Integrand size = 23, antiderivative size = 221 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {49 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {13 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {49 \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]
-1/5*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-8/15*sec(d*x+c)^(5/2 )*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-13/6*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a^ 3+a^3*sec(d*x+c))+49/10*sin(d*x+c)*sec(d*x+c)^(1/2)/a^3/d-49/10*(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^3/d-13/6*(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^3/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.85 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.64 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (147 \left (1+e^{2 i (c+d x)}\right )+147 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-65 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {1}{32} \left (1284 \cos \left (\frac {1}{2} (c-d x)\right )+921 \cos \left (\frac {1}{2} (3 c+d x)\right )+1243 \cos \left (\frac {1}{2} (c+3 d x)\right )+374 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+670 \cos \left (\frac {1}{2} (3 c+5 d x)\right )+65 \cos \left (\frac {1}{2} (7 c+5 d x)\right )+147 \cos \left (\frac {1}{2} (5 c+7 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}\right )}{15 a^3 d (1+\cos (c+d x))^3} \]
(2*Cos[(c + d*x)/2]^6*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)* (c + d*x)))]*(147*(1 + E^((2*I)*(c + d*x))) + 147*(-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))] - 65*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))) + ((1284*Cos[(c - d*x)/2] + 921*Cos[(3*c + d*x)/2] + 1243*Cos[(c + 3*d*x)/2] + 374*Cos[(5*c + 3*d*x)/2] + 670*Cos[(3*c + 5*d* x)/2] + 65*Cos[(7*c + 5*d*x)/2] + 147*Cos[(5*c + 7*d*x)/2])*Csc[c/2]*Sec[c /2]*Sec[(c + d*x)/2]^5*Sqrt[Sec[c + d*x]])/32))/(15*a^3*d*(1 + Cos[c + d*x ])^3)
Time = 1.35 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.06, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.826, Rules used = {3042, 3717, 3042, 4303, 27, 3042, 4507, 3042, 4507, 27, 3042, 4274, 3042, 4255, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3717 |
| \(\displaystyle \int \frac {\sec ^{\frac {9}{2}}(c+d x)}{(a \sec (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a-11 a \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a-11 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a-11 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (24 a^2-41 a^2 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (24 a^2-41 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} \left (65 a^3-147 a^3 \sec (c+d x)\right )dx}{a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\frac {\int \sqrt {\sec (c+d x)} \left (65 a^3-147 a^3 \sec (c+d x)\right )dx}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (65 a^3-147 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \int \sqrt {\sec (c+d x)}dx-147 a^3 \int \sec ^{\frac {3}{2}}(c+d x)dx}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-147 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {\frac {65 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}+\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {\frac {65 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {130 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-147 a^3 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}}{3 a^2}+\frac {16 a \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {\sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*(Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) - ((16*a *Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((65*a^2* Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((130*a^3*Sqrt [Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - 147*a^3*( (-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + ( 2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/(2*a^2))/(3*a^2))/(10*a^2)
3.4.32.3.1 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)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(554\) vs. \(2(245)=490\).
Time = 4.38 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.51
| method | result | size |
| default | \(-\frac {-2 \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}\, \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 (65 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\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 (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \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 (65 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \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}\, \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 (65 F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-147 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+588 \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 (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1634 \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 (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1488 \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 (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-439 \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 (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\) | \(555\) |
-1/60*(-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)*(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(65*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))-147*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4+4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+ 1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(65 *EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*EllipticE(cos(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*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x +1/2*c)^2)^(1/2)*(65*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-147*EllipticE(c os(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+588*(-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)^8-1634*(-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)^6+1488*(-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)^4-439*(-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)/a^3/cos( 1/2*d*x+1/2*c)^5/(-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 = 0.11 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {65 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 65 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 147 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 147 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (147 \, \cos \left (d x + c\right )^{3} + 376 \, \cos \left (d x + c\right )^{2} + 295 \, \cos \left (d x + c\right ) + 60\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
-1/60*(65*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sq rt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 65*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x + c)^ 2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(d*x + c)) + 147*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*co s(d*x + c)^2 + 3*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))) + 147*(-I*sqr t(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*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*(147*cos(d*x + c)^3 + 376*cos(d*x + c)^2 + 295* cos(d*x + c) + 60)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Integral(sec(c + d*x)**(3/2)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos( c + d*x) + 1), x)/a**3
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]