3.308 \(\int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=109 \[ \frac{2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}} \]

[Out]

(64*a^3*c^6*Cos[e + f*x]^7)/(693*f*(c - c*Sin[e + f*x])^(7/2)) + (16*a^3*c^5*Cos[e + f*x]^7)/(99*f*(c - c*Sin[
e + f*x])^(5/2)) + (2*a^3*c^4*Cos[e + f*x]^7)/(11*f*(c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.268288, 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} \[ \frac{2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac{64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(64*a^3*c^6*Cos[e + f*x]^7)/(693*f*(c - c*Sin[e + f*x])^(7/2)) + (16*a^3*c^5*Cos[e + f*x]^7)/(99*f*(c - c*Sin[
e + f*x])^(5/2)) + (2*a^3*c^4*Cos[e + f*x]^7)/(11*f*(c - c*Sin[e + f*x])^(3/2))

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 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))^3 (c-c \sin (e+f x))^{5/2} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac{1}{11} \left (8 a^3 c^4\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac{2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}+\frac{1}{99} \left (32 a^3 c^5\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac{64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}}+\frac{16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac{2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [B]  time = 6.4651, size = 1105, normalized size = 10.14 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(5*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (5*Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e +
 f*x])^(5/2))/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (Cos[(5
*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(16*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^
5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (5*Cos[(7*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x
])^(5/2))/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (Cos[(9*(e
 + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(144*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (Cos[(11*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^
(5/2))/(176*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*Sin[(e + f
*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e
 + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(3*(e + f*x))/2])
/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((a + a*Sin[e + f*x]
)^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(5*(e + f*x))/2])/(16*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2])^6) + (5*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(7*(e + f*x))/2])/(11
2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((a + a*Sin[e + f*x])^3
*(c - c*Sin[e + f*x])^(5/2)*Sin[(9*(e + f*x))/2])/(144*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x
)/2] + Sin[(e + f*x)/2])^6) + ((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)*Sin[(11*(e + f*x))/2])/(176*f
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6)

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Maple [A]  time = 0.525, 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 ) ^{4}{a}^{3} \left ( 63\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}-182\,\sin \left ( fx+e \right ) +151 \right ) }{693\,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))^(5/2),x)

[Out]

-2/693*(-1+sin(f*x+e))*c^3*(1+sin(f*x+e))^4*a^3*(63*sin(f*x+e)^2-182*sin(f*x+e)+151)/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}{\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))^3*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3*(-c*sin(f*x + e) + c)^(5/2), x)

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Fricas [B]  time = 1.09341, size = 567, normalized size = 5.2 \begin{align*} -\frac{2 \,{\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{6} - 7 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 10 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} - 16 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 32 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} - 128 \, a^{3} c^{2} \cos \left (f x + e\right ) - 256 \, a^{3} c^{2} -{\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 70 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} + 80 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 96 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} + 128 \, a^{3} c^{2} \cos \left (f x + e\right ) + 256 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{693 \,{\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))^(5/2),x, algorithm="fricas")

[Out]

-2/693*(63*a^3*c^2*cos(f*x + e)^6 - 7*a^3*c^2*cos(f*x + e)^5 + 10*a^3*c^2*cos(f*x + e)^4 - 16*a^3*c^2*cos(f*x
+ e)^3 + 32*a^3*c^2*cos(f*x + e)^2 - 128*a^3*c^2*cos(f*x + e) - 256*a^3*c^2 - (63*a^3*c^2*cos(f*x + e)^5 + 70*
a^3*c^2*cos(f*x + e)^4 + 80*a^3*c^2*cos(f*x + e)^3 + 96*a^3*c^2*cos(f*x + e)^2 + 128*a^3*c^2*cos(f*x + e) + 25
6*a^3*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))**3*(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 )}^{3}{\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))^3*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^3*(-c*sin(f*x + e) + c)^(5/2), x)