Integrand size = 23, antiderivative size = 149 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
-1/3*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-sin(d*x+c)*sec(d*x+c )^(1/2)/a^2/d/(1+sec(d*x+c))+(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+2/3*(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
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.08 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (-i e^{-i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+16 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )+i (5+14 \cos (c+d x)+5 \cos (2 (c+d x))+i \sin (2 (c+d x)))\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{6 a^2 d (1+\cos (c+d x))^2} \]
(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(((-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*(c + d*x)) + 16*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) + I*(5 + 14*Cos[c + d*x] + 5*Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(6*a^2*d*E^(I*d*x)*(1 + Cos[c + d*x])^2)
Time = 0.89 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3717, 3042, 4303, 27, 3042, 4507, 25, 3042, 4274, 3042, 4258, 3042, 3119, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3717 |
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a \sec (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sqrt {\sec (c+d x)} (a-5 a \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {\sec (c+d x)} (a-5 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a-5 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int -\frac {2 \sec (c+d x) a^2+3 a^2}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {2 \sec (c+d x) a^2+3 a^2}{\sqrt {\sec (c+d x)}}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {2 \csc \left (c+d x+\frac {\pi }{2}\right ) a^2+3 a^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 \int \frac {1}{\sqrt {\sec (c+d x)}}dx+2 a^2 \int \sqrt {\sec (c+d x)}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a^2 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {4 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*(Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - (-(((6 *a^2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d) /a^2) + (6*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec[c + d*x])))/(6*a^2 )
3.4.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Time = 2.58 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.72
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-16 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(257\) |
1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d* x+1/2*c)^6-4*(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+6*(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*E llipticE(cos(1/2*d*x+1/2*c),2^(1/2))-16*cos(1/2*d*x+1/2*c)^4+3*cos(1/2*d*x +1/2*c)^2+1)/a^2/cos(1/2*d*x+1/2*c)^3/(-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
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/6*(2*(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)) + 2*(-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)) + 3*(-I*sqrt(2)*cos(d*x + c)^2 - 2* I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*cos(d*x + c)^2 + 4* 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)
\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {\sqrt {\sec {\left (c + d x \right )}}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]