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

Optimal. Leaf size=140 \[ -\frac{2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac{a \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 c f} \]

[Out]

(-2*a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(5/2))/(5*c*f)

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Rubi [A]  time = 0.520556, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac{2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt{a \sin (e+f x)+a}}-\frac{\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac{a \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*c*f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(5*c*f) - (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e +
f*x])^(5/2))/(5*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) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx &=\frac{\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}+\frac{4 \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx}{5 c}\\ &=-\frac{a \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}+\frac{(2 a) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx}{5 c}\\ &=-\frac{2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt{a+a \sin (e+f x)}}-\frac{a \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac{\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}\\ \end{align*}

Mathematica [A]  time = 0.587113, size = 82, normalized size = 0.59 \[ -\frac{c (\sin (e+f x)-1) (150 \sin (e+f x)+25 \sin (3 (e+f x))+3 \sin (5 (e+f x))) \sec ^3(e+f x) (a (\sin (e+f x)+1))^{3/2} \sqrt{c-c \sin (e+f x)}}{240 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Sec[e + f*x]^3*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(150*Sin[e + f*x]
+ 25*Sin[3*(e + f*x)] + 3*Sin[5*(e + f*x)]))/(240*f)

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Maple [A]  time = 0.183, size = 67, normalized size = 0.5 \begin{align*}{\frac{ \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8 \right ) \sin \left ( fx+e \right ) }{15\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{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))^(3/2)*(c-c*sin(f*x+e))^(3/2),x)

[Out]

1/15/f*(3*cos(f*x+e)^4+4*cos(f*x+e)^2+8)*(-c*(-1+sin(f*x+e)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)/cos(f*
x+e)^3

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

[Out]

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

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Fricas [A]  time = 1.68888, size = 190, normalized size = 1.36 \begin{align*} \frac{{\left (3 \, a c \cos \left (f x + e\right )^{4} + 4 \, a c \cos \left (f x + e\right )^{2} + 8 \, a c\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 )}{15 \, 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))^(3/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*a*c*cos(f*x + e)^4 + 4*a*c*cos(f*x + e)^2 + 8*a*c)*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(-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))**(3/2)*(c-c*sin(f*x+e))**(3/2),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*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError