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

Optimal. Leaf size=151 \[ \frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{10 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{6 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

(6*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (10*a*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*a*Sin[c + d*x
])/(5*d*Sec[c + d*x]^(3/2)) + (10*a*Sin[c + d*x])/(21*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.112773, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3238, 3787, 3769, 3771, 2641, 2639} \[ \frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{10 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{6 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (10*a*Sqrt[Cos[c + d*x]]*Ellipti
cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (2*a*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (2*a*Sin[c + d*x
])/(5*d*Sec[c + d*x]^(3/2)) + (10*a*Sin[c + d*x])/(21*d*Sqrt[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 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{a+a \cos (c+d x)}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{a+a \sec (c+d x)}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\\ &=a \int \frac{1}{\sec ^{\frac{7}{2}}(c+d x)} \, dx+a \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{5} (3 a) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{7} (5 a) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{10 a \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} (5 a) \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (3 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{6 a \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{10 a \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}+\frac{1}{21} \left (5 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{6 a \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{10 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{2 a \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 a \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{10 a \sin (c+d x)}{21 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.15109, size = 198, normalized size = 1.31 \[ \frac{a e^{-4 i (c+d x)} \sqrt{\sec (c+d x)} (\cos (4 (c+d x))+i \sin (4 (c+d x))) \left (504 i e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i (c+d x)}\right )-200 i \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-e^{2 i (c+d x)}\right )+42 \sin (c+d x)+130 \sin (2 (c+d x))+42 \sin (3 (c+d x))+15 \sin (4 (c+d x))-504 i \cos (c+d x)\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[Sec[c + d*x]]*(Cos[4*(c + d*x)] + I*Sin[4*(c + d*x)])*((-504*I)*Cos[c + d*x] + ((504*I)*Sqrt[1 + E^((2
*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) - (200*I)*Sqrt[1 + E^
((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))] + 42*Sin[c + d*x] + 130*Sin[2*(c + d
*x)] + 42*Sin[3*(c + d*x)] + 15*Sin[4*(c + d*x)]))/(420*d*E^((4*I)*(c + d*x)))

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Maple [A]  time = 2.421, size = 270, normalized size = 1.8 \begin{align*} -{\frac{2\,a}{105\,d}\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 ( 240\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-528\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +448\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +25\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -63\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -122\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ){\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((a+cos(d*x+c)*a)/sec(d*x+c)^(5/2),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(240*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-
528*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+25*(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))-63*(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))-122*sin(1/2*d*x+1/2*c)^2*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{a \cos \left (d x + c\right ) + a}{\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((a+a*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((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{a \cos \left (d x + c\right ) + a}{\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((a+a*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((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((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{a \cos \left (d x + c\right ) + a}{\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((a+a*cos(d*x+c))/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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