\(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [162]

Optimal result

Integrand size = 38, antiderivative size = 204 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(-1)^{3/4} a^{3/2} (12 A-11 i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}-\frac {(2+2 i) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a (4 i A+5 B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \] Output:

-1/4*(-1)^(3/4)*a^(3/2)*(12*A-11*I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c) 
^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-(2+2*I)*a^(3/2)*(A-I*B)*arctanh((1+I)*a 
^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d+1/4*a*(4*I*A+5*B)*tan( 
d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d+1/2*I*a*B*tan(d*x+c)^(3/2)*(a+I*a* 
tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 3.00 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.39 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {a \left (\frac {a (4 A-3 i B) \left ((-1)^{3/4} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )-\sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}\right ) (-i+\tan (c+d x))}{\sqrt {1+i \tan (c+d x)}}-2 a B \sqrt {\tan (c+d x)} (-i+\tan (c+d x))^2+\frac {8 (A-i B) \sqrt {i a \tan (c+d x)} \left (\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )}{\sqrt {\tan (c+d x)}}\right )}{4 d \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d 
*x]),x]
 

Output:

(a*((a*(4*A - (3*I)*B)*((-1)^(3/4)*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]] 
- Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])*(-I + Tan[c + d*x]))/Sqrt[1 
 + I*Tan[c + d*x]] - 2*a*B*Sqrt[Tan[c + d*x]]*(-I + Tan[c + d*x])^2 + (8*( 
A - I*B)*Sqrt[I*a*Tan[c + d*x]]*(Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sq 
rt[a]]*Sqrt[1 + I*Tan[c + d*x]] - Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c 
+ d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]]))/Sqrt[Tan 
[c + d*x]]))/(4*d*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4077, 27, 3042, 4080, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {1}{2} \int \frac {1}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (a (4 A-3 i B)+a (4 i A+5 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (a (4 A-3 i B)+a (4 i A+5 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (a (4 A-3 i B)+a (4 i A+5 B) \tan (c+d x))dx+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4080

\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (4 i A+5 B)-a^2 (12 A-11 i B) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a}+\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (4 i A+5 B)-a^2 (12 A-11 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (4 i A+5 B)-a^2 (12 A-11 i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {16 a^2 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a (11 B+12 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {16 a^2 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a (11 B+12 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {-\frac {32 i a^4 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-a (11 B+12 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(16-16 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-a (11 B+12 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(16-16 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^3 (11 B+12 i A) \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(16-16 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^3 (11 B+12 i A) \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} \left (\frac {a (5 B+4 i A) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {2 \sqrt [4]{-1} a^{5/2} (11 B+12 i A) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(16-16 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}\right )+\frac {i a B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

Input:

Int[Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x 
]
 

Output:

((I/2)*a*B*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d + (-1/2*((2*(- 
1)^(1/4)*a^(5/2)*((12*I)*A + 11*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + 
 d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((16 - 16*I)*a^(5/2)*(I*A + B)*Ar 
cTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d) 
/a + (a*((4*I)*A + 5*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d)/ 
4
 

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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 4027

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

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]
 

rule 4080

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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

rule 4082

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/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

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[(A*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (162 ) = 324\).

Time = 0.23 (sec) , antiderivative size = 565, normalized size of antiderivative = 2.77

method	result	size
derivativedivides	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-4 i B \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 i A \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +5 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -10 B \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -4 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -8 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(565\)
default	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-4 i B \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-8 i A \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-4 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +5 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -10 B \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -4 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -8 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(565\)
parts	\(\frac {A \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +2 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}-\frac {B \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )-11 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+4 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -10 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+16 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -4 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(766\)

Input:

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

Output:

-1/8/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-4*I*B*(a*tan(d*x+c) 
*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+4*I*A*ln(1/2* 
(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/( 
I*a)^(1/2))*a*(-I*a)^(1/2)-8*I*A*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1 
+I*tan(d*x+c)))^(1/2)-4*I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2) 
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I)) 
*a+5*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a-10*B*(-I*a)^(1/2)*(I*a)^(1/2)*(a* 
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+8*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d 
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*(-I*a)^(1/2)-4 
*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d 
*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a-8*ln(1/2*(2*I*a*tan(d* 
x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a 
*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(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. 829 vs. \(2 (150) = 300\).

Time = 0.11 (sec) , antiderivative size = 829, normalized size of antiderivative = 4.06 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, al 
gorithm="fricas")
 

Output:

1/8*(8*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I 
*c) + d)*log((sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*d*e^(I*d*x + 
 I*c) + sqrt(2)*((-I*A - B)*a*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a)*sqrt(a/( 
e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a)) - 8*sqrt(2)*sqrt(-(-I*A^2 - 
 2*A*B + I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-(sqrt(2)*sqrt(-( 
-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c) - sqrt(2)*((-I*A - B)*a 
*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c 
)/((-I*A - B)*a)) + 2*sqrt(2)*((4*I*A + 7*B)*a*e^(3*I*d*x + 3*I*c) + (4*I* 
A + 3*B)*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^( 
2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + sqrt((144*I*A^2 + 264*A 
*B - 121*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((12*I*A 
 + 11*B)*a*e^(2*I*d*x + 2*I*c) + (12*I*A + 11*B)*a)*sqrt(a/(e^(2*I*d*x + 2 
*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + 
 2*sqrt((144*I*A^2 + 264*A*B - 121*I*B^2)*a^3/d^2)*d*e^(I*d*x + I*c))*e^(- 
I*d*x - I*c)/((12*I*A + 11*B)*a)) - sqrt((144*I*A^2 + 264*A*B - 121*I*B^2) 
*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((12*I*A + 11*B)*a*e^(2 
*I*d*x + 2*I*c) + (12*I*A + 11*B)*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - 2*sqrt((144...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, al 
gorithm="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 [F(-1)]

Timed out. \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*a*( - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*a*i + int((sq 
rt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x),x)*a*d*i + int(sqr 
t(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2,x)*b*d*i + 3*int( 
sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*a*d*i + int(sq 
rt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*b*d))/d