\(\int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) [432]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{21 b^2 f}-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}+\frac {4 \sin (e+f x)}{21 b f \sqrt {b \sec (e+f x)}} \] Output:

4/21*cos(f*x+e)^(1/2)*InverseJacobiAM(1/2*f*x+1/2*e,2^(1/2))*(b*sec(f*x+e) 
)^(1/2)/b^2/f-2/7*b*sin(f*x+e)/f/(b*sec(f*x+e))^(5/2)+4/21*sin(f*x+e)/b/f/ 
(b*sec(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (16 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+2 \sin (2 (e+f x))-3 \sin (4 (e+f x))\right )}{84 f (b \sec (e+f x))^{3/2}} \] Input:

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

Output:

(Sec[e + f*x]^2*(16*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + 2*Sin[2 
*(e + f*x)] - 3*Sin[4*(e + f*x)]))/(84*f*(b*Sec[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3107, 3042, 4256, 3042, 4258, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc (e+f x)^2 (b \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3107

\(\displaystyle \frac {2}{7} \int \frac {1}{(b \sec (e+f x))^{3/2}}dx-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \int \frac {1}{\left (b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {2}{7} \left (\frac {\int \sqrt {b \sec (e+f x)}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \left (\frac {\int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2}{7} \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{7} \left (\frac {\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2}{7} \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 b^2 f}+\frac {2 \sin (e+f x)}{3 b f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}\)

Input:

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

Output:

(-2*b*Sin[e + f*x])/(7*f*(b*Sec[e + f*x])^(5/2)) + (2*((2*Sqrt[Cos[e + f*x 
]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*b^2*f) + (2*Sin[e + 
f*x])/(3*b*f*Sqrt[b*Sec[e + f*x]])))/7
 

Defintions of rubi rules used

rule 3042

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

rule 3107

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

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]
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right ) \left (3 \cos \left (f x +e \right )^{2}-2\right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (-2-2 \sec \left (f x +e \right )\right )\right )}{21 f \sqrt {b \sec \left (f x +e \right )}\, b}\)	\(104\)

Input:

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

Output:

-2/21/f/(b*sec(f*x+e))^(1/2)/b*(sin(f*x+e)*(3*cos(f*x+e)^2-2)+I*(1/(cos(f* 
x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cot(f*x+e)-c 
sc(f*x+e)),I)*(-2-2*sec(f*x+e)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left ({\left (3 \, \cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{21 \, b^{2} f} \] Input:

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

Output:

-2/21*((3*cos(f*x + e)^3 - 2*cos(f*x + e))*sqrt(b/cos(f*x + e))*sin(f*x + 
e) + I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x 
 + e)) - I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin 
(f*x + e)))/(b^2*f)
 

Sympy [F]

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

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

Output:

Integral(sin(e + f*x)**2/(b*sec(e + f*x))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(b)*int((sqrt(sec(e + f*x))*sin(e + f*x)**2)/sec(e + f*x)**2,x))/b**2