3.261 \(\int \frac{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=153 \[ -\frac{10 \sqrt{e \cos (c+d x)}}{77 d e \left (a^3 \sin (c+d x)+a^3\right )}+\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^3 d \sqrt{e \cos (c+d x)}}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a \sin (c+d x)+a)^2}-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3} \]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^3*d*Sqrt[e*Cos[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]])/(
11*d*e*(a + a*Sin[c + d*x])^3) - (10*Sqrt[e*Cos[c + d*x]])/(77*a*d*e*(a + a*Sin[c + d*x])^2) - (10*Sqrt[e*Cos[
c + d*x]])/(77*d*e*(a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.184146, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2681, 2683, 2642, 2641} \[ -\frac{10 \sqrt{e \cos (c+d x)}}{77 d e \left (a^3 \sin (c+d x)+a^3\right )}+\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^3 d \sqrt{e \cos (c+d x)}}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a \sin (c+d x)+a)^2}-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^3),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(77*a^3*d*Sqrt[e*Cos[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]])/(
11*d*e*(a + a*Sin[c + d*x])^3) - (10*Sqrt[e*Cos[c + d*x]])/(77*a*d*e*(a + a*Sin[c + d*x])^2) - (10*Sqrt[e*Cos[
c + d*x]])/(77*d*e*(a^3 + a^3*Sin[c + d*x]))

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{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^3} \, dx &=-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}+\frac{5 \int \frac{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx}{11 a}\\ &=-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}+\frac{15 \int \frac{1}{\sqrt{e \cos (c+d x)} (a+a \sin (c+d x))} \, dx}{77 a^2}\\ &=-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac{10 \sqrt{e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}+\frac{5 \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{77 a^3}\\ &=-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac{10 \sqrt{e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}+\frac{\left (5 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{77 a^3 \sqrt{e \cos (c+d x)}}\\ &=\frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{77 a^3 d \sqrt{e \cos (c+d x)}}-\frac{2 \sqrt{e \cos (c+d x)}}{11 d e (a+a \sin (c+d x))^3}-\frac{10 \sqrt{e \cos (c+d x)}}{77 a d e (a+a \sin (c+d x))^2}-\frac{10 \sqrt{e \cos (c+d x)}}{77 d e \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0491033, size = 66, normalized size = 0.43 \[ -\frac{\sqrt{e \cos (c+d x)} \, _2F_1\left (\frac{1}{4},\frac{15}{4};\frac{5}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{2\ 2^{3/4} a^3 d e \sqrt [4]{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^3),x]

[Out]

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

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Maple [B]  time = 3.264, size = 580, 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(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),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^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(160*(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)^10-400*(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+160*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+400*(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)^6-320*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-200*
(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+264*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+50*(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)^2-104*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*
c)+44*sin(1/2*d*x+1/2*c)^5-5*(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))+72*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-44*sin(1/2*d*x+1/2*c)^3-17*sin(1/2*d*x+1/2*c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \cos \left (d x + c\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^3), 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 )}}{3 \, a^{3} e \cos \left (d x + c\right )^{3} - 4 \, a^{3} e \cos \left (d x + c\right ) +{\left (a^{3} e \cos \left (d x + c\right )^{3} - 4 \, a^{3} e \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))/(3*a^3*e*cos(d*x + c)^3 - 4*a^3*e*cos(d*x + c) + (a^3*e*cos(d*x + c)^3 - 4*a^3*
e*cos(d*x + c))*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(1/(a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \cos \left (d x + c\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^3), x)