3.147 \(\int \frac{(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{2 g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a \sin (e+f x)+a)^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.842057, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2852, 2842, 2640, 2639} \[ -\frac{2 g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a \sin (e+f x)+a)^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2852

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

Rule 2842

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(g*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), In
t[(g*Cos[e + f*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2
, 0]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}} \, dx &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}+\frac{\int \frac{(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}} \, dx}{5 a}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{\int \frac{(g \cos (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx}{5 a^2}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{(g \cos (e+f x)) \int \sqrt{g \cos (e+f x)} \, dx}{5 a^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{\left (g \sqrt{\cos (e+f x)} \sqrt{g \cos (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{5 a^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{2 g \sqrt{\cos (e+f x)} \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 a^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.59954, size = 189, normalized size = 1.06 \[ -\frac{(g \cos (e+f x))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sqrt{\cos (e+f x)} \left (-4 \sin ^3\left (\frac{1}{2} (e+f x)\right )+3 \cos \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{3}{2} (e+f x)\right )\right )+2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{5 f \cos ^{\frac{3}{2}}(e+f x) (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((g*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(2*Elli
pticE[(e + f*x)/2, 2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + Sqrt[Cos[e + f*x]]*(3*Cos[(e + f*x)/2] + Cos[(
3*(e + f*x))/2] - 4*Sin[(e + f*x)/2]^3)))/(5*f*Cos[e + f*x]^(3/2)*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[
e + f*x]])

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Maple [C]  time = 0.341, size = 778, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x)

[Out]

2/5/f*(g*cos(f*x+e))^(3/2)*(sin(f*x+e)*cos(f*x+e)-sin(f*x+e)+cos(f*x+e)-1)*(-I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f
*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^4*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+I*(1/(cos(f*x+e)+1))^(1/2)*
(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^4*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)+I*EllipticE(I*(-1+cos
(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)^2*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-I*El
lipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)^2*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+
e)+1))^(1/2)+2*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^2*EllipticE(I*(-1+cos(f
*x+e))/sin(f*x+e),I)-2*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^2*EllipticF(I*(
-1+cos(f*x+e))/sin(f*x+e),I)-I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(
f*x+e))/sin(f*x+e),I)*sin(f*x+e)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+
cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)+I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)
/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)^2*sin(f*x+e)-cos(f*x+e)^2+sin(f*x+e)+2*cos(f*x+e)-1)*(cos(f*x+e)^2+2*cos(f*x
+e)+1)/(a*(1+sin(f*x+e)))^(5/2)/(-c*(-1+sin(f*x+e)))^(1/2)/sin(f*x+e)^5/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{g \cos \left (f x + e\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} g}{a^{3} c \cos \left (f x + e\right )^{3} - 2 \, a^{3} c \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c \cos \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*g/(a^3*c*cos(f*x + e)^3 - 2*
a^3*c*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*cos(f*x + e)), x)

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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((g*cos(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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