3.269 \(\int \frac{(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=154 \[ -\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^4 d \sqrt{e \cos (c+d x)}}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a \sin (c+d x)+a)^3} \]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^4*d*Sqrt[e*Cos[c + d*x]]) - (4*e*Sqrt[e*Cos[c + d*
x]])/(11*a*d*(a + a*Sin[c + d*x])^3) + (2*e*Sqrt[e*Cos[c + d*x]])/(77*d*(a^2 + a^2*Sin[c + d*x])^2) + (2*e*Sqr
t[e*Cos[c + d*x]])/(77*d*(a^4 + a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.18481, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2680, 2681, 2683, 2642, 2641} \[ -\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^4 d \sqrt{e \cos (c+d x)}}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^4 \sin (c+d x)+a^4\right )}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^4*d*Sqrt[e*Cos[c + d*x]]) - (4*e*Sqrt[e*Cos[c + d*
x]])/(11*a*d*(a + a*Sin[c + d*x])^3) + (2*e*Sqrt[e*Cos[c + d*x]])/(77*d*(a^2 + a^2*Sin[c + d*x])^2) + (2*e*Sqr
t[e*Cos[c + d*x]])/(77*d*(a^4 + a^4*Sin[c + d*x]))

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2681

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

Rule 2683

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(g*Cos[e
+ f*x])^(p + 1))/(a*f*g*(p - 1)*(a + b*Sin[e + f*x])), x] + Dist[p/(a*(p - 1)), Int[(g*Cos[e + f*x])^p, x], x
] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && IntegerQ[2*p]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{3/2}}{(a+a \sin (c+d x))^4} \, dx &=-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}-\frac{e^2 \int \frac{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx}{11 a^2}\\ &=-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac{\left (3 e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))} \, dx}{77 a^3}\\ &=-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac{e^2 \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{77 a^4}\\ &=-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac{\left (e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{77 a^4 \sqrt{e \cos (c+d x)}}\\ &=-\frac{2 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^4 d \sqrt{e \cos (c+d x)}}-\frac{4 e \sqrt{e \cos (c+d x)}}{11 a d (a+a \sin (c+d x))^3}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac{2 e \sqrt{e \cos (c+d x)}}{77 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0742038, size = 66, normalized size = 0.43 \[ -\frac{(e \cos (c+d x))^{5/2} \, _2F_1\left (\frac{5}{4},\frac{15}{4};\frac{9}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{10\ 2^{3/4} a^4 d e (\sin (c+d x)+1)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

-((e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[5/4, 15/4, 9/4, (1 - Sin[c + d*x])/2])/(10*2^(3/4)*a^4*d*e*(1 + Sin
[c + d*x])^(5/4))

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Maple [B]  time = 3.526, size = 583, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x)

[Out]

2/77/(32*sin(1/2*d*x+1/2*c)^10-80*sin(1/2*d*x+1/2*c)^8+80*sin(1/2*d*x+1/2*c)^6-40*sin(1/2*d*x+1/2*c)^4+10*sin(
1/2*d*x+1/2*c)^2-1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(32*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^10-80*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^8+3
2*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+80*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^6-64*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-40*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c
)^4+176*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+10*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-144*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-176*
sin(1/2*d*x+1/2*c)^5-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))-78*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+176*sin(1/2*d*x+1/2*c)^3+12*sin(1/2*d*x+1/2*c))*e^2/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(3/2)/(a*sin(d*x + c) + a)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e \cos \left (d x + c\right )} e \cos \left (d x + c\right )}{a^{4} \cos \left (d x + c\right )^{4} - 8 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} - 4 \,{\left (a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*e*cos(d*x + c)/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d
*x + c)^2 - 2*a^4)*sin(d*x + c)), 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((e*cos(d*x+c))**(3/2)/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(3/2)/(a*sin(d*x + c) + a)^4, x)