Integrand size = 41, antiderivative size = 60 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \] Output:
2*a*(I*A+B)*(c-I*c*tan(f*x+e))^(1/2)/f-2/3*a*B*(c-I*c*tan(f*x+e))^(3/2)/c/ f
Time = 0.74 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a (3 i A+2 B+i B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]],x]
Output:
(2*a*((3*I)*A + 2*B + I*B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(3*f)
Time = 0.50 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {3042, 4071, 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.
\(\displaystyle \int (a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {a c \int \left (\frac {A-i B}{\sqrt {c-i c \tan (e+f x)}}+\frac {i B \sqrt {c-i c \tan (e+f x)}}{c}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a c \left (\frac {2 (B+i A) \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 B (c-i c \tan (e+f x))^{3/2}}{3 c^2}\right )}{f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]] ,x]
Output:
(a*c*((2*(I*A + B)*Sqrt[c - I*c*Tan[e + f*x]])/c - (2*B*(c - I*c*Tan[e + f *x])^(3/2))/(3*c^2)))/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[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.50 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A \right )}{f c}\) | \(66\) |
default | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A \right )}{f c}\) | \(66\) |
parts | \(\frac {i A a \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{f}+\frac {a \left (i A +B \right ) \left (2 \sqrt {c -i c \tan \left (f x +e \right )}-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 a B \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f c}\) | \(156\) |
Input:
int((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2),x,method= _RETURNVERBOSE)
Output:
2*I/f*a/c*(1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)-I*(c-I*c*tan(f*x+e))^(1/2)*B*c +(c-I*c*tan(f*x+e))^(1/2)*c*A)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 \, \sqrt {2} {\left (3 \, {\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A - B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
-2/3*sqrt(2)*(3*(-I*A - B)*a*e^(2*I*f*x + 2*I*e) + (-3*I*A - B)*a)*sqrt(c/ (e^(2*I*f*x + 2*I*e) + 1))/(f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=i a \left (\int \left (- i A \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int A \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\, dx + \int B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- i B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(1/2),x)
Output:
I*a*(Integral(-I*A*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(A*sqrt(-I*c* tan(e + f*x) + c)*tan(e + f*x), x) + Integral(B*sqrt(-I*c*tan(e + f*x) + c )*tan(e + f*x)**2, x) + Integral(-I*B*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x))
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 i \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B a + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a c\right )}}{3 \, c f} \] Input:
integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
2/3*I*(I*(-I*c*tan(f*x + e) + c)^(3/2)*B*a + 3*sqrt(-I*c*tan(f*x + e) + c) *(A - I*B)*a*c)/(c*f)
Exception generated. \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(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 TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar gument Ty
Time = 0.80 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.70 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (A\,3{}\mathrm {i}+2\,B+A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,3{}\mathrm {i}+2\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,f\,{\cos \left (e+f\,x\right )}^2} \] Input:
int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^( 1/2),x)
Output:
(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2 )*(A*3i + 2*B + A*(2*cos(e + f*x)^2 - 1)*3i + 2*B*(2*cos(e + f*x)^2 - 1) + B*sin(2*e + 2*f*x)*1i))/(3*f*cos(e + f*x)^2)
\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {\sqrt {c}\, a \left (2 \sqrt {-\tan \left (f x +e \right ) i +1}\, a i +\left (\int \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}d x \right ) b f i +\left (\int \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )d x \right ) b f \right )}{f} \] Input:
int((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2),x)
Output:
(sqrt(c)*a*(2*sqrt( - tan(e + f*x)*i + 1)*a*i + int(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2,x)*b*f*i + int(sqrt( - tan(e + f*x)*i + 1)*tan(e + f *x),x)*b*f))/f