Integrand size = 30, antiderivative size = 80 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}+\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}} \] Output:
2/5*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)+4/5*I*(e*sec(d*x+c)) ^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)
Time = 0.73 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {e \sec (c+d x)} (3+2 i \tan (c+d x))}{5 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sqrt[e*Sec[c + d*x]]/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(2*Sqrt[e*Sec[c + d*x]]*(3 + (2*I)*Tan[c + d*x]))/(5*a*d*(-I + Tan[c + d*x ])*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3983, 3042, 3969}
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.
\(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {2 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{5 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{5 a}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}\) |
Input:
Int[Sqrt[e*Sec[c + d*x]]/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(((2*I)/5)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((4*I )/5)*Sqrt[e*Sec[c + d*x]])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])
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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 0]
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]
Time = 8.70 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78
method | result | size |
default | \(-\frac {2 i \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right )-\cos \left (d x +c \right )^{2}-2\right ) \sqrt {e \sec \left (d x +c \right )}}{5 d a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(62\) |
Input:
int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/5*I/d*(I*sin(d*x+c)*cos(d*x+c)-cos(d*x+c)^2-2)*(e*sec(d*x+c))^(1/2)/a/( a*(1+I*tan(d*x+c)))^(1/2)
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{5 \, a^{2} d} \] Input:
integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fric as")
Output:
1/5*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(5 *I*e^(4*I*d*x + 4*I*c) + 6*I*e^(2*I*d*x + 2*I*c) + I)*e^(-5/2*I*d*x - 5/2* I*c)/(a^2*d)
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(sqrt(e*sec(c + d*x))/(I*a*(tan(c + d*x) - I))**(3/2), x)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {e} {\left (i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 i \, \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )}}{5 \, a^{\frac {3}{2}} d} \] Input:
integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxi ma")
Output:
1/5*sqrt(e)*(I*cos(5/2*d*x + 5/2*c) + 5*I*cos(1/5*arctan2(sin(5/2*d*x + 5/ 2*c), cos(5/2*d*x + 5/2*c))) + sin(5/2*d*x + 5/2*c) + 5*sin(1/5*arctan2(si n(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))/(a^(3/2)*d)
Exception generated. \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 1.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+5{}\mathrm {i}\right )}{5\,a\,d\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:
int((e/cos(c + d*x))^(1/2)/(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 5i))/(5* a*d*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))
\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {e}\, \sqrt {a}\, \left (2 \sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )+\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, i -3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) \tan \left (d x +c \right )^{2} d i -3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) d i \right )}{3 a^{2} d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:
int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(2*sqrt(e)*sqrt(a)*(2*sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x) + sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*i - 3*int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*tan(c + d*x)**2*d*i - 3*int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x))/(tan(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*d*i))/(3*a**2*d*(tan(c + d*x)**2 + 1))