Integrand size = 30, antiderivative size = 80 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i \sqrt {e \cos (c+d x)}}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d} \] Output:
2/3*I*(e*cos(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(1/2)-4/3*I*(e*cos(d*x+c)) ^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d
Time = 0.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {e \cos (c+d x)} (-i+2 \tan (c+d x))}{3 d \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sqrt[e*Cos[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
(2*Sqrt[e*Cos[c + d*x]]*(-I + 2*Tan[c + d*x]))/(3*d*Sqrt[a + I*a*Tan[c + d *x]])
Time = 0.53 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3998, 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.
\(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)} \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )\) |
Input:
Int[Sqrt[e*Cos[c + d*x]]/Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
Sqrt[e*Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]*(((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]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Time = 9.75 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52
method | result | size |
default | \(-\frac {2 \sqrt {e \cos \left (d x +c \right )}\, \left (i-2 \tan \left (d x +c \right )\right )}{3 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(42\) |
risch | \(-\frac {i \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(72\) |
Input:
int((e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/3/d*(e*cos(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(I-2*tan(d*x+c))
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{3 \, a d} \] Input:
integrate((e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fric as")
Output:
1/3*sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(2*I*d*x + 2*I*c) + I)*e^(-3/2*I*d*x - 3/2*I*c)/(a*d )
\[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\sqrt {e \cos {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate((e*cos(d*x+c))**(1/2)/(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Integral(sqrt(e*cos(c + d*x))/sqrt(I*a*(tan(c + d*x) - I)), x)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e} {\left (i \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )}}{3 \, \sqrt {a} d} \] Input:
integrate((e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxi ma")
Output:
1/3*sqrt(e)*(I*cos(3/2*d*x + 3/2*c) - 3*I*cos(1/3*arctan2(sin(3/2*d*x + 3/ 2*c), cos(3/2*d*x + 3/2*c))) + sin(3/2*d*x + 3/2*c) + 3*sin(1/3*arctan2(si n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))/(sqrt(a)*d)
Exception generated. \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),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 = 1.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )-3{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{3\,a\,d} \] Input:
int((e*cos(c + d*x))^(1/2)/(a + a*tan(c + d*x)*1i)^(1/2),x)
Output:
((e*cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) - 3i)*((a* (cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2 ))/(3*a*d)
\[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, i \left (-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{\cos \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) \tan \left (d x +c \right )^{2} d -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}\, \sin \left (d x +c \right )}{\cos \left (d x +c \right ) \tan \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) d -4 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d -4 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) d \right )}{a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:
int((e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2),x)
Output:
(sqrt(e)*sqrt(a)*i*( - 2*sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x)) - int ((sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x))*sin(c + d*x))/(cos(c + d*x)* tan(c + d*x)**2 + cos(c + d*x)),x)*tan(c + d*x)**2*d - int((sqrt(tan(c + d *x)*i + 1)*sqrt(cos(c + d*x))*sin(c + d*x))/(cos(c + d*x)*tan(c + d*x)**2 + cos(c + d*x)),x)*d - 4*int((sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x))* tan(c + d*x))/(tan(c + d*x)**2 + 1),x)*tan(c + d*x)**2*d - 4*int((sqrt(tan (c + d*x)*i + 1)*sqrt(cos(c + d*x))*tan(c + d*x))/(tan(c + d*x)**2 + 1),x) *d))/(a*d*(tan(c + d*x)**2 + 1))