3.322 \(\int \frac{1}{(a+a \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)}+\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}-\frac{3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

(-3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (5*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(d*Sqr
t[Sec[c + d*x]]*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.161564, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3238, 3819, 3787, 3769, 3771, 2641, 2639} \[ \frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)}+\frac{5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}-\frac{3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (5*Sqrt[Cos[c + d*x]]*EllipticF[(
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) + (5*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]) - Sin[c + d*x]/(d*Sqr
t[Sec[c + d*x]]*(a + a*Sec[c + d*x]))

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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]

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}{(a+a \cos (c+d x)) \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{1}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\\ &=-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))}-\frac{\int \frac{-\frac{5 a}{2}+\frac{3}{2} a \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))}-\frac{3 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a}+\frac{5 \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a}\\ &=\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))}+\frac{5 \int \sqrt{\sec (c+d x)} \, dx}{6 a}-\frac{\left (3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a}\\ &=-\frac{3 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a d}+\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))}+\frac{\left (5 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}\\ &=-\frac{3 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a d}+\frac{5 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a d}+\frac{5 \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\sin (c+d x)}{d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 4.00192, size = 312, normalized size = 2.23 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (2 \sqrt{\sec (c+d x)} \left (\sin (2 c) \cos (2 d x)-6 \cos (c) \sin (d x)+\cos (2 c) \sin (2 d x)+3 (\cos (2 c)+2) \csc (c) \cos (d x)-3 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right )-3 \tan \left (\frac{c}{2}\right )\right )-\frac{2 i \sqrt{2} e^{-i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (9 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i (c+d x)}\right )+5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-e^{2 i (c+d x)}\right )+9 \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{3 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(9*(1 + E^((2*I)*(c + d*x
))) + 9*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x)
)] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((
2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + 2*Sqrt[Sec[c + d*x]]*(3*(2 + Cos[2*c])*Cos[d*x]*Csc[
c] + Cos[2*d*x]*Sin[2*c] - 3*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] - 6*Cos[c]*Sin[d*x] + Cos[2*c]*Sin[2*d*x]
- 3*Tan[c/2])))/(3*a*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 2.188, size = 215, normalized size = 1.5 \begin{align*} -{\frac{1}{3\,da}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 5\,{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +9\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) -8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+18\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-7\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2)))-8*sin(1/2*d*x+1/2*c)^6+18*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/((a*cos(d*x + c) + a)*sec(d*x + c)^(5/2)), 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*cos(d*x+c))/sec(d*x+c)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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