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

Optimal. Leaf size=69 \[ \frac{2 a c^2 \cos ^3(e+f x)}{5 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.157268, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2674, 2673} \[ \frac{2 a c^2 \cos ^3(e+f x)}{5 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx &=(a c) \int \cos ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{2 a c^2 \cos ^3(e+f x)}{5 f \sqrt{c-c \sin (e+f x)}}+\frac{1}{5} \left (4 a c^2\right ) \int \frac{\cos ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{8 a c^3 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac{2 a c^2 \cos ^3(e+f x)}{5 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.336405, size = 82, normalized size = 1.19 \[ -\frac{2 a c (3 \sin (e+f x)-7) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{15 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-7 + 3*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*f*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 0.506, size = 59, normalized size = 0.9 \begin{align*}{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}a \left ( 3\,\sin \left ( fx+e \right ) -7 \right ) }{15\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(-1+sin(f*x+e))*c^2*(1+sin(f*x+e))^2*a*(3*sin(f*x+e)-7)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.07368, size = 284, normalized size = 4.12 \begin{align*} \frac{2 \,{\left (3 \, a c \cos \left (f x + e\right )^{3} - a c \cos \left (f x + e\right )^{2} + 4 \, a c \cos \left (f x + e\right ) + 8 \, a c +{\left (3 \, a c \cos \left (f x + e\right )^{2} + 4 \, a c \cos \left (f x + e\right ) + 8 \, a c\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out