\(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [213]

Optimal result

Integrand size = 28, antiderivative size = 88 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {i \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i a \tan (c+d x)-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{2 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((2*I)*a*Tan[c + d*x] - Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/S 
qrt[a + I*a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x 
]])/(2*a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 4029, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((-1/2 - I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Ta 
n[c + d*x]]])/(Sqrt[a]*d) + (I*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + 
 d*x]])
 

Defintions of rubi rules used

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 4029

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^n/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 

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. \(2 (69 ) = 138\).

Time = 1.72 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.94

method	result	size
derivativedivides	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+4 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(347\)
default	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}+4 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(347\)

Input:

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

Output:

1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(2*I*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^(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+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-4*I*(-I*a)^(1/2)*(a*tan(d*x+c) 
*(1+I*tan(d*x+c)))^(1/2)+4*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d 
*x+c)+I)^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. 301 vs. \(2 (62) = 124\).

Time = 0.13 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.42 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] 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 = 4.16 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.75 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\ln \left (2+\frac {\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )}{\sqrt {a}\,d}-\frac {2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )\,\left (d\,1{}\mathrm {i}-\frac {a\,d\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )}-\frac {\sqrt {\frac {1}{8}{}\mathrm {i}}\,\ln \left (-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+\frac {2\,{\left (-1\right )}^{3/4}\,\sqrt {2}\,\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+1{}\mathrm {i}\right )}{\sqrt {a}\,d} \] Input:

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

Output:

(log((a*tan(c + d*x)*2i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (a^ 
(1/2)*tan(c + d*x)^(1/2)*(4 + 4i))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2 
)) + 2)*(1/4 + 1i/4))/(a^(1/2)*d) - (2*tan(c + d*x)^(1/2))/(((a + a*tan(c 
+ d*x)*1i)^(1/2) - a^(1/2))*(d*1i - (a*d*tan(c + d*x))/((a + a*tan(c + d*x 
)*1i)^(1/2) - a^(1/2))^2)) - ((1i/8)^(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))/(a^(1/2)*d)
 

Reduce [F]

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

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

Output:

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