3.251 \(\int \frac{1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=150 \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}} \]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(33*
a^2*d*e*(e*Cos[c + d*x])^(3/2)) - 2/(11*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2) - 2/(11*d*e*(e*Cos[
c + d*x])^(3/2)*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.161927, antiderivative size = 150, 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 = {2681, 2683, 2636, 2642, 2641} \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{33 a^2 d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^2),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(33*a^2*d*e^2*Sqrt[e*Cos[c + d*x]]) + (10*Sin[c + d*x])/(33*
a^2*d*e*(e*Cos[c + d*x])^(3/2)) - 2/(11*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2) - 2/(11*d*e*(e*Cos[
c + d*x])^(3/2)*(a^2 + a^2*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 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2} \, dx &=-\frac{2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}+\frac{7 \int \frac{1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx}{11 a}\\ &=-\frac{2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \int \frac{1}{(e \cos (c+d x))^{5/2}} \, dx}{11 a^2}\\ &=\frac{10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{5 \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{33 a^2 e^2}\\ &=\frac{10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (5 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{33 a^2 e^2 \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 )}{33 a^2 d e^2 \sqrt{e \cos (c+d x)}}+\frac{10 \sin (c+d x)}{33 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2}-\frac{2}{11 d e (e \cos (c+d x))^{3/2} \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0725248, size = 66, normalized size = 0.44 \[ \frac{(\sin (c+d x)+1)^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{15}{4};\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{6\ 2^{3/4} a^2 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^2),x]

[Out]

(Hypergeometric2F1[-3/4, 15/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(6*2^(3/4)*a^2*d*e*(e*Cos[
c + d*x])^(3/2))

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Maple [B]  time = 2.612, size = 557, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x)

[Out]

-2/33/(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^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^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*si
n(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)-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))+2
8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-6*sin(1/2*d*x+1/2*c))/d

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Maxima [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/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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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 )}}{a^{2} e^{3} \cos \left (d x + c\right )^{5} - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))/(a^2*e^3*cos(d*x + c)^5 - 2*a^2*e^3*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*e^3*cos
(d*x + c)^3), 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/(e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a)^2), x)