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

Optimal. Leaf size=67 \[ \frac{4 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f}-\frac{2 b \sin (e+f x)}{3 f \sqrt{b \sec (e+f x)}} \]

[Out]

(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]])

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Rubi [A]  time = 0.0582376, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2627, 3771, 2641} \[ \frac{4 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f}-\frac{2 b \sin (e+f x)}{3 f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[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 2627

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 3771

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]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{b \sec (e+f x)} \sin ^2(e+f x) \, dx &=-\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)} F\left (\left .\frac{1}{2} (e+f x)\right |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]  time = 0.0939842, size = 51, normalized size = 0.76 \[ -\frac{\sqrt{b \sec (e+f x)} \left (\sin (2 (e+f x))-4 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.143, size = 123, normalized size = 1.8 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( 2\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}-\cos \left ( fx+e \right ) \right ) \sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{b \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(cos(f*x + e)^2 - 1)*sqrt(b*sec(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)