\(\int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [744]

Optimal result

Integrand size = 41, antiderivative size = 58 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 a (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a B \sqrt {c-i c \tan (e+f x)}}{c f} \] Output:

-2*a*(I*A+B)/f/(c-I*c*tan(f*x+e))^(1/2)-2*a*B*(c-I*c*tan(f*x+e))^(1/2)/c/f
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a (-A+2 i B+B \tan (e+f x))}{f \sqrt {c-i c \tan (e+f x)}} \] 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*I)*a*(-A + (2*I)*B + B*Tan[e + f*x]))/(f*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 59, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {a c \int \left (\frac {A-i B}{(c-i c \tan (e+f x))^{3/2}}+\frac {i B}{c \sqrt {c-i c \tan (e+f x)}}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a c \left (-\frac {2 (B+i A)}{c \sqrt {c-i c \tan (e+f x)}}-\frac {2 B \sqrt {c-i c \tan (e+f x)}}{c^2}\right )}{f}\)

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))/(c*Sqrt[c - I*c*Tan[e + f*x]]) - (2*B*Sqrt[c - I*c*Ta 
n[e + f*x]])/c^2))/f
 

Defintions of rubi rules used

rule 53

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])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4071

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]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {2 i a \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B -\frac {c \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(53\)
default	\(\frac {2 i a \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B -\frac {c \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(53\)
parts	\(\frac {2 i A a c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}+\frac {a \left (i A +B \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {2 a B \left (\sqrt {c -i c \tan \left (f x +e \right )}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(194\)

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*(I*(c-I*c*tan(f*x+e))^(1/2)*B-c*(A-I*B)/(c-I*c*tan(f*x+e))^(1/2) 
)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - 3 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \] Input:

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, 
algorithm="fricas")
 

Output:

sqrt(2)*((-I*A - B)*a*e^(2*I*f*x + 2*I*e) + (-I*A - 3*B)*a)*sqrt(c/(e^(2*I 
*f*x + 2*I*e) + 1))/(c*f)
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=i a \left (\int \left (- \frac {i A}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan ^{2}{\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {i B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \] 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*tan(e + f* 
x)/sqrt(-I*c*tan(e + f*x) + c), x) + Integral(B*tan(e + f*x)**2/sqrt(-I*c* 
tan(e + f*x) + c), x) + Integral(-I*B*tan(e + f*x)/sqrt(-I*c*tan(e + f*x) 
+ c), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i \, {\left (i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a - \frac {{\left (A - i \, B\right )} a c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}}{c f} \] 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*I*(I*sqrt(-I*c*tan(f*x + e) + c)*B*a - (A - I*B)*a*c/sqrt(-I*c*tan(f*x + 
 e) + c))/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] 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 TypeDone
 

Mupad [B] (verification not implemented)

Time = 5.68 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.83 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {a\,\sqrt {\frac {2\,c}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,1{}\mathrm {i}+3\,B+A\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}}{2}\right )\,1{}\mathrm {i}-A\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )+B\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}}{2}\right )+B\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}\right )}{c\,f} \] 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*((2*c)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(A*1i + 3*B + A*(exp(- e*2i - f 
*x*2i)/2 + exp(e*2i + f*x*2i)/2)*1i - A*((exp(- e*2i - f*x*2i)*1i)/2 - (ex 
p(e*2i + f*x*2i)*1i)/2) + B*(exp(- e*2i - f*x*2i)/2 + exp(e*2i + f*x*2i)/2 
) + B*((exp(- e*2i - f*x*2i)*1i)/2 - (exp(e*2i + f*x*2i)*1i)/2)*1i))/(c*f)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, a \left (2 \sqrt {-\tan \left (f x +e \right ) i +1}\, a i -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} b f -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+1}d x \right ) b f -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} a f +2 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} b f i -\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) a f +2 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) b f i +5 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} a f i +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} b f +5 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) a f i +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+1}d x \right ) b f \right )}{c f \left (\tan \left (f x +e \right )^{2}+1\right )} \] 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)**3)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f - int 
((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**2 + 1),x)*b* 
f - 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*a*f + 2*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*b*f*i - int((sqrt( - tan(e + 
 f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*a*f + 2*int((sqrt( 
- tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*b*f*i + 5* 
int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*ta 
n(e + f*x)**2*a*f*i + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan( 
e + f*x)**2 + 1),x)*tan(e + f*x)**2*b*f + 5*int((sqrt( - tan(e + f*x)*i + 
1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*a*f*i + int((sqrt( - tan(e + f*x 
)*i + 1)*tan(e + f*x))/(tan(e + f*x)**2 + 1),x)*b*f))/(c*f*(tan(e + f*x)** 
2 + 1))