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

Optimal. Leaf size=109 \[ \frac{2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt{c-c \sin (e+f x)}}+\frac{16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}} \]

[Out]

(64*a^2*c^5*Cos[e + f*x]^5)/(315*f*(c - c*Sin[e + f*x])^(5/2)) + (16*a^2*c^4*Cos[e + f*x]^5)/(63*f*(c - c*Sin[
e + f*x])^(3/2)) + (2*a^2*c^3*Cos[e + f*x]^5)/(9*f*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.259677, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac{2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt{c-c \sin (e+f x)}}+\frac{16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a^2*c^5*Cos[e + f*x]^5)/(315*f*(c - c*Sin[e + f*x])^(5/2)) + (16*a^2*c^4*Cos[e + f*x]^5)/(63*f*(c - c*Sin[
e + f*x])^(3/2)) + (2*a^2*c^3*Cos[e + f*x]^5)/(9*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))^2 (c-c \sin (e+f x))^{5/2} \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt{c-c \sin (e+f x)}}+\frac{1}{9} \left (8 a^2 c^3\right ) \int \frac{\cos ^4(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac{2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt{c-c \sin (e+f x)}}+\frac{1}{63} \left (32 a^2 c^4\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{64 a^2 c^5 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}}+\frac{16 a^2 c^4 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}+\frac{2 a^2 c^3 \cos ^5(e+f x)}{9 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.38052, size = 96, normalized size = 0.88 \[ -\frac{a^2 c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5 (220 \sin (e+f x)+35 \cos (2 (e+f x))-249)}{315 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-249 + 35*Cos[2*(e + f*x)] + 220*Sin[e + f*x])*Sqrt[c - c*S
in[e + f*x]])/(315*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]  time = 0.478, size = 71, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{3}{a}^{2} \left ( 35\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}-110\,\sin \left ( fx+e \right ) +107 \right ) }{315\,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))^2*(c-c*sin(f*x+e))^(5/2),x)

[Out]

-2/315*(-1+sin(f*x+e))*c^3*(1+sin(f*x+e))^3*a^2*(35*sin(f*x+e)^2-110*sin(f*x+e)+107)/cos(f*x+e)/(c-c*sin(f*x+e
))^(1/2)/f

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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 )}^{2}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.07699, size = 486, normalized size = 4.46 \begin{align*} \frac{2 \,{\left (35 \, a^{2} c^{2} \cos \left (f x + e\right )^{5} - 5 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 8 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} - 16 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 64 \, a^{2} c^{2} \cos \left (f x + e\right ) + 128 \, a^{2} c^{2} +{\left (35 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 40 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} + 48 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 64 \, a^{2} c^{2} \cos \left (f x + e\right ) + 128 \, a^{2} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{315 \,{\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))^2*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/315*(35*a^2*c^2*cos(f*x + e)^5 - 5*a^2*c^2*cos(f*x + e)^4 + 8*a^2*c^2*cos(f*x + e)^3 - 16*a^2*c^2*cos(f*x +
e)^2 + 64*a^2*c^2*cos(f*x + e) + 128*a^2*c^2 + (35*a^2*c^2*cos(f*x + e)^4 + 40*a^2*c^2*cos(f*x + e)^3 + 48*a^2
*c^2*cos(f*x + e)^2 + 64*a^2*c^2*cos(f*x + e) + 128*a^2*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*
x + e) - f*sin(f*x + e) + f)

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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((a+a*sin(f*x+e))**2*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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