\(\int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx\) [380]

Optimal result

Integrand size = 21, antiderivative size = 67 \[ \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx=\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}} \]

[Out]

-2/3*b*sin(f*x+e)/f/(b*sec(f*x+e))^(1/2)+4/3*(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)/f

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2707, 3856, 2720} \[ \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx=\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}} \]

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^2,x]

[Out]

(4*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) - (2*b*Sin[e + f*x])/(3*f*Sqrt[b*S
ec[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 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)}{3 f \sqrt {b \sec (e+f x)}}+\frac {2}{3} \int \sqrt {b \sec (e+f x)} \, dx \\ & = -\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {1}{3} \left (2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, 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)}}{3 f}-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx=-\frac {\sqrt {b \sec (e+f x)} \left (-4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+\sin (2 (e+f x))\right )}{3 f} \]

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^2,x]

[Out]

-1/3*(Sqrt[b*Sec[e + f*x]]*(-4*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] + Sin[2*(e + f*x)]))/f

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {2 \left (2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{3 f}\)	\(147\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx=-\frac {2 \, {\left (\sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, f} \]

[In]

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

[Out]

-2/3*(sqrt(b/cos(f*x + e))*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)))/f

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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