3.252 \(\int \frac{1}{(e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=181 \[ \frac{42 \sin (c+d x)}{65 a^2 d e^3 \sqrt{e \cos (c+d x)}}-\frac{42 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{65 a^2 d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}-\frac{2}{13 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}-\frac{2}{13 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}} \]

[Out]

(-42*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(65*a^2*d*e^4*Sqrt[Cos[c + d*x]]) + (14*Sin[c + d*x])/(65
*a^2*d*e*(e*Cos[c + d*x])^(5/2)) + (42*Sin[c + d*x])/(65*a^2*d*e^3*Sqrt[e*Cos[c + d*x]]) - 2/(13*d*e*(e*Cos[c
+ d*x])^(5/2)*(a + a*Sin[c + d*x])^2) - 2/(13*d*e*(e*Cos[c + d*x])^(5/2)*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.183031, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2681, 2683, 2636, 2640, 2639} \[ \frac{42 \sin (c+d x)}{65 a^2 d e^3 \sqrt{e \cos (c+d x)}}-\frac{42 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{65 a^2 d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}-\frac{2}{13 d e \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}-\frac{2}{13 d e (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(7/2)*(a + a*Sin[c + d*x])^2),x]

[Out]

(-42*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(65*a^2*d*e^4*Sqrt[Cos[c + d*x]]) + (14*Sin[c + d*x])/(65
*a^2*d*e*(e*Cos[c + d*x])^(5/2)) + (42*Sin[c + d*x])/(65*a^2*d*e^3*Sqrt[e*Cos[c + d*x]]) - 2/(13*d*e*(e*Cos[c
+ d*x])^(5/2)*(a + a*Sin[c + d*x])^2) - 2/(13*d*e*(e*Cos[c + d*x])^(5/2)*(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 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{1}{(e \cos (c+d x))^{7/2} (a+a \sin (c+d x))^2} \, dx &=-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}+\frac{9 \int \frac{1}{(e \cos (c+d x))^{7/2} (a+a \sin (c+d x))} \, dx}{13 a}\\ &=-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{7 \int \frac{1}{(e \cos (c+d x))^{7/2}} \, dx}{13 a^2}\\ &=\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{21 \int \frac{1}{(e \cos (c+d x))^{3/2}} \, dx}{65 a^2 e^2}\\ &=\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}+\frac{42 \sin (c+d x)}{65 a^2 d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}-\frac{21 \int \sqrt{e \cos (c+d x)} \, dx}{65 a^2 e^4}\\ &=\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}+\frac{42 \sin (c+d x)}{65 a^2 d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\left (21 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{65 a^2 e^4 \sqrt{\cos (c+d x)}}\\ &=-\frac{42 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{65 a^2 d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{65 a^2 d e (e \cos (c+d x))^{5/2}}+\frac{42 \sin (c+d x)}{65 a^2 d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}-\frac{2}{13 d e (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.105818, size = 66, normalized size = 0.36 \[ \frac{(\sin (c+d x)+1)^{5/4} \, _2F_1\left (-\frac{5}{4},\frac{17}{4};-\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{20 \sqrt [4]{2} a^2 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(7/2)*(a + a*Sin[c + d*x])^2),x]

[Out]

(Hypergeometric2F1[-5/4, 17/4, -1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(5/4))/(20*2^(1/4)*a^2*d*e*(e*Co
s[c + d*x])^(5/2))

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Maple [B]  time = 3.316, size = 670, 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))^(7/2)/(a+a*sin(d*x+c))^2,x)

[Out]

-2/65/(64*sin(1/2*d*x+1/2*c)^12-192*sin(1/2*d*x+1/2*c)^10+240*sin(1/2*d*x+1/2*c)^8-160*sin(1/2*d*x+1/2*c)^6+60
*sin(1/2*d*x+1/2*c)^4-12*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^
3*(1344*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*si
n(1/2*d*x+1/2*c)^12-2688*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-4032*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icE(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+8064*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^12+5040*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^8-10304*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-3360*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))*(sin(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+7168*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+1260*EllipticE(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-2896*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-252*EllipticE(co
s(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)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+6
56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))-86*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*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))^(7/2)/(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^{4} \cos \left (d x + c\right )^{6} - 2 \, a^{2} e^{4} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} e^{4} \cos \left (d x + c\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))/(a^2*e^4*cos(d*x + c)^6 - 2*a^2*e^4*cos(d*x + c)^4*sin(d*x + c) - 2*a^2*e^4*cos
(d*x + c)^4), 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))**(7/2)/(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{7}{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))^(7/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(7/2)*(a*sin(d*x + c) + a)^2), x)