Integrand size = 28, antiderivative size = 155 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}} \]
10/77*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(1/2)+10/77*a^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)*(e*sec(d*x+c))^(1/2)/d/e^6-2/11*I*(a+I*a*tan(d*x+c))^3/d/(e*s ec(d*x+c))^(11/2)-20/77*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(7/2 )
Time = 2.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {a^3 \sqrt {e \sec (c+d x)} \left (-46 i \cos (c+d x)-22 i \cos (3 (c+d x))-15 \sin (c+d x)+20 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-15 \sin (3 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{154 d e^6 (\cos (d x)+i \sin (d x))^3} \]
(a^3*Sqrt[e*Sec[c + d*x]]*((-46*I)*Cos[c + d*x] - (22*I)*Cos[3*(c + d*x)] - 15*Sin[c + d*x] + 20*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3 *(c + d*x)] - I*Sin[3*(c + d*x)]) - 15*Sin[3*(c + d*x)])*(Cos[3*(c + 2*d*x )] + I*Sin[3*(c + 2*d*x)]))/(154*d*e^6*(Cos[d*x] + I*Sin[d*x])^3)
Time = 0.79 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3978, 3042, 3977, 3042, 4256, 3042, 4258, 3042, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}}dx\) |
| \(\Big \downarrow \) 3978 |
| \(\displaystyle \frac {5 a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{7/2}}dx}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{7/2}}dx}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {5 a \left (\frac {3 a^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{7 d (e \sec (c+d x))^{7/2}}\right )}{11 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}\) |
(((-2*I)/11)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(11/2)) + (5*a* ((3*a^2*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d* x]])/(3*d*e^2) + (2*Sin[c + d*x])/(3*d*e*Sqrt[e*Sec[c + d*x]])))/(7*e^2) - (((4*I)/7)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(7/2))))/(11*e ^2)
3.3.10.3.1 Defintions of rubi rules used
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[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( a*f*m)), x] + Simp[a*((m + n)/(m*d^2)) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b ^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
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] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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]
Time = 20.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(-\frac {2 a^{3} \left (28 i \left (\cos ^{5}\left (d x +c \right )\right )-28 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+5 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-11 i \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 \sin \left (d x +c \right )\right )}{77 e^{5} d \sqrt {e \sec \left (d x +c \right )}}\) | \(201\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (7 \,{\mathrm e}^{4 i \left (d x +c \right )}+24 \,{\mathrm e}^{2 i \left (d x +c \right )}+37\right ) a^{3} \sqrt {2}}{308 d \,e^{5} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {10 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{3} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{77 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(260\) |
| parts | \(-\frac {2 a^{3} \left (-7 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-9 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-15 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {2 i a^{3} \left (7 \left (\cos ^{5}\left (d x +c \right )\right )-11 \left (\cos ^{3}\left (d x +c \right )\right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {6 i a^{3}}{11 d \left (e \sec \left (d x +c \right )\right )^{\frac {11}{2}}}+\frac {2 a^{3} \left (21 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+10 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+10 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}\) | \(420\) |
-2/77*a^3/e^5/d/(e*sec(d*x+c))^(1/2)*(28*I*cos(d*x+c)^5-28*sin(d*x+c)*cos( d*x+c)^4+5*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot (d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)-11*I*cos(d*x+c)^3+5*I*sec(d*x+c)*(cos (d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(c os(d*x+c)+1))^(1/2)-3*cos(d*x+c)^2*sin(d*x+c)-5*sin(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {-40 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-7 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 31 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 61 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 37 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{308 \, d e^{6}} \]
1/308*(-40*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c )) + sqrt(2)*(-7*I*a^3*e^(6*I*d*x + 6*I*c) - 31*I*a^3*e^(4*I*d*x + 4*I*c) - 61*I*a^3*e^(2*I*d*x + 2*I*c) - 37*I*a^3)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1 ))*e^(1/2*I*d*x + 1/2*I*c))/(d*e^6)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]
\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}} \,d x \]