3.4 \(\int \cos ^2(e+f x) \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=92 \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 c f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(3*c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin[e
+ f*x]]*(c - c*Sin[e + f*x])^(3/2))/(3*c*f)

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Rubi [A]  time = 0.371708, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 c f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(3*c*f*Sqrt[a + a*Sin[e + f*x]]) - (Cos[e + f*x]*Sqrt[a + a*Sin[e
+ f*x]]*(c - c*Sin[e + f*x])^(3/2))/(3*c*f)

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int \cos ^2(e+f x) \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)} \, dx &=\frac{\int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 c f}+\frac{2 \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx}{3 c}\\ &=-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 c f \sqrt{a+a \sin (e+f x)}}-\frac{\cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 c f}\\ \end{align*}

Mathematica [A]  time = 0.169298, size = 59, normalized size = 0.64 \[ \frac{(9 \sin (e+f x)+\sin (3 (e+f x))) \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)}}{12 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(9*Sin[e + f*x] + Sin[3*(e + f*x)]))/(12*f)

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Maple [A]  time = 0.241, size = 55, normalized size = 0.6 \begin{align*}{\frac{ \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2 \right ) \sin \left ( fx+e \right ) }{3\,f\cos \left ( fx+e \right ) }\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) }\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(1/2),x)

[Out]

1/3/f*(cos(f*x+e)^2+2)*(-c*(-1+sin(f*x+e)))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(1/2)/cos(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e)^2, x)

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Fricas [A]  time = 1.72177, size = 144, normalized size = 1.57 \begin{align*} \frac{{\left (\cos \left (f x + e\right )^{2} + 2\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{3 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/3*(cos(f*x + e)^2 + 2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \sqrt{- c \left (\sin{\left (e + f x \right )} - 1\right )} \cos ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(1/2)*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*sqrt(-c*(sin(e + f*x) - 1))*cos(e + f*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e)^2, x)