\(\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx\) [457]

Optimal result

Integrand size = 25, antiderivative size = 91 \[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}+\frac {a^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{2 f \sqrt {a \sin (e+f x)}} \] Output:

-a*b*(a*sin(f*x+e))^(1/2)/f/(b*sec(f*x+e))^(1/2)+1/2*a^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)/f/(a*sin(f* 
x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 11.57 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {1}{2},\sec ^2(e+f x)\right ) (b \sec (e+f x))^{3/2} (a \sin (e+f x))^{5/2}}{a b f \left (-\tan ^2(e+f x)\right )^{5/4}} \] Input:

Integrate[Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2),x]
 

Output:

(Hypergeometric2F1[-1/2, -1/4, 1/2, Sec[e + f*x]^2]*(b*Sec[e + f*x])^(3/2) 
*(a*Sin[e + f*x])^(5/2))/(a*b*f*(-Tan[e + f*x]^2)^(5/4))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 91, 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, 3063, 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.

\(\displaystyle \int (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}dx\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3065

\(\displaystyle \frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\)

Input:

Int[Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2),x]
 

Output:

-((a*b*Sqrt[a*Sin[e + f*x]])/(f*Sqrt[b*Sec[e + f*x]])) + (a^2*EllipticF[e 
- Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(2*f*Sqrt[a* 
Sin[e + f*x]])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053

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]
 

rule 3063

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*b*(a*Sin[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n 
- 1)/(f*(m - n))), x] + Simp[a^2*((m - 1)/(m - n))   Int[(a*Sin[e + f*x])^( 
m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
 

rule 3065

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]
 

rule 3120

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]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {\sqrt {a \sin \left (f x +e \right )}\, a \sqrt {b \sec \left (f x +e \right )}\, \left (\sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \left (-\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )+2 \cos \left (f x +e \right )\right )}{2 f}\)	\(133\)

Input:

int((b*sec(f*x+e))^(1/2)*(a*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/f*(a*sin(f*x+e))^(1/2)*a*(b*sec(f*x+e))^(1/2)*((-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)*(-cot(f*x+e)-cs 
c(f*x+e))+2*cos(f*x+e))
 

Fricas [F]

\[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((b*sec(f*x+e))^(1/2)*(a*sin(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))*a*sin(f*x + e), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate((b*sec(f*x+e))**(1/2)*(a*sin(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((b*sec(f*x+e))^(1/2)*(a*sin(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*sec(f*x + e))*(a*sin(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((b*sec(f*x+e))^(1/2)*(a*sin(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate(sqrt(b*sec(f*x + e))*(a*sin(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\int {\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \] Input:

int((a*sin(e + f*x))^(3/2)*(b/cos(e + f*x))^(1/2),x)
 

Output:

int((a*sin(e + f*x))^(3/2)*(b/cos(e + f*x))^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2} \, dx=\sqrt {b}\, \sqrt {a}\, \left (\int \sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )d x \right ) a \] Input:

int((b*sec(f*x+e))^(1/2)*(a*sin(f*x+e))^(3/2),x)
 

Output:

sqrt(b)*sqrt(a)*int(sqrt(sin(e + f*x))*sqrt(sec(e + f*x))*sin(e + f*x),x)* 
a