3.298 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=145 \[ \frac{256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac{8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f} \]

[Out]

(256*a^2*c^6*Cos[e + f*x]^5)/(1155*f*(c - c*Sin[e + f*x])^(5/2)) + (64*a^2*c^5*Cos[e + f*x]^5)/(231*f*(c - c*S
in[e + f*x])^(3/2)) + (8*a^2*c^4*Cos[e + f*x]^5)/(33*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a^2*c^3*Cos[e + f*x]^5*S
qrt[c - c*Sin[e + f*x]])/(11*f)

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Rubi [A]  time = 0.327226, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac{256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac{8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(256*a^2*c^6*Cos[e + f*x]^5)/(1155*f*(c - c*Sin[e + f*x])^(5/2)) + (64*a^2*c^5*Cos[e + f*x]^5)/(231*f*(c - c*S
in[e + f*x])^(3/2)) + (8*a^2*c^4*Cos[e + f*x]^5)/(33*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a^2*c^3*Cos[e + f*x]^5*S
qrt[c - c*Sin[e + f*x]])/(11*f)

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))^{7/2} \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{11} \left (12 a^2 c^3\right ) \int \cos ^4(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{33} \left (32 a^2 c^4\right ) \int \frac{\cos ^4(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac{8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}+\frac{1}{231} \left (128 a^2 c^5\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac{8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a^2 c^3 \cos ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{11 f}\\ \end{align*}

Mathematica [B]  time = 6.43329, size = 1105, normalized size = 7.62 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(7*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) - (Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f
*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*Cos[(5
*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(7*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])
^(7/2))/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(9*(e +
f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(48*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(11*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/
2))/(176*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (7*Sin[(e + f*x)
/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^4) + ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(3*(e + f*x))/2])/(8*f
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*(a + a*Sin[e + f*x])^2
*(c - c*Sin[e + f*x])^(7/2)*Sin[(5*(e + f*x))/2])/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^4) - ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(7*(e + f*x))/2])/(112*f*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + ((a + a*Sin[e + f*x])^2*(c -
c*Sin[e + f*x])^(7/2)*Sin[(9*(e + f*x))/2])/(48*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^4) - ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)*Sin[(11*(e + f*x))/2])/(176*f*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)

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Maple [A]  time = 0.632, size = 81, normalized size = 0.6 \begin{align*}{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{4} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{3}{a}^{2} \left ( 105\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}-455\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+755\,\sin \left ( fx+e \right ) -533 \right ) }{1155\,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))^(7/2),x)

[Out]

2/1155*(-1+sin(f*x+e))*c^4*(1+sin(f*x+e))^3*a^2*(105*sin(f*x+e)^3-455*sin(f*x+e)^2+755*sin(f*x+e)-533)/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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2*(-c*sin(f*x + e) + c)^(7/2), x)

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Fricas [A]  time = 1.09928, size = 576, normalized size = 3.97 \begin{align*} \frac{2 \,{\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 245 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 20 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 32 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 64 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 256 \, a^{2} c^{3} \cos \left (f x + e\right ) + 512 \, a^{2} c^{3} -{\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 140 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} - 160 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 192 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 256 \, a^{2} c^{3} \cos \left (f x + e\right ) - 512 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{1155 \,{\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))^(7/2),x, algorithm="fricas")

[Out]

2/1155*(105*a^2*c^3*cos(f*x + e)^6 + 245*a^2*c^3*cos(f*x + e)^5 - 20*a^2*c^3*cos(f*x + e)^4 + 32*a^2*c^3*cos(f
*x + e)^3 - 64*a^2*c^3*cos(f*x + e)^2 + 256*a^2*c^3*cos(f*x + e) + 512*a^2*c^3 - (105*a^2*c^3*cos(f*x + e)^5 -
140*a^2*c^3*cos(f*x + e)^4 - 160*a^2*c^3*cos(f*x + e)^3 - 192*a^2*c^3*cos(f*x + e)^2 - 256*a^2*c^3*cos(f*x +
e) - 512*a^2*c^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(-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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2*(-c*sin(f*x + e) + c)^(7/2), x)