Integrand size = 30, antiderivative size = 80 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{3 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}} \] Output:
2/3*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-4/3*I*(a+I*a*tan(d*x +c))^(1/2)/a/d/(e*sec(d*x+c))^(1/2)
Time = 0.58 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-2 i+4 \tan (c+d x)}{3 d \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[1/(Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]),x]
Output:
(-2*I + 4*Tan[c + d*x])/(3*d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x ]])
Time = 0.38 (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.
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\) |
Input:
Int[1/(Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]),x]
Output:
((2*I)/3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3) *Sqrt[a + I*a*Tan[c + d*x]])/(a*d*Sqrt[e*Sec[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 = 6.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {2 \left (i-2 \tan \left (d x +c \right )\right )}{3 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}}\) | \(42\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{3 \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(85\) |
Input:
int(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOS E)
Output:
-2/3/d/(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)*(I-2*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{3 \, a d e} \] Input:
integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fr icas")
Output:
1/3*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(- 3*I*e^(4*I*d*x + 4*I*c) - 2*I*e^(2*I*d*x + 2*I*c) + I)*e^(-3/2*I*d*x - 3/2 *I*c)/(a*d*e)
\[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(1/(e*sec(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(1/(sqrt(e*sec(c + d*x))*sqrt(I*a*(tan(c + d*x) - I))), x)
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )}{3 \, \sqrt {a} d \sqrt {e}} \] Input:
integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="ma xima")
Output:
1/3*(I*cos(3/2*d*x + 3/2*c) - 3*I*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), co s(3/2*d*x + 3/2*c))) + sin(3/2*d*x + 3/2*c) + 3*sin(1/3*arctan2(sin(3/2*d* x + 3/2*c), cos(3/2*d*x + 3/2*c))))/(sqrt(a)*d*sqrt(e))
Exception generated. \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="gi ac")
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 = 1.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (-2\,\sin \left (c+d\,x\right )+\cos \left (c+d\,x\right )\,1{}\mathrm {i}\right )}{3\,d\,e\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:
int(1/((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)
Output:
-(2*(e/cos(c + d*x))^(1/2)*(cos(c + d*x)*1i - 2*sin(c + d*x)))/(3*d*e*((a* (cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2 ))
\[ \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, \left (-\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) i +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right )}{a e} \] Input:
int(1/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x)
Output:
(sqrt(e)*sqrt(a)*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan( c + d*x))/(sec(c + d*x)*tan(c + d*x)**2 + sec(c + d*x)),x)*i + int((sqrt(s ec(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(sec(c + d*x)*tan(c + d*x)**2 + sec (c + d*x)),x)))/(a*e)