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

Optimal. Leaf size=92 \[ \frac{a (A+B) \cos (e+f x)}{2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{a B \cos (e+f x)}{c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.344732, antiderivative size = 92, 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} \[ \frac{a (A+B) \cos (e+f x)}{2 f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac{a B \cos (e+f x)}{c f \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \]

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

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

Mathematica [A]  time = 0.550104, size = 101, normalized size = 1.1 \[ \frac{\sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (A+2 B \sin (e+f x)-B)}{2 c^3 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

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]

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

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Maple [A]  time = 0.352, size = 137, normalized size = 1.5 \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{a \left ( 1+\sin \left ( fx+e \right ) \right ) } \left ( -c \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))*(a+a*sin(f*x+e))^(1/2)/(c-c*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)*(a*(1+sin(f*x+e)))^(1/2)/(-1+cos(f*x+e)-sin(f*x+e))/(-c*(-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{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))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),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 = 1.70649, size = 224, normalized size = 2.43 \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 (c^{3} f \cos \left (f x + e\right )^{3} + 2 \, c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{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))*(a+a*sin(f*x+e))^(1/2)/(c-c*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)/(c^3*f*cos(f*x + e)^3 + 2*c
^3*f*cos(f*x + e)*sin(f*x + e) - 2*c^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))*(a+a*sin(f*x+e))**(1/2)/(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 \frac{{\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))*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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