Integrand size = 28, antiderivative size = 152 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 a^3 d}+\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {20 i e^2}{77 d (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )} \]
10/77*e*sin(d*x+c)/a^3/d/(e*sec(d*x+c))^(1/2)+10/77*(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)/a^3/d+2/11*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan (d*x+c))^3+20/77*I*e^2/d/(e*sec(d*x+c))^(3/2)/(a^3+I*a^3*tan(d*x+c))
Time = 1.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \sec ^3(c+d x) \sqrt {e \sec (c+d x)} \left (46 i \cos (c+d x)+22 i \cos (3 (c+d x))-15 \sin (c+d x)+20 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-15 \sin (3 (c+d x))\right )}{154 a^3 d (-i+\tan (c+d x))^3} \]
((I/154)*Sec[c + d*x]^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)]) )/(a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.72 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3983, 3042, 3981, 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.
| \(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {5 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3981 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {5 \left (\frac {3 e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}\right )}{11 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}\) |
(((2*I)/11)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^3) + (5*((3*e^ 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*a^2) + (((4* I)/7)*e^2)/(d*(e*Sec[c + d*x])^(3/2)*(a^2 + I*a^2*Tan[c + d*x]))))/(11*a)
3.3.52.3.1 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 = 6.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.93
| method | result | size |
| default | \(-\frac {2 i \left (\left (-5 \cos \left (d x +c \right )-5\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (28 \left (\cos ^{4}\left (d x +c \right )\right )+3 \left (\cos ^{2}\left (d x +c \right )\right )+5\right )+\left (\cos ^{4}\left (d x +c \right )\right ) \left (-28 \left (\cos ^{2}\left (d x +c \right )\right )+11\right )\right ) \sqrt {e \sec \left (d x +c \right )}}{77 a^{3} d}\) | \(142\) |
-2/77*I/a^3/d*((-5*cos(d*x+c)-5)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos (d*x+c)+1))^(1/2)*EllipticF(I*(-csc(d*x+c)+cot(d*x+c)),I)+I*sin(d*x+c)*cos (d*x+c)*(28*cos(d*x+c)^4+3*cos(d*x+c)^2+5)+cos(d*x+c)^4*(-28*cos(d*x+c)^2+ 11))*(e*sec(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (37 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 61 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 31 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 40 i \, \sqrt {2} \sqrt {e} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{308 \, a^{3} d} \]
1/308*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(37*I*e^(6*I*d*x + 6*I*c) + 61*I*e^(4*I*d*x + 4*I*c) + 31*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(1/2*I*d*x + 1/2*I*c) - 40*I*sqrt(2)*sqrt(e)*e^(6*I*d*x + 6*I*c)*weierstrassPInverse (-4, 0, e^(I*d*x + I*c)))*e^(-6*I*d*x - 6*I*c)/(a^3*d)
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]
I*Integral(sqrt(e*sec(c + d*x))/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3 *tan(c + d*x) + I), x)/a**3
Exception generated. \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\sqrt {e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]