\(\int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [164]

Optimal result

Integrand size = 38, antiderivative size = 146 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt [4]{-1} a^{3/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2+2 i) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(301\) vs. \(2(146)=292\).

Time = 4.86 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.06 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a \left (\sqrt [4]{-1} a A \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (1+i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {i a \tan (c+d x)}+a^{3/2} (A-i B) \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \tan (c+d x) (-i+\tan (c+d x))+a \sqrt {1+i \tan (c+d x)} \left (A (-1-i \tan (c+d x)) \sqrt {i a \tan (c+d x)}+\sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan (c+d x) \sqrt {a+i a \tan (c+d x)}\right )\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)} \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4076, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(2*(-1)^(1/4)*a^(3/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + I*a*Tan[c + d*x]]])/d + ((2 - 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I 
)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a*A*Sqrt 
[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])
 

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 4076

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

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. 520 vs. \(2 (118 ) = 236\).

Time = 0.25 (sec) , antiderivative size = 521, normalized size of antiderivative = 3.57

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+2 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+2 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )-\sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-2 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(521\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (4 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+2 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+2 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )-\sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-2 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\)	\(521\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (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 ) \tan \left (d x +c \right ) a +\sqrt {2}\, \sqrt {i a}\, \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 ) \tan \left (d x +c \right ) 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 ) \sqrt {-i a}\, a \tan \left (d x +c \right )+4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}+\frac {B \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}\, a^{2} \left (i \sqrt {2}\, \sqrt {i a}\, \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 )+4 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}-\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}-2 \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 )\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}}\)	\(644\)

Input:

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

Output:

1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(4*I*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*ta 
n(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*tan(d*x+c)-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*tan(d*x+ 
c)+2*B*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*tan(d*x+c)+2*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*tan(d*x+c)-(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*tan(d*x+c)-4*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a 
)^(1/2)-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))*(-I*a)^(1/2)*a*tan(d*x+c))/tan(d*x+c)^(1/2)/( 
I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*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. 748 vs. \(2 (110) = 220\).

Time = 0.10 (sec) , antiderivative size = 748, normalized size of antiderivative = 5.12 \[ \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*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)) - 2*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))*sq 
rt((-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)) + 4*sqrt(2)*(I*A*a*e^(3*I*d*x + 3*I*c) + I*A*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(-4*I*B^2*a^3/d^2)*(d*e^(2*I*d*x + 2*I 
*c) - d)*log((sqrt(2)*(B*a*e^(2*I*d*x + 2*I*c) + 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)) 
 + sqrt(-4*I*B^2*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(B*a)) - sqr 
t(-4*I*B^2*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((sqrt(2)*(B*a*e^(2*I*d 
*x + 2*I*c) + 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)) - sqrt(-4*I*B^2*a^3/d^2)*d*e^(I*d* 
x + I*c))*e^(-I*d*x - I*c)/(B*a)))/(d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*sqrt(a)*a*(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*b + 
 sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*a + int((sqrt(tan(c + d*x))*s 
qrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)*a*d - int(sqrt(ta 
n(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*tan(c + d*x)*b*d))/(t 
an(c + d*x)*d)