3.569 \(\int \frac{\sqrt{a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{4 a \cos (e+f x)}{3 f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a \cos (e+f x)}{3 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.192069, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2772, 2771} \[ -\frac{4 a \cos (e+f x)}{3 f (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{2 a \cos (e+f x)}{3 f (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2771

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[(-2*b^2*Cos[e + f*x])/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.273571, size = 100, normalized size = 1.05 \[ -\frac{2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (3 c+2 d \sin (e+f x)+d)}{3 f (c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.215, size = 222, normalized size = 2.3 \begin{align*}{\frac{4\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{3}+2\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}c{d}^{2}+2\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{3}+8\,{c}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}d+2\,c \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{3}+6\,{c}^{3}\sin \left ( fx+e \right ) +10\,{c}^{2}d\sin \left ( fx+e \right ) +2\,\sin \left ( fx+e \right ){d}^{2}c-2\,{d}^{3}\sin \left ( fx+e \right ) -6\,{c}^{3}-10\,{c}^{2}d-2\,c{d}^{2}+2\,{d}^{3}}{3\,f \left ( c+d \right ) ^{2}\cos \left ( fx+e \right ) \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}+{c}^{2}-{d}^{2} \right ) ^{2}}\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{c+d\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))^(1/2)/(c+d*sin(f*x+e))^(5/2),x)

[Out]

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

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Maxima [B]  time = 1.95327, size = 459, normalized size = 4.83 \begin{align*} -\frac{2 \,{\left ({\left (3 \, c^{2} + c d\right )} \sqrt{a} - \frac{{\left (3 \, c^{2} - 9 \, c d - 2 \, d^{2}\right )} \sqrt{a} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \,{\left (3 \, c^{2} - 4 \, c d + 3 \, d^{2}\right )} \sqrt{a} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \,{\left (3 \, c^{2} - 4 \, c d + 3 \, d^{2}\right )} \sqrt{a} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{{\left (3 \, c^{2} - 9 \, c d - 2 \, d^{2}\right )} \sqrt{a} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{{\left (3 \, c^{2} + c d\right )} \sqrt{a} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{2}}{3 \,{\left (c^{2} + 2 \, c d + d^{2} + \frac{2 \,{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}{\left (c + \frac{2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac{5}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/3*((3*c^2 + c*d)*sqrt(a) - (3*c^2 - 9*c*d - 2*d^2)*sqrt(a)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(3*c^2 - 4*c
*d + 3*d^2)*sqrt(a)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*(3*c^2 - 4*c*d + 3*d^2)*sqrt(a)*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + (3*c^2 - 9*c*d - 2*d^2)*sqrt(a)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (3*c^2 + c*d)*sqrt(a
)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^2/((c^2 + 2*c*d + d^2 + 2*(c^
2 + 2*c*d + d^2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (c^2 + 2*c*d + d^2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)^(5/2)*f)

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Fricas [B]  time = 2.36704, size = 703, normalized size = 7.4 \begin{align*} \frac{2 \,{\left (2 \, d \cos \left (f x + e\right )^{2} +{\left (3 \, c + d\right )} \cos \left (f x + e\right ) +{\left (2 \, d \cos \left (f x + e\right ) - 3 \, c + d\right )} \sin \left (f x + e\right ) + 3 \, c - d\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \,{\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{3} +{\left (2 \, c^{3} d + 5 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} -{\left (c^{4} + 2 \, c^{3} d + 2 \, c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right ) -{\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f +{\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - 2 \,{\left (c^{3} d + 2 \, c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) -{\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/3*(2*d*cos(f*x + e)^2 + (3*c + d)*cos(f*x + e) + (2*d*cos(f*x + e) - 3*c + d)*sin(f*x + e) + 3*c - d)*sqrt(a
*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2
+ 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e) - (c^4 + 4*c^3
*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x + e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3
)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin(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))**(1/2)/(c+d*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{\sqrt{a \sin \left (f x + e\right ) + a}}{{\left (d \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))^(1/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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