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

Optimal. Leaf size=182 \[ \frac{6 a g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{6 a (g \cos (e+f x))^{5/2}}{5 c f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{4 a (g \cos (e+f x))^{5/2}}{5 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[Out]

(4*a*(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)) - (6*a*(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)) + (6*a*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 = 0.878637, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.119, Rules used = {2850, 2852, 2842, 2640, 2639} \[ \frac{6 a g \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 c^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{6 a (g \cos (e+f x))^{5/2}}{5 c f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac{4 a (g \cos (e+f x))^{5/2}}{5 f g \sqrt{a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

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]

(4*a*(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)) - (6*a*(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)) + (6*a*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 2850

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

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{4 a (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{(3 a) \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{4 a (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{6 a (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{(3 a) \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{4 a (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{6 a (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{(3 a 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{4 a (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{6 a (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 (3 a 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{4 a (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{6 a (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{6 a 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 [C]  time = 2.14511, size = 229, normalized size = 1.26 \[ \frac{4 i g \sqrt{g e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right )} \left (e^{i (e+f x)} \left (e^{i (e+f x)}-i\right )^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (e+f x)}\right )+\left (4 i e^{i (e+f x)}-3 e^{2 i (e+f x)}+5\right ) \sqrt{1+e^{2 i (e+f x)}}\right ) \sqrt{a (\sin (e+f x)+1)}}{5 c f \left (e^{i (e+f x)}+i\right ) \sqrt{1+e^{2 i (e+f x)}} \left (i c e^{-i (e+f x)} \left (e^{i (e+f x)}-i\right )^2\right )^{3/2}} \]

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]

(((4*I)/5)*g*Sqrt[((1 + E^((2*I)*(e + f*x)))*g)/E^(I*(e + f*x))]*((5 + (4*I)*E^(I*(e + f*x)) - 3*E^((2*I)*(e +
f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))] + E^(I*(e + f*x))*(-I + E^(I*(e + f*x)))^3*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(e + f*x))])*Sqrt[a*(1 + Sin[e + f*x])])/(c*((I*c*(-I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/
2)*(I + E^(I*(e + f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))]*f)

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Maple [C]  time = 0.342, size = 2040, normalized size = 11.2 \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))^(1/2)/(c-c*sin(f*x+e))^(5/2),x)

[Out]

-1/10/f*(g*cos(f*x+e))^(3/2)*(a*(1+sin(f*x+e)))^(1/2)*(-1+cos(f*x+e))^3*(-1+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)+20*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-5*ln(-2*(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)*cos(f*x+e)^3+5*ln(-(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)*cos(f*x+e)^3+5*ln(-2*(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)*cos(f
*x+e)-5*cos(f*x+e)*ln(-(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)+8*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-20*cos(f*x+e
)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+8*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-8*cos(f*x+e)^2*(-cos(f*x+e)/(c
os(f*x+e)+1)^2)^(1/2)-5*ln(-2*(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)*cos(f*x+e)*sin(f*x+e)+5*cos(f*x+e)*ln(-(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)-12*sin(f*x+e)*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+24*I*sin(f*x+e)*cos(f*x+e)
*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(
f*x+e)/(cos(f*x+e)+1))^(1/2)-24*I*sin(f*x+e)*cos(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)
/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+12*I*sin(f*x+e)*cos(f*x+e)
^2*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)-12*I*sin(f*x+e)*cos(f*x+e)^2*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*
x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-24*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)/(cos(f*x+e)+1
)^2)^(1/2)*cos(f*x+e)-12*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)/(cos(f*x+e)+1)^2)^(1/2)+12*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+24*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)*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)*cos(f*x+e)+12*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)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)-12*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+
12*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(
cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^4-12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(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)*cos(f*x+e)^4+24*I*cos(f*x+e)^3*
EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f
*x+e)/(cos(f*x+e)+1))^(1/2)-24*I*cos(f*x+e)^3*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2))/(1+sin(f*x+e))/(-cos(f*x+e)/(cos(f*
x+e)+1)^2)^(3/2)/(-c*(-1+sin(f*x+e)))^(5/2)/sin(f*x+e)^7

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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)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),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 \cos \left (f x + e\right )}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \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))^(1/2)/(c-c*sin(f*x+e))^(5/2),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*cos(f*x + e)/(3*c^3*cos(f*
x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(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))**(1/2)/(c-c*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{\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)*(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),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)