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 b^2}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx}{2 b^2}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3065 |
\(\displaystyle \frac {\sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{2 b^2}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx}{2 b^2}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 b^2 \sqrt {a \sin (e+f x)}}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 b^2 \sqrt {a \sin (e+f x)}}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{2 b^2 f \sqrt {a \sin (e+f x)}}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}\) |
Input:
Int[1/((b*Sec[e + f*x])^(3/2)*Sqrt[a*Sin[e + f*x]]),x]
Output:
Sqrt[a*Sin[e + f*x]]/(a*b*f*Sqrt[b*Sec[e + f*x]]) + (EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(2*b^2*f*Sqrt[a*Sin[e + f*x]])
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n + 1)/(a *b*f*(m - n))), x] - Simp[(n + 1)/(b^2*(m - n)) Int[(a*Sin[e + f*x])^m*(b *Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] & & NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int egerQ[m - 1/2] && IntegerQ[n - 1/2]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Time = 1.77 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.34
method | result | size |
default | \(\frac {\sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \left (\sec \left (f x +e \right )+1\right )+2 \sin \left (f x +e \right )}{2 f \sqrt {a \sin \left (f x +e \right )}\, \sqrt {b \sec \left (f x +e \right )}\, b}\) | \(126\) |
Input:
int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2/f/(a*sin(f*x+e))^(1/2)/(b*sec(f*x+e))^(1/2)/b*((-2*csc(f*x+e)+2*cot(f* x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*EllipticF((csc(f*x+e)-cot(f*x +e)+1)^(1/2),1/2*2^(1/2))*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(sec(f*x+e)+1)+2 *sin(f*x+e))
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {a \sin \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="fricas ")
Output:
integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))/(a*b^2*sec(f*x + e)^2*s in(f*x + e)), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {a \sin {\left (e + f x \right )}} \left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b*sec(f*x+e))**(3/2)/(a*sin(f*x+e))**(1/2),x)
Output:
Integral(1/(sqrt(a*sin(e + f*x))*(b*sec(e + f*x))**(3/2)), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {a \sin \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="maxima ")
Output:
integrate(1/((b*sec(f*x + e))^(3/2)*sqrt(a*sin(f*x + e))), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\int { \frac {1}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {a \sin \left (f x + e\right )}} \,d x } \] Input:
integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate(1/((b*sec(f*x + e))^(3/2)*sqrt(a*sin(f*x + e))), x)
Timed out. \[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\int \frac {1}{\sqrt {a\,\sin \left (e+f\,x\right )}\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:
int(1/((a*sin(e + f*x))^(1/2)*(b/cos(e + f*x))^(3/2)),x)
Output:
int(1/((a*sin(e + f*x))^(1/2)*(b/cos(e + f*x))^(3/2)), x)
\[ \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{2} \sin \left (f x +e \right )}d x \right )}{a \,b^{2}} \] Input:
int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(1/2),x)
Output:
(sqrt(b)*sqrt(a)*int((sqrt(sin(e + f*x))*sqrt(sec(e + f*x)))/(sec(e + f*x) **2*sin(e + f*x)),x))/(a*b**2)