\(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [128]

Optimal result

Integrand size = 36, antiderivative size = 171 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {8 a^3 (i A+B) \sqrt {\tan (c+d x)}}{d}-\frac {8 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{105 d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d} \] Output:

8*(-1)^(1/4)*a^3*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+8*a^3*(I*A+ 
B)*tan(d*x+c)^(1/2)/d-8/105*a^3*(21*A-23*I*B)*tan(d*x+c)^(3/2)/d+2/7*I*a*B 
*tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^2/d-2/35*(7*A-11*I*B)*tan(d*x+c)^(3/2 
)*(a^3+I*a^3*tan(d*x+c))/d
 

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.65 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 a^3 \left (420 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} \left (420 (i A+B)-35 (3 A-4 i B) \tan (c+d x)-21 i (A-3 i B) \tan ^2(c+d x)-15 i B \tan ^3(c+d x)\right )\right )}{105 d} \] Input:

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]) 
,x]
 

Output:

(2*a^3*(420*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + S 
qrt[Tan[c + d*x]]*(420*(I*A + B) - 35*(3*A - (4*I)*B)*Tan[c + d*x] - (21*I 
)*(A - (3*I)*B)*Tan[c + d*x]^2 - (15*I)*B*Tan[c + d*x]^3)))/(105*d)
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4077, 27, 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.

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {2}{7} \int \frac {1}{2} \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^2 (a (7 A-3 i B)+a (7 i A+11 B) \tan (c+d x))dx+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^2 (a (7 A-3 i B)+a (7 i A+11 B) \tan (c+d x))dx+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^2 (a (7 A-3 i B)+a (7 i A+11 B) \tan (c+d x))dx+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int 2 \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) \left (2 (7 A-6 i B) a^2+(21 i A+23 B) \tan (c+d x) a^2\right )dx-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) \left (2 (7 A-6 i B) a^2+(21 i A+23 B) \tan (c+d x) a^2\right )dx-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a) \left (2 (7 A-6 i B) a^2+(21 i A+23 B) \tan (c+d x) a^2\right )dx-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\int \sqrt {\tan (c+d x)} \left (35 (A-i B) a^3+35 (i A+B) \tan (c+d x) a^3\right )dx-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\int \sqrt {\tan (c+d x)} \left (35 (A-i B) a^3+35 (i A+B) \tan (c+d x) a^3\right )dx-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\int \frac {35 a^3 (A-i B) \tan (c+d x)-35 a^3 (i A+B)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {70 a^3 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\int \frac {35 a^3 (A-i B) \tan (c+d x)-35 a^3 (i A+B)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {70 a^3 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 4016

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\frac {2450 a^6 (B+i A)^2 \int \frac {1}{-35 (i A+B) a^3-35 (A-i B) \tan (c+d x) a^3}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {70 a^3 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{7} \left (\frac {4}{5} \left (\frac {70 \sqrt [4]{-1} a^3 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^3 (21 A-23 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {70 a^3 (B+i A) \sqrt {\tan (c+d x)}}{d}\right )-\frac {2 (7 A-11 i B) \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\right )+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}{7 d}\)

Input:

Int[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 

Output:

(((2*I)/7)*a*B*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2)/d + ((-2*(7*A 
- (11*I)*B)*Tan[c + d*x]^(3/2)*(a^3 + I*a^3*Tan[c + d*x]))/(5*d) + (4*((70 
*(-1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (70*a 
^3*(I*A + B)*Sqrt[Tan[c + d*x]])/d - (2*a^3*(21*A - (23*I)*B)*Tan[c + d*x] 
^(3/2))/(3*d)))/5)/7
 

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 [A] (verified)

Time = 0.08 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.61

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {2 i B \tan \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {2 i A \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {6 B \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 i B \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-2 A \tan \left (d x +c \right )^{\frac {3}{2}}+8 i A \sqrt {\tan \left (d x +c \right )}+8 B \sqrt {\tan \left (d x +c \right )}+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(276\)
default	\(\frac {a^{3} \left (-\frac {2 i B \tan \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {2 i A \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {6 B \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 i B \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-2 A \tan \left (d x +c \right )^{\frac {3}{2}}+8 i A \sqrt {\tan \left (d x +c \right )}+8 B \sqrt {\tan \left (d x +c \right )}+\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(276\)
parts	\(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \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}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\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 (3 i B \,a^{3}-3 A \,a^{3}\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^{3} \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 {i B \,a^{3} \left (\frac {2 \tan \left (d x +c \right )^{\frac {7}{2}}}{7}-\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}\)	\(561\)

Input:

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

1/d*a^3*(-2/7*I*B*tan(d*x+c)^(7/2)-2/5*I*A*tan(d*x+c)^(5/2)-6/5*B*tan(d*x+ 
c)^(5/2)+8/3*I*B*tan(d*x+c)^(3/2)-2*A*tan(d*x+c)^(3/2)+8*I*A*tan(d*x+c)^(1 
/2)+8*B*tan(d*x+c)^(1/2)+1/4*(-4*I*A-4*B)*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*(-4*I*B 
+4*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. 498 vs. \(2 (137) = 274\).

