\(\int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx\) [477]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {a^3 \sqrt {a \sin (e+f x)}}{12 b f \sqrt {b \sec (e+f x)}}-\frac {a (a \sin (e+f x))^{5/2}}{30 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}+\frac {a^4 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{24 b^2 f \sqrt {a \sin (e+f x)}} \] Output:

-1/12*a^3*(a*sin(f*x+e))^(1/2)/b/f/(b*sec(f*x+e))^(1/2)-1/30*a*(a*sin(f*x+ 
e))^(5/2)/b/f/(b*sec(f*x+e))^(1/2)+1/5*(a*sin(f*x+e))^(9/2)/a/b/f/(b*sec(f 
*x+e))^(1/2)+1/24*a^4*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)
 

Mathematica [C] (verified)

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

Time = 1.64 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.60 \[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {a^5 \left (-4+17 \cos (2 (e+f x))-16 \cos (4 (e+f x))+3 \cos (6 (e+f x))-20 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{480 b f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}} \] Input:

Integrate[(a*Sin[e + f*x])^(7/2)/(b*Sec[e + f*x])^(3/2),x]
 

Output:

-1/480*(a^5*(-4 + 17*Cos[2*(e + f*x)] - 16*Cos[4*(e + f*x)] + 3*Cos[6*(e + 
 f*x)] - 20*Hypergeometric2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x 
]^2)^(3/4)))/(b*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3062, 3042, 3063, 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.

\(\displaystyle \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3062

\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{7/2}dx}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{7/2}dx}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {\frac {5}{6} a^2 \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} a^2 \int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3065

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {5}{6} a^2 \left (\frac {a^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}\right )-\frac {a b (a \sin (e+f x))^{5/2}}{3 f \sqrt {b \sec (e+f x)}}}{10 b^2}+\frac {(a \sin (e+f x))^{9/2}}{5 a b f \sqrt {b \sec (e+f x)}}\)

Input:

Int[(a*Sin[e + f*x])^(7/2)/(b*Sec[e + f*x])^(3/2),x]
 

Output:

(a*Sin[e + f*x])^(9/2)/(5*a*b*f*Sqrt[b*Sec[e + f*x]]) + (-1/3*(a*b*(a*Sin[ 
e + f*x])^(5/2))/(f*Sqrt[b*Sec[e + f*x]]) + (5*a^2*(-((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]])))/6)/(1 
0*b^2)
 

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 3062

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]
 

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 = 2.64 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\sqrt {a \sin \left (f x +e \right )}\, a^{3} \left (24 \cos \left (f x +e \right )^{4}-44 \cos \left (f x +e \right )^{2}+10+\sqrt {-2 \csc \left (f x +e \right )+2 \cot \left (f x +e \right )+2}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )+1}\, \left (5 \csc \left (f x +e \right )+5 \sec \left (f x +e \right ) \csc \left (f x +e \right )\right )\right )}{120 f \sqrt {b \sec \left (f x +e \right )}\, b}\)	\(157\)

Input:

int((a*sin(f*x+e))^(7/2)/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/120/f*(a*sin(f*x+e))^(1/2)*a^3/(b*sec(f*x+e))^(1/2)/b*(24*cos(f*x+e)^4-4 
4*cos(f*x+e)^2+10+(-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)*(5*csc(f*x+e)+5*sec(f*x+e)*csc(f*x+e)))
 

Fricas [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a*sin(f*x+e))**(7/2)/(b*sec(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e))^(7/2)/(b*sec(f*x + e))^(3/2), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (a \sin \left (f x + e\right )\right )^{\frac {7}{2}}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*sin(f*x+e))^(7/2)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e))^(7/2)/(b*sec(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((a*sin(e + f*x))^(7/2)/(b/cos(e + f*x))^(3/2),x)
 

Output:

int((a*sin(e + f*x))^(7/2)/(b/cos(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a \sin (e+f x))^{7/2}}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sin \left (f x +e \right )^{3}}{\sec \left (f x +e \right )^{2}}d x \right ) a^{3}}{b^{2}} \] Input:

int((a*sin(f*x+e))^(7/2)/(b*sec(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)** 
3)/sec(e + f*x)**2,x)*a**3)/b**2