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

Optimal. Leaf size=121 \[ -\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.56, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2852, 2842, 2640, 2639} \[ -\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

\begin {align*} \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 &=-\frac {2 (g \cos (e+f x))^{5/2}}{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}{a}\\ &=-\frac {2 (g \cos (e+f x))^{5/2}}{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}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {2 (g \cos (e+f x))^{5/2}}{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}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {2 (g \cos (e+f x))^{5/2}}{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 )}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.67, size = 170, normalized size = 1.40 \[ -\frac {2 (g \cos (e+f x))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\sqrt {\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \cos ^{\frac {3}{2}}(e+f x) (a (\sin (e+f x)+1))^{3/2} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g}{a^{2} c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} c \cos \left (f x + e\right )}, x\right ) \]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

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

________________________________________________________________________________________

maple [C]  time = 0.59, size = 925, normalized size = 7.64 \[ \text {result too large to display} \]

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))^(3/2)/(c-c*sin(f*x+e))^(1/2),x)

[Out]

1/2/f*(g*cos(f*x+e))^(3/2)*(-1+cos(f*x+e))^2*(1+sin(f*x+e))*(4*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x
+e),I)-4*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)*(-cos(f*x+e)/(co
s(f*x+e)+1)^2)^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+8*I*sin(f*x+e)*cos(f*x+e)*(1/(cos(f*x+e)+1))^(1
/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x
+e),I)-8*I*sin(f*x+e)*cos(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)+4*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x
+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-4*I*(1
/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*EllipticE(I*(-1+
cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)+4*sin(f*x+e)*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)*l
n(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)-1)/sin(f*x+e)^2)*sin(f*x+e)+cos(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-c
os(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*sin(f*x+e)+4*cos(f*x+e)^2*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-4*(-cos(f*x+e)/(cos(f*x+e)
+1)^2)^(1/2))/(cos(f*x+e)+1)/sin(f*x+e)^5/(a*(1+sin(f*x+e)))^(3/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)/(-c*(s
in(f*x+e)-1))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(3/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________