Time = 0.11 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.91 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (105 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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^{3}}\right ) - 105 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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^{3}}\right ) + 2 \, {\left ({\left (-273 i \, A - 319 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (-336 i \, A - 323 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-567 i \, A - 551 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (-42 i \, A - 41 \, B\right )} a^{3}\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 )}}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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))^3*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

-2/105*(105*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) 
 + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B 
)*a^3*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/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^3)) - 105*sqrt(-(-I*A^2 - 2*A 
*B + I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3* 
d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqr 
t(-(-I*A^2 - 2*A*B + I*B^2)*a^6/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^3)) + 2*((-273*I*A - 319*B)*a^3*e^(6*I*d*x + 6*I*c) + 2*(-33 
6*I*A - 323*B)*a^3*e^(4*I*d*x + 4*I*c) + (-567*I*A - 551*B)*a^3*e^(2*I*d*x 
 + 2*I*c) + 4*(-42*I*A - 41*B)*a^3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^( 
2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 
 3*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

\[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- i a^{3} \left (\int \left (- 3 A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {9}{2}}{\left (c + d x \right )}\, dx + \int i A \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \left (- 3 i A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int i B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 i B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx\right ) \] Input:

integrate(tan(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
 

Output:

-I*a**3*(Integral(-3*A*tan(c + d*x)**(3/2), x) + Integral(A*tan(c + d*x)** 
(7/2), x) + Integral(-3*B*tan(c + d*x)**(5/2), x) + Integral(B*tan(c + d*x 
)**(9/2), x) + Integral(I*A*sqrt(tan(c + d*x)), x) + Integral(-3*I*A*tan(c 
 + d*x)**(5/2), x) + Integral(I*B*tan(c + d*x)**(3/2), x) + Integral(-3*I* 
B*tan(c + d*x)**(7/2), x))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.26 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {30 i \, B a^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 42 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 70 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} + 840 \, {\left (-i \, A - B\right )} a^{3} \sqrt {\tan \left (d x + c\right )} + 105 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3}}{105 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

-1/105*(30*I*B*a^3*tan(d*x + c)^(7/2) + 42*(I*A + 3*B)*a^3*tan(d*x + c)^(5 
/2) + 70*(3*A - 4*I*B)*a^3*tan(d*x + c)^(3/2) + 840*(-I*A - B)*a^3*sqrt(ta 
n(d*x + c)) + 105*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(s 
qrt(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^3)/d
 

Giac [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (15 i \, B a^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 21 i \, A a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 63 \, B a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 105 \, A a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} - 140 i \, B a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} - 420 i \, A a^{3} \sqrt {\tan \left (d x + c\right )} - 420 \, B a^{3} \sqrt {\tan \left (d x + c\right )} + 210 \, \sqrt {2} {\left (\left (i - 1\right ) \, A a^{3} + \left (i + 1\right ) \, B a^{3}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )\right )}}{105 \, d} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algori 
thm="giac")
 

Output:

-2/105*(15*I*B*a^3*tan(d*x + c)^(7/2) + 21*I*A*a^3*tan(d*x + c)^(5/2) + 63 
*B*a^3*tan(d*x + c)^(5/2) + 105*A*a^3*tan(d*x + c)^(3/2) - 140*I*B*a^3*tan 
(d*x + c)^(3/2) - 420*I*A*a^3*sqrt(tan(d*x + c)) - 420*B*a^3*sqrt(tan(d*x 
+ c)) + 210*sqrt(2)*((I - 1)*A*a^3 + (I + 1)*B*a^3)*arctan(-(1/2*I - 1/2)* 
sqrt(2)*sqrt(tan(d*x + c))))/d
 

Mupad [B] (verification not implemented)

Time = 6.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.71 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {A\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,8{}\mathrm {i}}{d}-\frac {2\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{d}-\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {8\,B\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,8{}\mathrm {i}}{3\,d}-\frac {6\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,2{}\mathrm {i}}{7\,d}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \] Input:

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

Output:

(A*a^3*tan(c + d*x)^(1/2)*8i)/d - (2*A*a^3*tan(c + d*x)^(3/2))/d - (A*a^3* 
tan(c + d*x)^(5/2)*2i)/(5*d) + (8*B*a^3*tan(c + d*x)^(1/2))/d + (B*a^3*tan 
(c + d*x)^(3/2)*8i)/(3*d) - (6*B*a^3*tan(c + d*x)^(5/2))/(5*d) - (B*a^3*ta 
n(c + d*x)^(7/2)*2i)/(7*d) + (2^(1/2)*A*a^3*log(8*A*a^3*d - 2^(1/2)*A*a^3* 
d*tan(c + d*x)^(1/2)*(4 + 4i))*(2 + 2i))/d - (16i^(1/2)*A*a^3*log(8*A*a^3* 
d + 2*16i^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^3*log(- B*a^ 
3*d*8i - 2^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((-16i 
)^(1/2)*B*a^3*log(2*(-16i)^(1/2)*B*a^3*d*tan(c + d*x)^(1/2) - B*a^3*d*8i)) 
/d
 

Reduce [F]

\[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^{3} \left (-2 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} a i -6 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} b +40 \sqrt {\tan \left (d x +c \right )}\, a i +40 \sqrt {\tan \left (d x +c \right )}\, b -20 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a d i -20 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b d +5 \left (\int \sqrt {\tan \left (d x +c \right )}d x \right ) a d -5 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}d x \right ) b d i -15 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a d +15 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) b d i \right )}{5 d} \] Input:

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)
 

Output:

(a**3*( - 2*sqrt(tan(c + d*x))*tan(c + d*x)**2*a*i - 6*sqrt(tan(c + d*x))* 
tan(c + d*x)**2*b + 40*sqrt(tan(c + d*x))*a*i + 40*sqrt(tan(c + d*x))*b - 
20*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a*d*i - 20*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)**4,x)*b*d*i - 15*int(sqrt(tan(c + d*x))*tan(c + d*x) 
**2,x)*a*d + 15*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*b*d*i))/(5*d)