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

Optimal. Leaf size=89 \[ \frac{c^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}}{4 f} \]

[Out]

(c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*f*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e +
f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(4*f)

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Rubi [A]  time = 0.17159, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2740, 2738} \[ \frac{c^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*f*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e +
f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(4*f)

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx &=\frac{c \cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}{4 f}+\frac{1}{2} c \int (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{c^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f \sqrt{c-c \sin (e+f x)}}+\frac{c \cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.664354, size = 133, normalized size = 1.49 \[ -\frac{c (\sin (e+f x)-1) (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)} (8 (9 \sin (e+f x)+\sin (3 (e+f x)))-12 \cos (2 (e+f x))-3 \cos (4 (e+f x)))}{96 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(-12*Cos[2*(e + f*x)] - 3*Cos[4*
(e + f*x)] + 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/(96*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.141, size = 90, normalized size = 1. \begin{align*} -{\frac{\sin \left ( fx+e \right ) \left ( -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\,\sin \left ( fx+e \right ) -5 \right ) }{12\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}} \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/f*(-c*(-1+sin(f*x+e)))^(3/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)*(-3*cos(f*x+e)^4+cos(f*x+e)^2*sin(f*x+e
)-4*cos(f*x+e)^2+5*sin(f*x+e)-5)/cos(f*x+e)^5

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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 )}^{\frac{5}{2}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.088, size = 216, normalized size = 2.43 \begin{align*} -\frac{{\left (3 \, a^{2} c \cos \left (f x + e\right )^{4} - 3 \, a^{2} c - 4 \,{\left (a^{2} c \cos \left (f x + e\right )^{2} + 2 \, a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a^2*c*cos(f*x + e)^4 - 3*a^2*c - 4*(a^2*c*cos(f*x + e)^2 + 2*a^2*c)*sin(f*x + e))*sqrt(a*sin(f*x + e)
+ a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError