Integrand size = 33, antiderivative size = 58 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {4 i a^2}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 i a^2 \sqrt {c-i c \tan (e+f x)}}{c f} \] Output:
-4*I*a^2/f/(c-I*c*tan(f*x+e))^(1/2)-2*I*a^2*(c-I*c*tan(f*x+e))^(1/2)/c/f
Time = 1.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 a^2 (3 i+\tan (e+f x))}{f \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^2/Sqrt[c - I*c*Tan[e + f*x]],x]
Output:
(-2*a^2*(3*I + Tan[e + f*x]))/(f*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4005, 3042, 3968, 53, 2009}
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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^2 c^2 \int \frac {\sec ^4(e+f x)}{(c-i c \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \int \frac {\sec (e+f x)^4}{(c-i c \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^2 \int \frac {i \tan (e+f x) c+c}{(c-i c \tan (e+f x))^{3/2}}d(-i c \tan (e+f x))}{c f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {i a^2 \int \left (\frac {2 c}{(c-i c \tan (e+f x))^{3/2}}-\frac {1}{\sqrt {c-i c \tan (e+f x)}}\right )d(-i c \tan (e+f x))}{c f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^2 \left (-\frac {4 c}{\sqrt {c-i c \tan (e+f x)}}-2 \sqrt {c-i c \tan (e+f x)}\right )}{c f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^2/Sqrt[c - I*c*Tan[e + f*x]],x]
Output:
(I*a^2*((-4*c)/Sqrt[c - I*c*Tan[e + f*x]] - 2*Sqrt[c - I*c*Tan[e + f*x]])) /(c*f)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.35 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 i a^{2} \left (-\sqrt {c -i c \tan \left (f x +e \right )}-\frac {2 c}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\) | \(47\) |
default | \(\frac {2 i a^{2} \left (-\sqrt {c -i c \tan \left (f x +e \right )}-\frac {2 c}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\) | \(47\) |
parts | \(\frac {2 i a^{2} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}+\frac {2 i a^{2} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {2 i a^{2} \left (\sqrt {c -i c \tan \left (f x +e \right )}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\) | \(195\) |
Input:
int((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
2*I/f*a^2/c*(-(c-I*c*tan(f*x+e))^(1/2)-2*c/(c-I*c*tan(f*x+e))^(1/2))
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 \, \sqrt {2} {\left (i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \] Input:
integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fric as")
Output:
-2*sqrt(2)*(I*a^2*e^(2*I*f*x + 2*I*e) + 2*I*a^2)*sqrt(c/(e^(2*I*f*x + 2*I* e) + 1))/(c*f)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=- a^{2} \left (\int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {1}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e))**(1/2),x)
Output:
-a**2*(Integral(tan(e + f*x)**2/sqrt(-I*c*tan(e + f*x) + c), x) + Integral (-2*I*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-1/sqrt(-I*c *tan(e + f*x) + c), x))
Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, {\left (\sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} + \frac {2 \, a^{2} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}}{c f} \] Input:
integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxi ma")
Output:
-2*I*(sqrt(-I*c*tan(f*x + e) + c)*a^2 + 2*a^2*c/sqrt(-I*c*tan(f*x + e) + c ))/(c*f)
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(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 = 1.99 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.33 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2\,a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )+2{}\mathrm {i}\right )}{c\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^2/(c - c*tan(e + f*x)*1i)^(1/2),x)
Output:
-(2*a^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x ) + 1))^(1/2)*(cos(2*e + 2*f*x)*1i - sin(2*e + 2*f*x) + 2i))/(c*f)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, a^{2} \left (2 \sqrt {-\tan \left (f x +e \right ) i +1}\, i -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f i -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f i -3 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f -3 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f +6 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f i +6 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) f i \right )}{c f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:
int((a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(1/2),x)
Output:
(sqrt(c)*a**2*(2*sqrt( - tan(e + f*x)*i + 1)*i - int((sqrt( - tan(e + f*x) *i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f*i - in t((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*f *i - 3*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f - 3*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x )**2)/(tan(e + f*x)**2 + 1),x)*f + 6*int((sqrt( - tan(e + f*x)*i + 1)*tan( e + f*x))/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f*i + 6*int((sqrt( - ta n(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*f*i))/(c*f*(tan(e + f*x)**2 + 1))