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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2707, 3854, 3856, 2720} \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{21 b^2 f}+\frac {4 \sin (e+f x)}{21 b f \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(4*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(21*b^2*f) - (2*b*Sin[e + f*x])/(7*f*(b*
Sec[e + f*x])^(5/2)) + (4*Sin[e + f*x])/(21*b*f*Sqrt[b*Sec[e + f*x]])

Rule 2707

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] + Dist[(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 2720

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 3854

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
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 b \sin (e+f x)}{7 f (b \sec (e+f x))^{5/2}}+\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}}+\frac {4 \sin (e+f x)}{21 b f \sqrt {b \sec (e+f x)}}+\frac {2 \int \sqrt {b \sec (e+f x)} \, dx}{21 b^2} \\ & = -\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)}}+\frac {\left (2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{21 b^2} \\ & = \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\sec ^2(e+f x) \left (16 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+2 \sin (2 (e+f x))-3 \sin (4 (e+f x))\right )}{84 f (b \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(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))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.63

method	result	size
default	\(-\frac {2 \left (2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right )+2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), i\right ) \sec \left (f x +e \right )+3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right )\right )}{21 f \sqrt {b \sec \left (f x +e \right )}\, b}\)	\(160\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

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

[In]

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

[Out]

-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)*weierstrassPI
nverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) - I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*s
in(f*x + e)))/(b^2*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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