\(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [121]

Optimal result

Integrand size = 36, antiderivative size = 129 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {4 \sqrt [4]{-1} a^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \] Output:

4*(-1)^(1/4)*a^2*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+4*a^2*(I*A+ 
B)*tan(d*x+c)^(1/2)/d-2/15*a^2*(5*A-7*I*B)*tan(d*x+c)^(3/2)/d+2/5*I*B*tan( 
d*x+c)^(3/2)*(a^2+I*a^2*tan(d*x+c))/d
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.72 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 a^2 \left ((15+15 i) \sqrt {2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right )+\sqrt {\tan (c+d x)} \left (-30 i A-30 B+5 (A-2 i B) \tan (c+d x)+3 B \tan ^2(c+d x)\right )\right )}{15 d} \] Input:

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

Output:

(-2*a^2*((15 + 15*I)*Sqrt[2]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[Tan[c + d*x]] 
)/Sqrt[2]] + Sqrt[Tan[c + d*x]]*((-30*I)*A - 30*B + 5*(A - (2*I)*B)*Tan[c 
+ d*x] + 3*B*Tan[c + d*x]^2)))/(15*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 4077, 27, 3042, 4075, 3042, 4011, 3042, 4016, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4077

\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) (a (5 A-3 i B)+a (5 i A+7 B) \tan (c+d x))dx+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) (a (5 A-3 i B)+a (5 i A+7 B) \tan (c+d x))dx+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) (a (5 A-3 i B)+a (5 i A+7 B) \tan (c+d x))dx+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (10 (A-i B) a^2+10 (i A+B) \tan (c+d x) a^2\right )dx-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (10 (A-i B) a^2+10 (i A+B) \tan (c+d x) a^2\right )dx-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{5} \left (\int \frac {10 a^2 (A-i B) \tan (c+d x)-10 a^2 (i A+B)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {20 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \frac {10 a^2 (A-i B) \tan (c+d x)-10 a^2 (i A+B)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {20 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 4016

\(\displaystyle \frac {1}{5} \left (\frac {200 a^4 (B+i A)^2 \int \frac {1}{-10 (i A+B) a^2-10 (A-i B) \tan (c+d x) a^2}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {20 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{5} \left (\frac {20 \sqrt [4]{-1} a^2 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {20 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )+\frac {2 i B \tan ^{\frac {3}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\)

Input:

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

Output:

(((2*I)/5)*B*Tan[c + d*x]^(3/2)*(a^2 + I*a^2*Tan[c + d*x]))/d + ((20*(-1)^ 
(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (20*a^2*(I* 
A + B)*Sqrt[Tan[c + d*x]])/d - (2*a^2*(5*A - (7*I)*B)*Tan[c + d*x]^(3/2))/ 
(3*d))/5
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4016

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2*(c^2/f)   Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b 
*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
 

rule 4075

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B 
*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* 
x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1]
 

rule 4077

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan 
[e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - 
 a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && 
GtQ[m, 1] &&  !LtQ[n, -1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (108 ) = 216\).

Time = 0.11 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.95

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {2 B \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 i B \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 A \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+4 i A \sqrt {\tan \left (d x +c \right )}+4 B \sqrt {\tan \left (d x +c \right )}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(252\)
default	\(\frac {a^{2} \left (-\frac {2 B \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {4 i B \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 A \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+4 i A \sqrt {\tan \left (d x +c \right )}+4 B \sqrt {\tan \left (d x +c \right )}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(252\)
parts	\(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (2 \sqrt {\tan \left (d x +c \right )}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \left (\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {A \,a^{2} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}-\frac {B \,a^{2} \left (\frac {2 \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (d x +c \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(436\)

Input:

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

1/d*a^2*(-2/5*B*tan(d*x+c)^(5/2)+4/3*I*B*tan(d*x+c)^(3/2)-2/3*A*tan(d*x+c) 
^(3/2)+4*I*A*tan(d*x+c)^(1/2)+4*B*tan(d*x+c)^(1/2)+1/4*(-2*B-2*I*A)*2^(1/2 
)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)-2^(1/2)*tan(d*x+ 
c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan( 
d*x+c)^(1/2)))+1/4*(-2*I*B+2*A)*2^(1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c) 
^(1/2)+1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan( 
d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (103) = 206\).

