\(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\) [317]

Optimal result

Integrand size = 23, antiderivative size = 164 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

5/3*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))-3*sin(d*x+c)*sec(d*x+c)^(1/
2)/a/d+3*(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
)*sec(d*x+c)^(1/2)/a/d+5/3*(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)/a/d

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3317, 3903, 3872, 3853, 3856, 2719, 2720} \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {5 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {3 \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}+\frac {5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[In]

Int[Sec[c + d*x]^(5/2)/(a + a*Cos[c + d*x]),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) - (3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*d) + (5*Sec[c + d*x]^(3/2)*
Sin[c + d*x])/(3*a*d) - (Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*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 3853

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

Rule 3903

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{a+a \sec (c+d x)} \, dx \\ & = -\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a}{2}-\frac {5}{2} a \sec (c+d x)\right ) \, dx}{a^2} \\ & = -\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a}+\frac {5 \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a} \\ & = -\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {5 \int \sqrt {\sec (c+d x)} \, dx}{6 a}+\frac {3 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a} \\ & = -\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (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 {\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 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {5 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.51 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (9 \left (1+e^{2 i (c+d x)}\right )+9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}-\sqrt {\sec (c+d x)} \left (18 \cos (d x) \csc (c)+\sec (c+d x) \left (-5 \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {3}{2} (c+d x)\right )+\tan \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a d (1+\cos (c+d x))} \]

[In]

Integrate[Sec[c + d*x]^(5/2)/(a + a*Cos[c + d*x]),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))) - Sqrt[Sec[c + d*x]]*(18*Cos[d*x]*Csc[c] + Sec[c + d*x]
*(-5*Sec[(c + d*x)/2]*Sin[(3*(c + d*x))/2] + Tan[(c + d*x)/2]))))/(3*a*d*(1 + Cos[c + d*x]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(198)=396\).

Time = 5.17 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.52

method	result	size
default	\(\frac {\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 )}\, \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 )}\, \left (10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+44 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-11 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(413\)

[In]

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

[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)/a/cos(1/2*d*x+1/2*c)/sin(1/2*d*x+1/2*c)^3/(4*sin
(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*cos(1/2*d
*x+1/2*c)*EllipticF(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)^2-18*cos(1/2*d*x+1/2*c)*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)^2-36*sin(1/2*d*x+1/2*c)^6-5*cos(1/2*d*x+1/2*c)*(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))+9*cos(1/2*d*x+1/2*c)*
(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))+44*sin(1/2
*d*x+1/2*c)^4-11*sin(1/2*d*x+1/2*c)^2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/6*(5*(I*sqrt(2)*cos(d*x + c)^2 + I*sqrt(2)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*
x + c)) + 5*(-I*sqrt(2)*cos(d*x + c)^2 - I*sqrt(2)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*s
in(d*x + c)) + 9*(-I*sqrt(2)*cos(d*x + c)^2 - I*sqrt(2)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInver
se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(I*sqrt(2)*cos(d*x + c)^2 + I*sqrt(2)*cos(d*x + c))*weierstrassZ
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(9*cos(d*x + c)^2 + 4*cos(d*x + c) -
2)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{a+a\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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