3.132 \(\int \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=94 \[ \frac{a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt{a \sin (e+f x)+a}}-\frac{a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt{a \sin (e+f x)+a}} \]

[Out]

-(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*f*Sqrt[a + a*Sin[e + f*x]]) + (a*B*Cos[e + f*x]*(c - c
*Sin[e + f*x])^(7/2))/(4*c*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.337209, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {2971, 2738} \[ \frac{a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt{a \sin (e+f x)+a}}-\frac{a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*f*Sqrt[a + a*Sin[e + f*x]]) + (a*B*Cos[e + f*x]*(c - c
*Sin[e + f*x])^(7/2))/(4*c*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2971

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 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 \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=(A+B) \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx-\frac{B \int \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2} \, dx}{c}\\ &=-\frac{a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{a B \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 c f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.841178, size = 102, normalized size = 1.09 \[ \frac{c^2 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (16 (7 A-2 B) \sin (e+f x)-4 \cos (2 (e+f x)) (4 (A-2 B) \sin (e+f x)-12 A+9 B)+3 B \cos (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(3*B*Cos[4*(e + f*x)] + 16*(7*A - 2*B)*S
in[e + f*x] - 4*Cos[2*(e + f*x)]*(-12*A + 9*B + 4*(A - 2*B)*Sin[e + f*x])))/(96*f)

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Maple [A]  time = 0.378, size = 129, normalized size = 1.4 \begin{align*}{\frac{ \left ( 3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +4\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}-8\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+12\,A\sin \left ( fx+e \right ) -9\,B\sin \left ( fx+e \right ) -16\,A+8\,B \right ) \sin \left ( fx+e \right ) }{12\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sin \left ( fx+e \right ) -2 \right ) \cos \left ( fx+e \right ) } \left ( -c \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{5}{2}}}\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/12/f*(3*B*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)^2-8*B*cos(f*x+e)^2+12*A*sin(f*x+e)-9*B*sin(f*x+e)-16*A+8*B)
*(-c*(-1+sin(f*x+e)))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(1/2)/(cos(f*x+e)^2+2*sin(f*x+e)-2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}{\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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.71983, size = 293, normalized size = 3.12 \begin{align*} \frac{{\left (3 \, B c^{2} \cos \left (f x + e\right )^{4} + 12 \,{\left (A - B\right )} c^{2} \cos \left (f x + e\right )^{2} - 3 \,{\left (4 \, A - 3 \, B\right )} c^{2} - 4 \,{\left ({\left (A - 2 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (2 \, A - B\right )} c^{2}\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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*B*c^2*cos(f*x + e)^4 + 12*(A - B)*c^2*cos(f*x + e)^2 - 3*(4*A - 3*B)*c^2 - 4*((A - 2*B)*c^2*cos(f*x +
e)^2 - 2*(2*A - B)*c^2)*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+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)*(a+a*sin(f*x+e))**(1/2),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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError