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

Optimal. Leaf size=137 \[ \frac{256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a c^4 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}+\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[Out]

(256*a*c^5*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*c^4*Cos[e + f*x]^3)/(105*f*Sqrt[c - c*Si
n[e + f*x]]) + (8*a*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) + (2*a*c^2*Cos[e + f*x]^3*(c - c*Sin[e
+ f*x])^(3/2))/(9*f)

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Rubi [A]  time = 0.294873, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2674, 2673} \[ \frac{256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a c^4 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}+\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(256*a*c^5*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (64*a*c^4*Cos[e + f*x]^3)/(105*f*Sqrt[c - c*Si
n[e + f*x]]) + (8*a*c^3*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) + (2*a*c^2*Cos[e + f*x]^3*(c - c*Sin[e
+ f*x])^(3/2))/(9*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)) (c-c \sin (e+f x))^{7/2} \, dx &=(a c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx\\ &=\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{3} \left (4 a c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{8 a c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}+\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{21} \left (32 a c^3\right ) \int \cos ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{64 a c^4 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}+\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{105} \left (128 a c^4\right ) \int \frac{\cos ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{256 a c^5 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{64 a c^4 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{8 a c^3 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}+\frac{2 a c^2 \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\\ \end{align*}

Mathematica [A]  time = 0.840671, size = 104, normalized size = 0.76 \[ \frac{a c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 (-1389 \sin (e+f x)+35 \sin (3 (e+f x))-330 \cos (2 (e+f x))+1606)}{630 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(1606 - 330*Cos[2*(e + f*x)] - 1389*Si
n[e + f*x] + 35*Sin[3*(e + f*x)]))/(630*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]  time = 0.573, size = 79, 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 ) ^{2}a \left ( 35\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}-165\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+321\,\sin \left ( fx+e \right ) -319 \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))*(c-c*sin(f*x+e))^(7/2),x)

[Out]

2/315*(-1+sin(f*x+e))*c^4*(1+sin(f*x+e))^2*a*(35*sin(f*x+e)^3-165*sin(f*x+e)^2+321*sin(f*x+e)-319)/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 )}{\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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.1593, size = 466, normalized size = 3.4 \begin{align*} -\frac{2 \,{\left (35 \, a c^{3} \cos \left (f x + e\right )^{5} - 95 \, a c^{3} \cos \left (f x + e\right )^{4} - 226 \, a c^{3} \cos \left (f x + e\right )^{3} + 32 \, a c^{3} \cos \left (f x + e\right )^{2} - 128 \, a c^{3} \cos \left (f x + e\right ) - 256 \, a c^{3} +{\left (35 \, a c^{3} \cos \left (f x + e\right )^{4} + 130 \, a c^{3} \cos \left (f x + e\right )^{3} - 96 \, a c^{3} \cos \left (f x + e\right )^{2} - 128 \, a c^{3} \cos \left (f x + e\right ) - 256 \, a c^{3}\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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-2/315*(35*a*c^3*cos(f*x + e)^5 - 95*a*c^3*cos(f*x + e)^4 - 226*a*c^3*cos(f*x + e)^3 + 32*a*c^3*cos(f*x + e)^2
- 128*a*c^3*cos(f*x + e) - 256*a*c^3 + (35*a*c^3*cos(f*x + e)^4 + 130*a*c^3*cos(f*x + e)^3 - 96*a*c^3*cos(f*x
+ e)^2 - 128*a*c^3*cos(f*x + e) - 256*a*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))*(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 )}{\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))*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

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