\(\int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [194]

Optimal result

Integrand size = 28, antiderivative size = 217 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {11 \sqrt [4]{-1} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 d}-\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \tan ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}+\frac {5 a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 5.84 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.29 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a \left (3 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)} (-i+\tan (c+d x))+8 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)} (-i+\tan (c+d x))+\sqrt {1+i \tan (c+d x)} \left (8 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+a \tan (c+d x) \left (5+7 i \tan (c+d x)-2 \tan ^2(c+d x)\right )\right )\right )}{4 d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(a*(3*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[Tan[c + d*x 
]]*(-I + Tan[c + d*x]) + 8*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]] 
*Sqrt[I*a*Tan[c + d*x]]*(-I + Tan[c + d*x]) + Sqrt[1 + I*Tan[c + d*x]]*((8 
*I)*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
d*x]]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + a*Tan[c + d*x]* 
(5 + (7*I)*Tan[c + d*x] - 2*Tan[c + d*x]^2))))/(4*d*Sqrt[1 + I*Tan[c + d*x 
]]*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4039, 27, 3042, 4078, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

\(\displaystyle \frac {1}{2} a \int \frac {\tan ^{\frac {3}{2}}(c+d x) (7 i \tan (c+d x) a+9 a)}{2 \sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} a \int \frac {\tan ^{\frac {3}{2}}(c+d x) (7 i \tan (c+d x) a+9 a)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} a \int \frac {\tan (c+d x)^{3/2} (7 i \tan (c+d x) a+9 a)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{4} a \left (\frac {2 i a \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {\frac {(16-16 i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-11 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}-\frac {5 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{4} a \left (\frac {2 i a \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {\frac {(16-16 i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {11 a^4 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}-\frac {5 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {1}{4} a \left (\frac {2 i a \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {\frac {(16-16 i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {22 a^4 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}-\frac {5 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} a \left (\frac {2 i a \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {\frac {22 \sqrt [4]{-1} a^{7/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(16-16 i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}-\frac {5 a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

Input:

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

Output:

-1/2*(a^2*Tan[c + d*x]^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (a*(((2*I)* 
a*Tan[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((22*(-1)^(1/4)*a^ 
(7/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
d*x]]])/d + ((16 - 16*I)*a^(7/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x 
]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/(2*a) - (5*a^2*Sqrt[Tan[c + d*x]]*Sqrt 
[a + I*a*Tan[c + d*x]])/d)/a^2))/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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4078

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

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. 404 vs. \(2 (171 ) = 342\).

Time = 1.63 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.87

method	result	size
derivativedivides	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-4 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-10 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-11 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+4 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +16 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -4 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(405\)
default	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-4 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-10 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-11 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+4 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +16 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -4 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{8 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(405\)

Input:

int(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-4*I*(I*a)^(1/2)*(-I 
*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-10*(-I*a)^(1/2) 
*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-11*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*(I*a)^(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))*2^(1/2)*a+16*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))*(-I*a)^(1/2)*a-4*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))*( 
I*a)^(1/2)*a)/(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. 600 vs. \(2 (161) = 322\).

Time = 0.12 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.76 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/4*(sqrt(2)*(7*a*e^(3*I*d*x + 3*I*c) + 3*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)) + 2*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-121/16*I*a^3/d^2)*log(1/11*( 
11*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s 
qrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + 8*sqrt(-121/ 
16*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - 2*(d*e^(2*I*d*x + 2 
*I*c) + d)*sqrt(-121/16*I*a^3/d^2)*log(1/11*(11*sqrt(2)*(a*e^(2*I*d*x + 2* 
I*c) + 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)) - 8*sqrt(-121/16*I*a^3/d^2)*d*e^(I*d*x + I* 
c))*e^(-I*d*x - I*c)/a) - 2*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I*a^3/d^2) 
*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + 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)) + sqr 
t(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + 2*(d*e^(2*I*d*x + 
 2*I*c) + d)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) 
+ 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)) - sqrt(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d* 
x - I*c)/a))/(d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(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 [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 
,x)*i + int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x))