Integrand size = 28, antiderivative size = 186 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{11 a^3 d e^2}+\frac {6 e \sin (c+d x)}{55 a^3 d (e \sec (c+d x))^{5/2}}+\frac {2 \sin (c+d x)}{11 a^3 d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{15 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3}+\frac {12 i e^2}{55 d (e \sec (c+d x))^{7/2} \left (a^3+i a^3 \tan (c+d x)\right )} \]
6/55*e*sin(d*x+c)/a^3/d/(e*sec(d*x+c))^(5/2)+2/11*sin(d*x+c)/a^3/d/e/(e*se c(d*x+c))^(1/2)+2/11*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/a^3 /d/e^2+2/15*I/d/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^3+12/55*I*e^2/d/(e *sec(d*x+c))^(7/2)/(a^3+I*a^3*tan(d*x+c))
Time = 1.67 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^5(c+d x) \left (-332 \cos (c+d x)-154 \cos (3 (c+d x))+22 \cos (5 (c+d x))-114 i \sin (c+d x)+240 i \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-81 i \sin (3 (c+d x))+33 i \sin (5 (c+d x))\right )}{1320 a^3 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^3} \]
(Sec[c + d*x]^5*(-332*Cos[c + d*x] - 154*Cos[3*(c + d*x)] + 22*Cos[5*(c + d*x)] - (114*I)*Sin[c + d*x] + (240*I)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d *x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - (81*I)*Sin[3*(c + d*x) ] + (33*I)*Sin[5*(c + d*x)]))/(1320*a^3*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan [c + d*x])^3)
Time = 0.93 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3983, 3042, 3981, 3042, 4256, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {3 \int \frac {1}{(e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {1}{(e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2}dx}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \int \frac {1}{(e \sec (c+d x))^{7/2}}dx}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {3 \left (\frac {7 e^2 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 a^2}+\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}\right )}{5 a}+\frac {2 i}{15 d (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}}\) |
((2*I)/15)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3) + (3*((7*e^ 2*((2*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (5*((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)))/(11*a^2) + (((4*I)/11)*e^ 2)/(d*(e*Sec[c + d*x])^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))))/(5*a)
3.3.54.3.1 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*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ {a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x ] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && 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 = 9.31 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.17
| method | result | size |
| default | \(\frac {\frac {8 i \left (\cos ^{7}\left (d x +c \right )\right )}{15}+\frac {8 \sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{15}-\frac {2 i \left (\cos ^{5}\left (d x +c \right )\right )}{11}+\frac {14 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{165}+\frac {2 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{11}+\frac {6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{55}+\frac {2 i \sec \left (d x +c \right ) F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{11}+\frac {2 \sin \left (d x +c \right )}{11}}{a^{3} d \sqrt {e \sec \left (d x +c \right )}\, e}\) | \(217\) |
2/165/a^3/d/(e*sec(d*x+c))^(1/2)/e*(44*I*cos(d*x+c)^7+44*sin(d*x+c)*cos(d* x+c)^6-15*I*cos(d*x+c)^5+7*sin(d*x+c)*cos(d*x+c)^4+15*I*EllipticF(I*(-csc( d*x+c)+cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1)) ^(1/2)+9*cos(d*x+c)^2*sin(d*x+c)+15*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)+ 15*sin(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-55 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 235 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 446 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 218 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 73 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 480 i \, \sqrt {2} \sqrt {e} e^{\left (8 i \, d x + 8 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{2640 \, a^{3} d e^{2}} \]
1/2640*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-55*I*e^(10*I*d*x + 10* I*c) + 235*I*e^(8*I*d*x + 8*I*c) + 446*I*e^(6*I*d*x + 6*I*c) + 218*I*e^(4* I*d*x + 4*I*c) + 73*I*e^(2*I*d*x + 2*I*c) + 11*I)*e^(1/2*I*d*x + 1/2*I*c) - 480*I*sqrt(2)*sqrt(e)*e^(8*I*d*x + 8*I*c)*weierstrassPInverse(-4, 0, e^( I*d*x + I*c)))*e^(-8*I*d*x - 8*I*c)/(a^3*d*e^2)
\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (c + d x \right )} - 3 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )} - 3 \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )} + i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{3}} \]
I*Integral(1/((e*sec(c + d*x))**(3/2)*tan(c + d*x)**3 - 3*I*(e*sec(c + d*x ))**(3/2)*tan(c + d*x)**2 - 3*(e*sec(c + d*x))**(3/2)*tan(c + d*x) + I*(e* sec(c + d*x))**(3/2)), x)/a**3
Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]