Time = 0.11 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.42 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 2 \, {\left ({\left (-35 i \, A - 43 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (-10 i \, A - 9 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-25 i \, A - 23 \, B\right )} a^{2}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

-1/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 
 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + 
sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(( 
-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I* 
c)/((-I*A - B)*a^2)) - 15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(4* 
I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^2*e^(2*I 
*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2* 
I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^ 
(-2*I*d*x - 2*I*c)/((-I*A - B)*a^2)) + 2*((-35*I*A - 43*B)*a^2*e^(4*I*d*x 
+ 4*I*c) + 6*(-10*I*A - 9*B)*a^2*e^(2*I*d*x + 2*I*c) + (-25*I*A - 23*B)*a^ 
2)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I 
*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

\[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- a^{2} \left (\int \left (- A \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int \left (- B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx\right ) \] Input:

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

Output:

-a**2*(Integral(-A*sqrt(tan(c + d*x)), x) + Integral(A*tan(c + d*x)**(5/2) 
, x) + Integral(-B*tan(c + d*x)**(3/2), x) + Integral(B*tan(c + d*x)**(7/2 
), x) + Integral(-2*I*A*tan(c + d*x)**(3/2), x) + Integral(-2*I*B*tan(c + 
d*x)**(5/2), x))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.50 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {12 \, B a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} + 20 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 120 \, {\left (-i \, A - B\right )} a^{2} \sqrt {\tan \left (d x + c\right )} + 15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2}}{30 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

-1/30*(12*B*a^2*tan(d*x + c)^(5/2) + 20*(A - 2*I*B)*a^2*tan(d*x + c)^(3/2) 
 + 120*(-I*A - B)*a^2*sqrt(tan(d*x + c)) + 15*(2*sqrt(2)*((I - 1)*A + (I + 
 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(( 
I - 1)*A + (I + 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))) 
) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan( 
d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqrt(tan(d*x 
 + c)) + tan(d*x + c) + 1))*a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (3 \, B a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} + 5 \, A a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 10 i \, B a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 30 i \, A a^{2} \sqrt {\tan \left (d x + c\right )} - 30 \, B a^{2} \sqrt {\tan \left (d x + c\right )} + 15 \, \sqrt {2} {\left (\left (i - 1\right ) \, A a^{2} + \left (i + 1\right ) \, B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )\right )}}{15 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algori 
thm="giac")
 

Output:

-2/15*(3*B*a^2*tan(d*x + c)^(5/2) + 5*A*a^2*tan(d*x + c)^(3/2) - 10*I*B*a^ 
2*tan(d*x + c)^(3/2) - 30*I*A*a^2*sqrt(tan(d*x + c)) - 30*B*a^2*sqrt(tan(d 
*x + c)) + 15*sqrt(2)*((I - 1)*A*a^2 + (I + 1)*B*a^2)*arctan(-(1/2*I - 1/2 
)*sqrt(2)*sqrt(tan(d*x + c))))/d
 

Mupad [B] (verification not implemented)

Time = 4.89 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.98 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {A\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,4{}\mathrm {i}}{d}-\frac {2\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {4\,B\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}-\frac {2\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (-B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \] Input:

int(tan(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)
 

Output:

(A*a^2*tan(c + d*x)^(1/2)*4i)/d - (2*A*a^2*tan(c + d*x)^(3/2))/(3*d) + (4* 
B*a^2*tan(c + d*x)^(1/2))/d + (B*a^2*tan(c + d*x)^(3/2)*4i)/(3*d) - (2*B*a 
^2*tan(c + d*x)^(5/2))/(5*d) + (2^(1/2)*A*a^2*log(4*A*a^2*d - 2^(1/2)*A*a^ 
2*d*tan(c + d*x)^(1/2)*(2 + 2i))*(1 + 1i))/d - (4i^(1/2)*A*a^2*log(4*A*a^2 
*d + 2*4i^(1/2)*A*a^2*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^2*log(- B*a^ 
2*d*4i - 2^(1/2)*B*a^2*d*tan(c + d*x)^(1/2)*(2 - 2i))*(1 - 1i))/d - ((-4i) 
^(1/2)*B*a^2*log(2*(-4i)^(1/2)*B*a^2*d*tan(c + d*x)^(1/2) - B*a^2*d*4i))/d
 

Reduce [F]

\[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^{2} \left (-2 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} b +20 \sqrt {\tan \left (d x +c \right )}\, a i +20 \sqrt {\tan \left (d x +c \right )}\, b -10 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a d i -10 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b d +5 \left (\int \sqrt {\tan \left (d x +c \right )}d x \right ) a d -5 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a d +10 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) b d i \right )}{5 d} \] Input:

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

Output:

(a**2*( - 2*sqrt(tan(c + d*x))*tan(c + d*x)**2*b + 20*sqrt(tan(c + d*x))*a 
*i + 20*sqrt(tan(c + d*x))*b - 10*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a 
*d*i - 10*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*b*d + 5*int(sqrt(tan(c + 
d*x)),x)*a*d - 5*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*a*d + 10*int(sq 
rt(tan(c + d*x))*tan(c + d*x)**2,x)*b*d*i))/(5*d)