\(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [753]

Optimal result

Integrand size = 43, antiderivative size = 101 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {4 a^2 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 (i A+3 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \] Output:

-4/3*a^2*(I*A+B)/f/(c-I*c*tan(f*x+e))^(3/2)+2*a^2*(I*A+3*B)/c/f/(c-I*c*tan 
(f*x+e))^(1/2)+2*a^2*B*(c-I*c*tan(f*x+e))^(1/2)/c^2/f
 

Mathematica [A] (verified)

Time = 4.58 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 a^2 \left (-A+10 i B+3 (i A+5 B) \tan (e+f x)-3 i B \tan ^2(e+f x)\right )}{3 c f (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[((a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + 
 f*x])^(3/2),x]
 

Output:

(2*a^2*(-A + (10*I)*B + 3*(I*A + 5*B)*Tan[e + f*x] - (3*I)*B*Tan[e + f*x]^ 
2))/(3*c*f*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3042, 4071, 27, 86, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {a (i \tan (e+f x)+1) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 c \int \frac {(i \tan (e+f x)+1) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {a^2 c \int \left (\frac {2 (A-i B)}{(c-i c \tan (e+f x))^{5/2}}-\frac {i B}{c^2 \sqrt {c-i c \tan (e+f x)}}+\frac {3 i B-A}{c (c-i c \tan (e+f x))^{3/2}}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 c \left (\frac {2 (3 B+i A)}{c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {4 (B+i A)}{3 c (c-i c \tan (e+f x))^{3/2}}+\frac {2 B \sqrt {c-i c \tan (e+f x)}}{c^3}\right )}{f}\)

Input:

Int[((a + I*a*Tan[e + f*x])^2*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x]) 
^(3/2),x]
 

Output:

(a^2*c*((-4*(I*A + B))/(3*c*(c - I*c*Tan[e + f*x])^(3/2)) + (2*(I*A + 3*B) 
)/(c^2*Sqrt[c - I*c*Tan[e + f*x]]) + (2*B*Sqrt[c - I*c*Tan[e + f*x]])/c^3) 
)/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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 = 0.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-i B \sqrt {c -i c \tan \left (f x +e \right )}-\frac {2 c^{2} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {c \left (-3 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(79\)
default	\(\frac {2 i a^{2} \left (-i B \sqrt {c -i c \tan \left (f x +e \right )}-\frac {2 c^{2} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {c \left (-3 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}}\)	\(79\)
parts	\(\frac {2 i A \,a^{2} c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}+\frac {a^{2} \left (2 i A +B \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}-\frac {2 B \,a^{2} \left (-\sqrt {c -i c \tan \left (f x +e \right )}+\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8}-\frac {5 c}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c^{2}}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}-\frac {2 i a^{2} \left (-2 i B +A \right ) \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}-\frac {3}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f c}\)	\(342\)

Input:

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A + 7 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, {\left (-i \, A - 7 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \] Input:

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x 
, algorithm="fricas")
 

Output:

1/3*sqrt(2)*((-I*A - B)*a^2*e^(4*I*f*x + 4*I*e) + (I*A + 7*B)*a^2*e^(2*I*f 
*x + 2*I*e) - 2*(-I*A - 7*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f 
)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=- a^{2} \left (\int \left (- \frac {A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \] Input:

integrate((a+I*a*tan(f*x+e))**2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(3/2) 
,x)
 

Output:

-a**2*(Integral(-A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt 
(-I*c*tan(e + f*x) + c)), x) + Integral(A*tan(e + f*x)**2/(-I*c*sqrt(-I*c* 
tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Inte 
gral(-B*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sq 
rt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)**3/(-I*c*sqrt(-I* 
c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + In 
tegral(-2*I*A*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) 
+ c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(-2*I*B*tan(e + f*x)**2/(-I 
*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c 
)), x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 i \, {\left (\frac {3 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a^{2}}{c} - \frac {3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} - 2 \, {\left (A - i \, B\right )} a^{2} c}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, c f} \] Input:

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x 
, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),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 = 6.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,2{}\mathrm {i}+14\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+7\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,7{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,c^2\,f} \] Input:

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^2)/(c - c*tan(e + f*x)*1 
i)^(3/2),x)
 

Output:

(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)*(A*2i + 14*B + A*cos(2*e + 2*f*x)*1i - A*cos(4*e + 4*f*x)*1i + 
7*B*cos(2*e + 2*f*x) - B*cos(4*e + 4*f*x) - A*sin(2*e + 2*f*x) + A*sin(4*e 
 + 4*f*x) + B*sin(2*e + 2*f*x)*7i - B*sin(4*e + 4*f*x)*1i))/(3*c^2*f)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2),x)
 

Output:

(sqrt(c)*a**2*(2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)*a - 6*sqrt( - ta 
n(e + f*x)*i + 1)*tan(e + f*x)*b*i + 2*sqrt( - tan(e + f*x)*i + 1)*b + int 
(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e 
+ f*x)*i - 1),x)*tan(e + f*x)**2*a*f - 7*int(sqrt( - tan(e + f*x)*i + 1)/( 
tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)* 
*2*b*f*i + int(sqrt( - tan(e + f*x)*i + 1)/(tan(e + f*x)**3*i - tan(e + f* 
x)**2 + tan(e + f*x)*i - 1),x)*a*f - 7*int(sqrt( - tan(e + f*x)*i + 1)/(ta 
n(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*b*f*i + int((sq 
rt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**3*i - tan(e + f* 
x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*b*f*i + int((sqrt( - tan(e 
+ f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan( 
e + f*x)*i - 1),x)*b*f*i - 3*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x) 
**2)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + 
 f*x)**2*a*f - 6*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e 
+ f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*b*f 
*i - 3*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3* 
i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*a*f - 6*int((sqrt( - tan(e + 
f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e 
+ f*x)*i - 1),x)*b*f*i - 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/ 
(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f...