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

Optimal. Leaf size=143 \[ \frac{14 \sin (c+d x)}{15 a d e^3 \sqrt{e \cos (c+d x)}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 a d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}-\frac{2}{9 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{5/2}} \]

[Out]

(-14*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*a*d*e^4*Sqrt[Cos[c + d*x]]) + (14*Sin[c + d*x])/(45*a
*d*e*(e*Cos[c + d*x])^(5/2)) + (14*Sin[c + d*x])/(15*a*d*e^3*Sqrt[e*Cos[c + d*x]]) - 2/(9*d*e*(e*Cos[c + d*x])
^(5/2)*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.114257, antiderivative size = 143, 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 = {2683, 2636, 2640, 2639} \[ \frac{14 \sin (c+d x)}{15 a d e^3 \sqrt{e \cos (c+d x)}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 a d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}-\frac{2}{9 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*a*d*e^4*Sqrt[Cos[c + d*x]]) + (14*Sin[c + d*x])/(45*a
*d*e*(e*Cos[c + d*x])^(5/2)) + (14*Sin[c + d*x])/(15*a*d*e^3*Sqrt[e*Cos[c + d*x]]) - 2/(9*d*e*(e*Cos[c + d*x])
^(5/2)*(a + a*Sin[c + d*x]))

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))} \, dx &=-\frac{2}{9 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))}+\frac{7 \int \frac{1}{(e \cos (c+d x))^{7/2}} \, dx}{9 a}\\ &=\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}-\frac{2}{9 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))}+\frac{7 \int \frac{1}{(e \cos (c+d x))^{3/2}} \, dx}{15 a e^2}\\ &=\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 a d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))}-\frac{7 \int \sqrt{e \cos (c+d x)} \, dx}{15 a e^4}\\ &=\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 a d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))}-\frac{\left (7 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 a e^4 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 a d e^4 \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{45 a d e (e \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 a d e^3 \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.082537, size = 66, normalized size = 0.46 \[ \frac{(\sin (c+d x)+1)^{5/4} \, _2F_1\left (-\frac{5}{4},\frac{13}{4};-\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{10 \sqrt [4]{2} a d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[-5/4, 13/4, -1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(5/4))/(10*2^(1/4)*a*d*e*(e*Cos[
c + d*x])^(5/2))

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Maple [B]  time = 2.212, size = 488, normalized size = 3.4 \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)),x)

[Out]

-2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/a/sin
(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^3*(336*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)^8-672*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+
1/2*c)-672*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+1344*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+504*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)^4-1064*sin(1/2*d*x+1/2*c)^6*co
s(1/2*d*x+1/2*c)-168*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)^2+392*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-66*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c
)+5*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}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}}\,{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)),x, algorithm="maxima")

[Out]

integrate(1/((e*cos(d*x + c))^(7/2)*(a*sin(d*x + c) + a)), 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 )}}{a e^{4} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a 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)),x, algorithm="fricas")

[Out]

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

[Out]

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