3.3.97 \(\int \frac {a+a \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [297]

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

[Out]

2/7*a*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*a*sin(d*x+c)/d/sec(d*x+c)^(3/2)+10/21*a*sin(d*x+c)/d/sec(d*x+c)^(1/2)+
6/5*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*s
ec(d*x+c)^(1/2)/d+10/21*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)
)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 = {3317, 3872, 3854, 3856, 2720, 2719} \begin {gather*} \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} \end {gather*}

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3317

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 3854

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 3856

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 3872

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]

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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.12, size = 198, normalized size = 1.31 \begin {gather*} \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 \cos (c+d x)+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))\right )}{420 d} \end {gather*}

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)))

________________________________________________________________________________________

Maple [A]
time = 0.18, size = 270, normalized size = 1.79

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (240 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-528 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+448 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-122 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-122*sin(1/2*d*x+1/2*c)
^2*cos(1/2*d*x+1/2*c)+25*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))-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)))/(-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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 156, normalized size = 1.03 \begin {gather*} \frac {-25 i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{3} + 21 \, a \cos \left (d x + c\right )^{2} + 25 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \end {gather*}

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]

1/105*(-25*I*sqrt(2)*a*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*I*sqrt(2)*a*weierstrassP
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 63*I*sqrt(2)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(2)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) + 2*(15*a*cos(d*x + c)^3 + 21*a*cos(d*x + c)^2 + 25*a*cos(d*x + c))*sin(d*x + c)/sqrt(
cos(d*x + c)))/d

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+a\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________