3.310 \(\int (a+a \sin (e+f x))^3 \sqrt{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=36 \[ \frac{2 a^3 c^4 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

(2*a^3*c^4*Cos[e + f*x]^7)/(7*f*(c - c*Sin[e + f*x])^(7/2))

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Rubi [A]  time = 0.127129, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {2736, 2673} \[ \frac{2 a^3 c^4 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*c^4*Cos[e + f*x]^7)/(7*f*(c - c*Sin[e + f*x])^(7/2))

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 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))^3 \sqrt{c-c \sin (e+f x)} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac{2 a^3 c^4 \cos ^7(e+f x)}{7 f (c-c \sin (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 0.373951, size = 73, normalized size = 2.03 \[ \frac{2 a^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^7}{7 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.383, size = 49, normalized size = 1.4 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ) c \left ( 1+\sin \left ( fx+e \right ) \right ) ^{4}{a}^{3}}{7\,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))^3*(c-c*sin(f*x+e))^(1/2),x)

[Out]

-2/7*(-1+sin(f*x+e))*c*(1+sin(f*x+e))^4*a^3/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 )}^{3} \sqrt{-c \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.05076, size = 342, normalized size = 9.5 \begin{align*} \frac{2 \,{\left (a^{3} \cos \left (f x + e\right )^{4} - 3 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} + 4 \, a^{3} \cos \left (f x + e\right ) + 8 \, a^{3} -{\left (a^{3} \cos \left (f x + e\right )^{3} + 4 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) - 8 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{7 \,{\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))^3*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/7*(a^3*cos(f*x + e)^4 - 3*a^3*cos(f*x + e)^3 - 8*a^3*cos(f*x + e)^2 + 4*a^3*cos(f*x + e) + 8*a^3 - (a^3*cos(
f*x + e)^3 + 4*a^3*cos(f*x + e)^2 - 4*a^3*cos(f*x + e) - 8*a^3)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int 3 \sqrt{- c \sin{\left (e + f x \right )} + c} \sin{\left (e + f x \right )}\, dx + \int 3 \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx + \int \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\, dx + \int \sqrt{- c \sin{\left (e + f x \right )} + c}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(3*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral(3*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x
)**2, x) + Integral(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3, x) + Integral(sqrt(-c*sin(e + f*x) + c), x))

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

[Out]

Timed out