3.150 \(\int \frac{(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=350 \[ \frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac{14 g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{15 a^2 c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}} \]

[Out]

(-2*(g*Cos[e + f*x])^(5/2))/(5*f*g*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2)) - (14*(g*Cos[e + f*x
])^(5/2))/(5*a*f*g*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2)) + (14*(g*Cos[e + f*x])^(5/2))/(9*a^2
*f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2)) + (14*(g*Cos[e + f*x])^(5/2))/(15*a^2*c*f*g*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + (14*(g*Cos[e + f*x])^(5/2))/(15*a^2*c^2*f*g*Sqrt[a + a*Sin[e +
f*x]]*(c - c*Sin[e + f*x])^(3/2)) - (14*g*Sqrt[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(
15*a^2*c^3*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.79434, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2852, 2842, 2640, 2639} \[ \frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac{14 g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{15 a^2 c^3 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(3/2)/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2)),x]

[Out]

(-2*(g*Cos[e + f*x])^(5/2))/(5*f*g*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2)) - (14*(g*Cos[e + f*x
])^(5/2))/(5*a*f*g*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2)) + (14*(g*Cos[e + f*x])^(5/2))/(9*a^2
*f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2)) + (14*(g*Cos[e + f*x])^(5/2))/(15*a^2*c*f*g*Sqrt[a +
a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + (14*(g*Cos[e + f*x])^(5/2))/(15*a^2*c^2*f*g*Sqrt[a + a*Sin[e +
f*x]]*(c - c*Sin[e + f*x])^(3/2)) - (14*g*Sqrt[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(
15*a^2*c^3*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} (c-c \sin (e+f x))^{7/2}} \, dx &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}+\frac{7 \int \frac{(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}} \, dx}{5 a}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{7 \int \frac{(g \cos (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \, dx}{a^2}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{7 \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}{3 a^2 c}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{7 \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}{15 a^2 c^2}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac{7 \int \frac{(g \cos (e+f x))^{3/2}}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx}{15 a^2 c^3}\\ &=-\frac{2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac{(7 g \cos (e+f x)) \int \sqrt{g \cos (e+f x)} \, dx}{15 a^2 c^3 \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} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac{\left (7 g \sqrt{\cos (e+f x)} \sqrt{g \cos (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{15 a^2 c^3 \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} (c-c \sin (e+f x))^{7/2}}-\frac{14 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{9 a^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac{14 (g \cos (e+f x))^{5/2}}{15 a^2 c^2 f g \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac{14 g \sqrt{\cos (e+f x)} \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 a^2 c^3 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.16182, size = 171, normalized size = 0.49 \[ -\frac{\sqrt{\cos (e+f x)} (g \cos (e+f x))^{3/2} \left (\sqrt{\cos (e+f x)} (98 \sin (e+f x)+42 \sin (3 (e+f x))+28 \cos (2 (e+f x))+21 \cos (4 (e+f x))-9)+42 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \left (-3 \cos (e+f x)-\cos (3 (e+f x))+4 \sin (e+f x) \cos ^3(e+f x)\right )\right )}{180 c^3 f (\sin (e+f x)-1)^3 (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^(3/2)/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2)),x]

[Out]

-(Sqrt[Cos[e + f*x]]*(g*Cos[e + f*x])^(3/2)*(42*EllipticE[(e + f*x)/2, 2]*(-3*Cos[e + f*x] - Cos[3*(e + f*x)]
+ 4*Cos[e + f*x]^3*Sin[e + f*x]) + Sqrt[Cos[e + f*x]]*(-9 + 28*Cos[2*(e + f*x)] + 21*Cos[4*(e + f*x)] + 98*Sin
[e + f*x] + 42*Sin[3*(e + f*x)])))/(180*c^3*f*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Si
n[e + f*x]])

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Maple [C]  time = 0.391, size = 947, normalized size = 2.7 \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))^(7/2),x)

[Out]

2/45/f*(cos(f*x+e)+1)^2*(g*cos(f*x+e))^(3/2)*(-1+cos(f*x+e))^2*(1+sin(f*x+e))*(-1+sin(f*x+e))*(21*I*cos(f*x+e)
^2*(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)+21*I*E
llipticF(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)+21*I*cos(f*x+e)^5*(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)-21*I*cos(f*x+e)^2*(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)-21*I*cos(f*x+e)^5*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*E
llipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)-21*I*sin(f*x+e)*cos(f*x+e)^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+21*I*sin(f*x+e)*cos(f*x+e)^3*(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)+21*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)*cos(f*x+e)^4-21*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)-
21*I*cos(f*x+e)^3*(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)+21*I*cos(f*x+e)^3*(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)-21*I*cos(f*x+e)^4*(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)-21*sin(f*x+e)*cos(f*x+e)^3+14*cos(f*x+e)^2*sin(f*x+e)+21*cos(f*x+e)^3-14*cos(f*x+e)
^2+2*sin(f*x+e)-7)/(a*(1+sin(f*x+e)))^(5/2)/(-c*(-1+sin(f*x+e)))^(7/2)/cos(f*x+e)/sin(f*x+e)^5

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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}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(3/2)/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2)), 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^{3} c^{4} \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - a^{3} c^{4} \cos \left (f x + e\right )^{5}}, 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))^(7/2),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^4*cos(f*x + e)^5*si
n(f*x + e) - a^3*c^4*cos(f*x + e)^5), 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))**(7/2),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}}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{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))^(7/2),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(3/2)/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2)), x)