\(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\) [751]

Optimal result

Integrand size = 43, antiderivative size = 103 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {4 a^2 (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{3/2}}{3 c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2 f} \] Output:

4*a^2*(I*A+B)*(c-I*c*tan(f*x+e))^(1/2)/f-2/3*a^2*(I*A+3*B)*(c-I*c*tan(f*x+ 
e))^(3/2)/c/f+2/5*a^2*B*(c-I*c*tan(f*x+e))^(5/2)/c^2/f
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a^2 c (i+\tan (e+f x)) \left (25 A-18 i B+(5 i A+9 B) \tan (e+f x)+3 i B \tan ^2(e+f x)\right )}{15 f \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

(2*a^2*c*(I + Tan[e + f*x])*(25*A - (18*I)*B + ((5*I)*A + 9*B)*Tan[e + f*x 
] + (3*I)*B*Tan[e + f*x]^2))/(15*f*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 96, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4071

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 c \left (-\frac {2 (3 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c^2}+\frac {4 (B+i A) \sqrt {c-i c \tan (e+f x)}}{c}+\frac {2 B (c-i c \tan (e+f x))^{5/2}}{5 c^3}\right )}{f}\)

Input:

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

Output:

(a^2*c*((4*(I*A + B)*Sqrt[c - I*c*Tan[e + f*x]])/c - (2*(I*A + 3*B)*(c - I 
*c*Tan[e + f*x])^(3/2))/(3*c^2) + (2*B*(c - I*c*Tan[e + f*x])^(5/2))/(5*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 = 0.83 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (3 i B c -c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {c -i c \tan \left (f x +e \right )}\, \left (-i B c +c A \right ) c \right )}{f \,c^{2}}\)	\(84\)
default	\(\frac {2 i a^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (3 i B c -c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 \sqrt {c -i c \tan \left (f x +e \right )}\, \left (-i B c +c A \right ) c \right )}{f \,c^{2}}\)	\(84\)
parts	\(\frac {i A \,a^{2} \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{f}+\frac {a^{2} \left (2 i A +B \right ) \left (2 \sqrt {c -i c \tan \left (f x +e \right )}-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 B \,a^{2} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}+\frac {c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f \,c^{2}}-\frac {2 i a^{2} \left (-2 i B +A \right ) \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f c}\)	\(264\)

Input:

int((a+I*a*tan(f*x+e))^2*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(1/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

2*I/f*a^2/c^2*(-1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)+1/3*(3*I*B*c-c*A)*(c-I*c* 
tan(f*x+e))^(3/2)+2*(c-I*c*tan(f*x+e))^(1/2)*(-I*B*c+c*A)*c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {4 \, \sqrt {2} {\left (15 \, {\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 5 \, {\left (-5 i \, A - 3 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-5 i \, A - 3 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:

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

Output:

-4/15*sqrt(2)*(15*(-I*A - B)*a^2*e^(4*I*f*x + 4*I*e) + 5*(-5*I*A - 3*B)*a^ 
2*e^(2*I*f*x + 2*I*e) + 2*(-5*I*A - 3*B)*a^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) 
+ 1))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a^{2} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 3 i \, B\right )} a^{2} c - 30 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{15 \, c^{2} f} \] Input:

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

Output:

-2/15*I*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*a^2 + 5*(-I*c*tan(f*x + e) + 
c)^(3/2)*(A - 3*I*B)*a^2*c - 30*sqrt(-I*c*tan(f*x + e) + c)*(A - I*B)*a^2* 
c^2)/(c^2*f)
 

Giac [F(-2)]

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

integrate((a+I*a*tan(f*x+e))^2*(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
 

Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.34 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2\,a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,250{}\mathrm {i}+174\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,375{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,150{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,25{}\mathrm {i}+267\,B\,\cos \left (2\,e+2\,f\,x\right )+114\,B\,\cos \left (4\,e+4\,f\,x\right )+21\,B\,\cos \left (6\,e+6\,f\,x\right )-25\,A\,\sin \left (2\,e+2\,f\,x\right )-20\,A\,\sin \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,45{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,36{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,9{}\mathrm {i}\right )}{15\,f\,\left (15\,\cos \left (2\,e+2\,f\,x\right )+6\,\cos \left (4\,e+4\,f\,x\right )+\cos \left (6\,e+6\,f\,x\right )+10\right )} \] Input:

int((A + B*tan(e + f*x))*(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)*(A*250i + 174*B + A*cos(2*e + 2*f*x)*375i + A*cos(4*e + 4*f*x 
)*150i + A*cos(6*e + 6*f*x)*25i + 267*B*cos(2*e + 2*f*x) + 114*B*cos(4*e + 
 4*f*x) + 21*B*cos(6*e + 6*f*x) - 25*A*sin(2*e + 2*f*x) - 20*A*sin(4*e + 4 
*f*x) - 5*A*sin(6*e + 6*f*x) + B*sin(2*e + 2*f*x)*45i + B*sin(4*e + 4*f*x) 
*36i + B*sin(6*e + 6*f*x)*9i))/(15*f*(15*cos(2*e + 2*f*x) + 6*cos(4*e + 4* 
f*x) + cos(6*e + 6*f*x) + 10))
 

Reduce [F]

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

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

Output:

(sqrt(c)*a**2*(2*sqrt( - tan(e + f*x)*i + 1)*a*i - int(sqrt( - tan(e + f*x 
)*i + 1)*tan(e + f*x)**3,x)*b*f - int(sqrt( - tan(e + f*x)*i + 1)*tan(e + 
f*x)**2,x)*a*f + 2*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)*a*f*i + int(sqrt( - ta 
n(e + f*x)*i + 1)*tan(e + f*x),x)*b*f))/f