\(\int \frac {1}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [329]

Optimal result

Integrand size = 23, antiderivative size = 152 \[ \int \frac {1}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {5 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-5/3*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(1+sec(d*x+c))-1/3*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^2+4*(
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^2/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^2/d

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3317, 3902, 4105, 3872, 3856, 2719, 2720} \[ \int \frac {1}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 a^2 d (\sec (c+d x)+1)}-\frac {5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2} \]

[In]

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

[Out]

(4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) - (5*Sqrt[Cos[c + d*x]]*EllipticF[
(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - (5*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x]
)) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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 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 3902

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

Rule 4105

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx \\ & = -\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {7 a}{2}+\frac {3}{2} a \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{3 a^2} \\ & = -\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {-6 a^2+\frac {5}{2} a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{3 a^4} \\ & = -\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {5 \int \sqrt {\sec (c+d x)} \, dx}{6 a^2}+\frac {2 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{a^2} \\ & = -\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\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^2}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{a^2} \\ & = \frac {4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 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^2 d}-\frac {5 \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 1.66 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \sin (c) (\cos (d x)+i \sin (d x)) \left (-24 i \cos \left (\frac {1}{2} (c+d x)\right )-18 i \cos \left (\frac {3}{2} (c+d x)\right )-6 i \cos \left (\frac {5}{2} (c+d x)\right )+20 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 i e^{-\frac {1}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+\sin \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{6 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

-1/6*(Cos[(c + d*x)/2]*Csc[c/2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c]*(Cos[d*x] + I*Sin[d*x])*((-24*I)*Cos[(c + d
*x)/2] - (18*I)*Cos[(3*(c + d*x))/2] - (6*I)*Cos[(5*(c + d*x))/2] + 20*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2] + ((2*I)*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2,
3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((I/2)*(c + d*x)) + Sin[(c + d*x)/2] + 2*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(
c + d*x))/2]))/(a^2*d*E^(I*d*x)*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 4.53 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.69

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (24 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-38 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}{6 a^{2} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(257\)

[In]

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

[Out]

1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*cos(1/2*d*x+1/2*c)^6+10*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+24*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^3*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-38*cos(1/2*d*x+1/2*c)^4+15*cos(1/2*d*x+1/2*c)^2-1)/a^2/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)/cos(1/2*d*x+1/2*c)^3/sin(1/2*d*x+1/2*c)/(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 = 0.10 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.82 \[ \int \frac {1}{(a+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 12 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (6 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-1/6*(5*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x
+ c) + I*sin(d*x + c)) + 5*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInver
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 12*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2
))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 12*(I*sqrt(2)*cos(d*x +
c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) -
I*sin(d*x + c))) + 2*(6*cos(d*x + c)^2 + 5*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)
^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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