3.22 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=92 \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a f \sqrt{c-c \sin (e+f x)}} \]

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*a*f*Sqrt[c - c*Sin[e + f*x]]) + (Cos[e + f*x]*(a + a*Sin[e + f
*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(5*a*f)

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Rubi [A]  time = 0.395089, 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} \[ \frac{\cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 a f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*a*f*Sqrt[c - c*Sin[e + f*x]]) + (Cos[e + f*x]*(a + a*Sin[e + f
*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(5*a*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) (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)} \, dx &=\frac{\int (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{5 a f}+\frac{2 \int (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)} \, dx}{5 a}\\ &=\frac{c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a f \sqrt{c-c \sin (e+f x)}}+\frac{\cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt{c-c \sin (e+f x)}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.493871, size = 92, normalized size = 1. \[ -\frac{a^2 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (-70 \sin (e+f x)-5 \sin (3 (e+f x))+\sin (5 (e+f x))+20 \cos (2 (e+f x))+5 \cos (4 (e+f x)))}{80 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

-(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(20*Cos[2*(e + f*x)] + 5*Cos[4*(e + f*x
)] - 70*Sin[e + f*x] - 5*Sin[3*(e + f*x)] + Sin[5*(e + f*x)]))/(80*f)

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Maple [A]  time = 0.227, size = 106, normalized size = 1.2 \begin{align*} -{\frac{\sin \left ( fx+e \right ) \left ( -2\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +6\,\sin \left ( fx+e \right ) -6 \right ) }{10\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}}\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}}} \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))^(5/2)*(c-c*sin(f*x+e))^(1/2),x)

[Out]

-1/10/f*(-c*(-1+sin(f*x+e)))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(-2*cos(f*x+e)^6+sin(f*x+e)*cos(f*x+e)^
4-2*cos(f*x+e)^4+3*cos(f*x+e)^2*sin(f*x+e)+6*sin(f*x+e)-6)/cos(f*x+e)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \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))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e)^2, x)

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Fricas [A]  time = 2.29749, size = 236, normalized size = 2.57 \begin{align*} -\frac{{\left (5 \, a^{2} \cos \left (f x + e\right )^{4} - 5 \, a^{2} + 2 \,{\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{10 \, 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))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/10*(5*a^2*cos(f*x + e)^4 - 5*a^2 + 2*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 - 4*a^2)*sin(f*x + e))*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*(a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \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))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

sage2