3.3 \(\int \cos ^2(e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=92 \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(6*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])^(5/2))/(4*c*f)

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Rubi [A]  time = 0.390051, 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} \[ -\frac{\cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 c f}-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(6*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])^(5/2))/(4*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)} (c-c \sin (e+f x))^{3/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 c f}+\frac{\int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx}{2 c}\\ &=-\frac{a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 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))^{5/2}}{4 c f}\\ \end{align*}

Mathematica [A]  time = 0.403659, size = 83, normalized size = 0.9 \[ \frac{c \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (8 (9 \sin (e+f x)+\sin (3 (e+f x)))+12 \cos (2 (e+f x))+3 \cos (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(12*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)]
+ 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/(96*f)

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Maple [A]  time = 0.243, size = 90, normalized size = 1. \begin{align*}{\frac{\sin \left ( fx+e \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\,\sin \left ( fx+e \right ) +5 \right ) }{12\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}\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*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/12/f*(-c*(-1+sin(f*x+e)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(1/2)*(3*cos(f*x+e)^4+cos(f*x+e)^2*sin(f*x+e)+
4*cos(f*x+e)^2+5*sin(f*x+e)+5)/cos(f*x+e)^3

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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}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}} \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*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.68568, size = 193, normalized size = 2.1 \begin{align*} \frac{{\left (3 \, c \cos \left (f x + e\right )^{4} + 4 \,{\left (c \cos \left (f x + e\right )^{2} + 2 \, c\right )} \sin \left (f x + e\right ) - 3 \, c\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{12 \, 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*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*c*cos(f*x + e)^4 + 4*(c*cos(f*x + e)^2 + 2*c)*sin(f*x + e) - 3*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin
(f*x + e) + c)/(f*cos(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError