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

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

[Out]

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

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Rubi [A]  time = 0.333082, 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} \[ -\frac{c (A-B) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{B c \cos (e+f x)}{a f (a \sin (e+f x)+a)^{3/2} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A - B)*c*Cos[e + f*x])/(2*f*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]) - (B*c*Cos[e + f*x])/(a*f*
(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*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 \frac{(A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx &=\frac{B \int \frac{\sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a}-(-A+B) \int \frac{\sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\\ &=-\frac{(A-B) c \cos (e+f x)}{2 f (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{B c \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.505045, size = 99, normalized size = 1.05 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (A+2 B \sin (e+f x)+B)}{2 a^3 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a*(1 + Sin[e + f*x])]*(A + B + 2*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(2*a^3*f*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [A]  time = 0.32, size = 135, normalized size = 1.4 \begin{align*} -{\frac{ \left ( A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+A\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+B\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +2\,A\cos \left ( fx+e \right ) -3\,A\sin \left ( fx+e \right ) -B\sin \left ( fx+e \right ) -3\,A-B \right ) \sin \left ( fx+e \right ) }{2\,f \left ( -1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) } \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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.9835, size = 223, normalized size = 2.37 \begin{align*} \frac{{\left (2 \, B \sin \left (f x + e\right ) + A + B\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{2 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\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))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2*(2*B*sin(f*x + e) + A + B)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^
3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*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))**(1/2)/(a+a*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 \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\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))^(1/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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