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

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))*2^(1/2)/a^(1/2)/(c-I*d)^(1/2)/f-(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)
 

Mathematica [A] (verified)

Time = 0.73 (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]])
 

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)
 

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]
 

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 = 0.78 (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*(-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*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan( 
f*x+e)))*c^2*d*tan(f*x+e)^2+2*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln((3*a*c+I*a*t 
an(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*(c+d*ta 
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*c^3*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*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/( 
I+tan(f*x+e)))*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*(c+d*tan(f*x+e)) 
*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*d^3-2^(1/2)*(-a*(I*d-c))^(1/2)*l 
n((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*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*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*(c+d*tan(f*x+e))*(1+I*tan(f*x 
+e)))^(1/2))/(I+tan(f*x+e)))*c*d^2*tan(f*x+e)^2-4*I*c*d^2*(a*(c+d*tan(f*x+ 
e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f* 
x+e)))^(1/2)*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*(c+d*tan(f*x+e))*(1+ 
I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*c^2*d*tan(f*x+e)+2*2^(1/2)*(-a*(I...
 

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 = 0.13 (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 
)
 

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)
 

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)
                                                                                    
                                                                                    
 

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 TypeRecursive ass 
umption s
 

Mupad [B] (verification not implemented)

Time = 17.58 (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*...
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx=\text {too large to display} \] Input:

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

Output:

(2*sqrt(a)*(2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f* 
x)*c**2*d*i + 4*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + 
f*x)*c*d**2 - 2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + 
f*x)*d**3*i - sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**3*i - 3 
*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**2*d + 3*sqrt(tan(e + 
 f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c*d**2*i + sqrt(tan(e + f*x)*i + 1)* 
sqrt(tan(e + f*x)*d + c)*d**3 + 6*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e 
 + f*x)*d + c)*tan(e + f*x))/(3*tan(e + f*x)**3*c**4*d*i + 8*tan(e + f*x)* 
*3*c**3*d**2 - 14*tan(e + f*x)**3*c**2*d**3*i - 25*tan(e + f*x)**3*d**5*i 
+ 3*tan(e + f*x)**2*c**5*i + 8*tan(e + f*x)**2*c**4*d - 14*tan(e + f*x)**2 
*c**3*d**2*i - 25*tan(e + f*x)**2*c*d**4*i + 3*tan(e + f*x)*c**4*d*i + 8*t 
an(e + f*x)*c**3*d**2 - 14*tan(e + f*x)*c**2*d**3*i - 25*tan(e + f*x)*d**5 
*i + 3*c**5*i + 8*c**4*d - 14*c**3*d**2*i - 25*c*d**4*i),x)*tan(e + f*x)** 
2*c**8*f - 28*int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e 
 + f*x))/(3*tan(e + f*x)**3*c**4*d*i + 8*tan(e + f*x)**3*c**3*d**2 - 14*ta 
n(e + f*x)**3*c**2*d**3*i - 25*tan(e + f*x)**3*d**5*i + 3*tan(e + f*x)**2* 
c**5*i + 8*tan(e + f*x)**2*c**4*d - 14*tan(e + f*x)**2*c**3*d**2*i - 25*ta 
n(e + f*x)**2*c*d**4*i + 3*tan(e + f*x)*c**4*d*i + 8*tan(e + f*x)*c**3*d** 
2 - 14*tan(e + f*x)*c**2*d**3*i - 25*tan(e + f*x)*d**5*i + 3*c**5*i + 8*c* 
*4*d - 14*c**3*d**2*i - 25*c*d**4*i),x)*tan(e + f*x)**2*c**7*d*f*i - 60...