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

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

[Out]

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

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Rubi [A]  time = 0.859745, 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} \[ -\frac{2 g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{2 (g \cos (e+f x))^{5/2}}{5 c f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.6232, size = 204, normalized size = 1.14 \[ \frac{(g \cos (e+f x))^{3/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sqrt{\cos (e+f x)} \left (4 \sin ^3\left (\frac{1}{2} (e+f x)\right )+3 \cos \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{3}{2} (e+f x)\right )\right )-2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{5 c^2 f (\sin (e+f x)-1)^2 \cos ^{\frac{3}{2}}(e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.346, 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)/(c-c*sin(f*x+e))^(5/2)/(a+a*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*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)*cos(f*x+e)^4*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+I*sin(f*x+e)*cos(f*x+e
)^2*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)-I*sin
(f*x+e)*cos(f*x+e)^2*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)-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)+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)-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*sin(f*x+e)*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)-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)-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)))^(1/2)/(-c*(-1+sin(f*x+e)))^(5/2)/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}}}{\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((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(3/2)/(sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(5/2)), 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 c^{3} \cos \left (f x + e\right )^{3} + 2 \, a c^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} \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)/(c-c*sin(f*x+e))^(5/2)/(a+a*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*c^3*cos(f*x + e)^3 + 2*
a*c^3*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*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)/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}}{\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((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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