3.12.58 \(\int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx\) [1158]

3.12.58.1 Optimal result

Integrand size = 32, antiderivative size = 174 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{(i c+d) f \sqrt {a+i a \tan (e+f x)}}+\frac {2 d \sqrt {c+d \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {a+i a \tan (e+f x)}} \]

output

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

3.12.58.2 Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {-a (c-i d)} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{(c+i d) f \sqrt {a+i a \tan (e+f x)}} \]

input

Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]
 

output

((-I)*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a* 
Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[2]*Sqrt[-(a*(c - I*d))]*f) + (I*Sqrt[c + 
 d*Tan[e + f*x]])/((c + I*d)*f*Sqrt[a + I*a*Tan[e + f*x]])
 

3.12.58.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4031, 3042, 4029, 3042, 4027, 221}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4031

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

input

Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]),x]
 

output

(2*d*Sqrt[c + d*Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[a + I*a*Tan[e + f*x]]) 
+ (((-I)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/ 
(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[2]*Sqrt[a]*f) + (I*Sqrt 
[c + d*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x]]))/(c - I*d)
 

3.12.58.3.1 Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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)]
 

rule 4031

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

3.12.58.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1737 vs. \(2 (143 ) = 286\).

Time = 1.74 (sec) , antiderivative size = 1738, normalized size of antiderivative = 9.99

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1738\)
default	\(\text {Expression too large to display}\)	\(1738\)

input

int(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERB 
OSE)
 

output

1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)/a*(-4*I*c*d^2*(a*( 
1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*tan(f*x+e)-4*I*(a*(1+I*tan(f*x+e)) 
*(c+d*tan(f*x+e)))^(1/2)*d^3-6*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a* 
tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*t 
an(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c*d^2*tan(f*x+e)+3*I*2 
^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e 
)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2) 
)/(tan(f*x+e)+I))*c^2*d-2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e 
)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e) 
)*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^3*tan(f*x+e)^2+3*2^(1/2)*(-a* 
(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2) 
*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+ 
e)+I))*c*d^2*tan(f*x+e)^2-4*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)* 
c^2*d-I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d* 
tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e 
)))^(1/2))/(tan(f*x+e)+I))*d^3-6*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a* 
tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*t 
an(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*c^2*d*tan(f*x+e)+2*2^( 
1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+ 
2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2...
 

3.12.58.5 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. 381 vs. \(2 (136) = 272\).

Time = 0.28 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.19 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {{\left ({\left (-i \, a c + a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left ({\left (i \, a c + a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + {\left (i \, a c - a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left ({\left (-i \, a c - a d\right )} f \sqrt {-\frac {2 i}{{\left (i \, a c + a d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + 2 \, \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, {\left (i \, a c - a d\right )} f} \]

input

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

output

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

3.12.58.6 Sympy [F]

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

input

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

output

Integral(1/(sqrt(I*a*(tan(e + f*x) - I))*sqrt(c + d*tan(e + f*x))), x)
 

3.12.58.7 Maxima [F]

\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (f x + e\right ) + a} \sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]

input

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

output

integrate(1/(sqrt(I*a*tan(f*x + e) + a)*sqrt(d*tan(f*x + e) + c)), x)
 

3.12.58.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(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 TypeError: Bad Argument TypeError: Bad Argument Typeindex.cc inde 
x_m i_lex
 

3.12.58.9 Mupad [B] (verification not implemented)

Time = 21.00 (sec) , antiderivative size = 1508, normalized size of antiderivative = 8.67 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

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

output

(2*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(d*f*((c + d*tan(e + f*x))^( 
1/2) - c^(1/2))*((a*1i)/d + ((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2/(( 
c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (a^(1/2)*c^(1/2)*((a + a*tan(e + 
f*x)*1i)^(1/2) - a^(1/2))*2i)/(d*((c + d*tan(e + f*x))^(1/2) - c^(1/2))))) 
 - (2^(1/2)*atan(((2^(1/2)*((4*d^7*f*(a*c - a*d*1i)*((a + a*tan(e + f*x)*1 
i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) - 4*a^(3/2)*c^ 
(1/2)*d^7*f + (2^(1/2)*(d^7*(a^2*c*f^2*1i - a^2*d*f^2) + (d^8*(5*a*c*f^2 - 
 a*d*f^2*3i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + 
f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(3/2)*c^(3/2)*f*2i + 6*a^(3/2)*c^(1/2 
)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1 
/2) - c^(1/2))))/(a^(1/2)*f*(d*1i - c)^(1/2)) + (a^(1/2)*c^(1/2)*d^8*f*((a 
 + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*4i)/((c + d*tan(e + f*x))^(1/2) - 
 c^(1/2))^2)*1i)/(a^(1/2)*f*(d*1i - c)^(1/2)) - (2^(1/2)*(4*a^(3/2)*c^(1/2 
)*d^7*f - (4*d^7*f*(a*c - a*d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2) 
))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (2^(1/2)*(d^7*(a^2*c*f^2*1i - 
a^2*d*f^2) + (d^8*(5*a*c*f^2 - a*d*f^2*3i)*((a + a*tan(e + f*x)*1i)^(1/2) 
- a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (d^7*f*(a^(3/2)*c 
^(3/2)*f*2i + 6*a^(3/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1 
/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2))))/(a^(1/2)*f*(d*1i - c)^(1/2) 
) - (a^(1/2)*c^(1/2)*d^8*f*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2*...