\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [186]

Optimal result

Integrand size = 38, antiderivative size = 150 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(i A+5 B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}} \] Output:

(-1/4-1/4*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x 
+c))^(1/2))/a^(3/2)/d+1/3*(I*A-B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(3 
/2)+1/6*(I*A+5*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\tan (c+d x)} \left (-\frac {3 i \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {6 (A-i B)+2 (i A+5 B) \tan (c+d x)}{(-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}\right )}{12 a d} \] Input:

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

Output:

(Sqrt[Tan[c + d*x]]*(((-3*I)*Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*T 
an[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/Sqrt[I*a*Tan[c + d*x]] + (6*(A 
- I*B) + 2*(I*A + 5*B)*Tan[c + d*x])/((-I + Tan[c + d*x])*Sqrt[a + I*a*Tan 
[c + d*x]])))/(12*a*d)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3042, 4078, 27, 3042, 4079, 27, 3042, 4027, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {a (i A-B)-2 a (A-2 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {a (i A-B)-2 a (A-2 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {3 a^2 (i A+B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {a (5 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {3}{2} (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {a (5 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {3}{2} (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {a (5 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {3 i a^2 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {a (5 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\left (\frac {3}{2}-\frac {3 i}{2}\right ) \sqrt {a} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a (5 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}\)

Input:

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

Output:

((I*A - B)*Sqrt[Tan[c + d*x]])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) - (((3/2 
 - (3*I)/2)*Sqrt[a]*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]]) 
/Sqrt[a + I*a*Tan[c + d*x]]])/d - (a*(I*A + 5*B)*Sqrt[Tan[c + d*x]])/(d*Sq 
rt[a + I*a*Tan[c + d*x]]))/(6*a^2)
 

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 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 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 4079

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 867 vs. \(2 (120 ) = 240\).

Time = 0.22 (sec) , antiderivative size = 868, normalized size of antiderivative = 5.79

method	result	size
derivativedivides	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-16 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 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+9 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}-3 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 )^{3}+3 i B \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 )^{3}-3 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 +9 B \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}-20 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+12 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+9 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 )-9 i B \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 )-3 B \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 -32 B \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 )-12 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{24 d \,a^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (i-\tan \left (d x +c \right )\right )^{3} \sqrt {-i a}}\)	\(868\)
default	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-16 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 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+9 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}-3 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 )^{3}+3 i B \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 )^{3}-3 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 +9 B \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}-20 i B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+12 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+9 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 )-9 i B \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 )-3 B \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 -32 B \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 )-12 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{24 d \,a^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (i-\tan \left (d x +c \right )\right )^{3} \sqrt {-i a}}\)	\(868\)
parts	\(\text {Expression too large to display}\)	\(921\)

Input:

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

Output:

-1/24/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(-16*I*A*(-I*a)^(1 
/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+4*A*(a*tan(d*x+c)*(1+ 
I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+9*I*A*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)^2-3*A*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)^3+3*I*B*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)^3-3*I 
*A*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+9*B*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)^2-20*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)) 
)^(1/2)*tan(d*x+c)^2+12*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 
1/2)+9*A*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)-9*I*B*2^(1/2)*l 
n(-(-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)-3*B*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-32*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+ 
c)-12*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+...
 

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. 443 vs. \(2 (110) = 220\).

Time = 0.12 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.95 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left (2 \, {\left (-i \, A - 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (-i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \] Input:

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

Output:

-1/12*(3*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))*e^(3*I*d* 
x + 3*I*c)*log((2*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^3*d^2))* 
e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 3*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 2*A*B - I 
*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-(2*sqrt(1/2)*a^2*d*sqrt((I*A^2 + 
 2*A*B - I*B^2)/(a^3*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x 
 + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + sqrt(2)*(2*(-I 
*A - 2*B)*e^(4*I*d*x + 4*I*c) + 3*(-I*A - B)*e^(2*I*d*x + 2*I*c) - I*A + B 
)*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^(-3*I*d*x - 3*I*c)/(a^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*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 TypeWarning, need 
 to choos
 

Mupad [F(-1)]

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

\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (16 \left (\int -\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{16 \tan \left (d x +c \right )^{3} i +16 \tan \left (d x +c \right )^{2}+16 \tan \left (d x +c \right ) i +16}d x \right ) a i -16 \left (\int -\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{16 \tan \left (d x +c \right )^{3} i +16 \tan \left (d x +c \right )^{2}+16 \tan \left (d x +c \right ) i +16}d x \right ) b -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) b i +\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3} i +\tan \left (d x +c \right )^{2}+\tan \left (d x +c \right ) i +1}d x \right ) a \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*(16*int(( - sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d 
*x))/(16*tan(c + d*x)**3*i + 16*tan(c + d*x)**2 + 16*tan(c + d*x)*i + 16), 
x)*a*i - 16*int(( - sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d* 
x))/(16*tan(c + d*x)**3*i + 16*tan(c + d*x)**2 + 16*tan(c + d*x)*i + 16),x 
)*b - int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2)/(t 
an(c + d*x)**3*i + tan(c + d*x)**2 + tan(c + d*x)*i + 1),x)*b*i + int((sqr 
t(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**3*i + tan(c + d*x 
)**2 + tan(c + d*x)*i + 1),x)*a))/a**2