\(\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx\) [188]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 (-1)^{3/4} \sqrt {a} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \] Output:

-2*(-1)^(3/4)*a^(1/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*ta 
n(d*x+c))^(1/2))/d-(1+I)*a^(1/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a 
+I*a*tan(d*x+c))^(1/2))/d
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {i a \tan (c+d x)} \left (2 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

(Sqrt[I*a*Tan[c + d*x]]*(2*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[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]]))/(d*Sqrt[Tan[c 
+ d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4046, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (\tan (c+d x) a+i a)}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {i a \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\)

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

Int[Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]
 

Output:

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

Defintions of rubi rules used

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 4046

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (82 ) = 164\).

Time = 1.71 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.34

method	result	size
derivativedivides	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+2 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right )-\sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right )+2 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(347\)
default	\(-\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+2 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right )-\sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right )+2 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(347\)

Input:

int(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*2^(1/2)*(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*I*(-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))*tan(d*x+c 
)-2^(1/2)*(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))*tan(d*x+c)+2*(-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*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-t 
an(d*x+c)+I)/(-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. 413 vs. \(2 (76) = 152\).

Time = 0.10 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.97 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {4 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {4 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

-1/2*sqrt(4*I*a/d^2)*log((sqrt(2)*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^(2*I*d*x + 2*I*c 
) + 1) + I*d*sqrt(4*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 1/2*sqrt 
(4*I*a/d^2)*log((sqrt(2)*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^(2*I*d*x + 2*I*c) + 1) - 
I*d*sqrt(4*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 1/2*sqrt(2*I*a/d^ 
2)*log((sqrt(2)*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^(2*I*d*x + 2*I*c) + 1) + I*d*sqrt( 
2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 1/2*sqrt(2*I*a/d^2)*log((s 
qrt(2)*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^(2*I*d*x + 2*I*c) + 1) - I*d*sqrt(2*I*a/d^2 
)*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
 

Sympy [F]

\[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sqrt {\tan {\left (c + d x \right )}}\, dx \] Input:

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

Output:

Integral(sqrt(I*a*(tan(c + d*x) - I))*sqrt(tan(c + d*x)), x)
 

Maxima [F]

\[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \sqrt {\tan \left (d x + c\right )} \,d x } \] Input:

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(I*a*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2),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 [B] (verification not implemented)

Time = 4.68 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.67 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {a}\,\ln \left (\frac {\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (2-2{}\mathrm {i}\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )}{d}-\frac {\sqrt {\frac {1}{2}{}\mathrm {i}}\,\sqrt {a}\,\ln \left (-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+\frac {2\,{\left (-1\right )}^{3/4}\,\sqrt {2}\,\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {a}\,\ln \left ({\left (-1\right )}^{3/4}+\frac {\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}\right )}{d}+\frac {\sqrt {2}\,\sqrt {a}\,\ln \left (\sqrt {2}\,\left (1-\mathrm {i}\right )+\frac {2\,\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}\right )\,\left (1+1{}\mathrm {i}\right )}{d} \] Input:

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

Output:

(a^(1/2)*log((a^(1/2)*tan(c + d*x)^(1/2)*(2 - 2i))/((a + a*tan(c + d*x)*1i 
)^(1/2) - a^(1/2)) - (a*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^( 
1/2))^2 + 1i)*(1/2 + 1i/2))/d - ((1i/2)^(1/2)*a^(1/2)*log((2*(-1)^(3/4)*2^ 
(1/2)*a^(1/2)*tan(c + d*x)^(1/2))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2) 
) - (a*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + 1i))/d 
- (4i^(1/2)*a^(1/2)*log((-1)^(3/4) + (a^(1/2)*tan(c + d*x)^(1/2))/((a + a* 
tan(c + d*x)*1i)^(1/2) - a^(1/2))))/d + (2^(1/2)*a^(1/2)*log(2^(1/2)*(1 - 
1i) + (2*a^(1/2)*tan(c + d*x)^(1/2))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1 
/2)))*(1 + 1i))/d
 

Reduce [F]

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

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

Output:

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