\(\int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1154]

Optimal result

Integrand size = 32, antiderivative size = 225 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {i (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(i c+d) (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \] Output:

-1/8*I*(c-I*d)^(5/2)*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^(5/2)/f+1/4*I*(c-I*d)^2*(c+d*t 
an(f*x+e))^(1/2)/a^2/f/(a+I*a*tan(f*x+e))^(1/2)+1/6*(I*c+d)*(c+d*tan(f*x+e 
))^(3/2)/a/f/(a+I*a*tan(f*x+e))^(3/2)+1/5*I*(c+d*tan(f*x+e))^(5/2)/f/(a+I* 
a*tan(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 3.46 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.36 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {\left (\frac {1}{120}-\frac {i}{120}\right ) \left ((12+12 i) a d (-i c+d) (-i+\tan (e+f x)) (c+d \tan (e+f x))^{5/2}-(12-12 i) a (c+i d) (c+d \tan (e+f x))^{7/2}+5 a \left (c^2+d^2\right ) (i-\tan (e+f x)) \left ((2+2 i) (c+d \tan (e+f x))^{5/2}+(i-\tan (e+f x)) \left ((2+2 i) d (c+d \tan (e+f x))^{3/2}+3 \left (c^2+d^2\right ) \left (\sqrt {\frac {i c+d}{a}} \arctan \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {i c+d}{a}} \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right ) \sqrt {a+i a \tan (e+f x)}+(1-i) \sqrt {c+d \tan (e+f x)}\right )\right )\right )\right )}{a^3 (c+i d)^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \] Input:

Integrate[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
 

Output:

((-1/120 + I/120)*((12 + 12*I)*a*d*((-I)*c + d)*(-I + Tan[e + f*x])*(c + d 
*Tan[e + f*x])^(5/2) - (12 - 12*I)*a*(c + I*d)*(c + d*Tan[e + f*x])^(7/2) 
+ 5*a*(c^2 + d^2)*(I - Tan[e + f*x])*((2 + 2*I)*(c + d*Tan[e + f*x])^(5/2) 
 + (I - Tan[e + f*x])*((2 + 2*I)*d*(c + d*Tan[e + f*x])^(3/2) + 3*(c^2 + d 
^2)*(Sqrt[(I*c + d)/a]*ArcTan[((1/2 + I/2)*Sqrt[(I*c + d)/a]*Sqrt[a + I*a* 
Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]]]*Sqrt[a + I*a*Tan[e + f*x]] + (1 - 
 I)*Sqrt[c + d*Tan[e + f*x]])))))/(a^3*(c + I*d)^2*f*(-I + Tan[e + f*x])^2 
*Sqrt[a + I*a*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4029, 3042, 4029, 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.

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4029

\(\displaystyle \frac {(c-i d) \int \frac {(c+d \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{3/2}}dx}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c-i d) \int \frac {(c+d \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{3/2}}dx}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4029

\(\displaystyle \frac {(c-i d) \left (\frac {(c-i d) \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}dx}{2 a}+\frac {i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c-i d) \left (\frac {(c-i d) \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}dx}{2 a}+\frac {i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4029

\(\displaystyle \frac {(c-i d) \left (\frac {(c-i d) \left (\frac {(c-i d) \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(c-i d) \left (\frac {(c-i d) \left (\frac {(c-i d) \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {(c-i d) \left (\frac {(c-i d) \left (\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {i a (c-i d) \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}}\right )}{2 a}+\frac {i (c+d \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\)

\(\Big \downarrow \) 221

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

Input:

Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
 

Output:

((I/5)*(c + d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(5/2)) + ((c 
- I*d)*(((I/3)*(c + d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^(3/2) 
) + ((c - I*d)*(((-I)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Ta 
n[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]])))/(2*a)) 
)/(2*a)
 

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)]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2558 vs. \(2 (178 ) = 356\).

Time = 0.53 (sec) , antiderivative size = 2559, normalized size of antiderivative = 11.37

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2559\)
default	\(\text {Expression too large to display}\)	\(2559\)

Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/240/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*(220*I*c^4*( 
a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2+136*(a*(c+d*tan(f* 
x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)^2+40*(a*(c+d*tan(f*x+e))*(1 
+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)^2+624*(a*(c+d*tan(f*x+e))*(1+I*tan( 
f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)-112*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)) 
)^(1/2)*c^2*d^2*tan(f*x+e)^3+308*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^( 
1/2)*d^4*tan(f*x+e)^2-112*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^ 
2*d^2+40*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)^3+ 
136*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)^3+15*I* 
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)))*2^(1/2)* 
(-a*(I*d-c))^(1/2)*c^4+15*I*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)))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^4+624*I*(a*(c+d*tan(f*x+e 
))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)^2+40*I*(a*(c+d*tan(f*x+e))*( 
1+I*tan(f*x+e)))^(1/2)*c^3*d*tan(f*x+e)+136*I*(a*(c+d*tan(f*x+e))*(1+I*tan 
(f*x+e)))^(1/2)*c*d^3*tan(f*x+e)+60*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)))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^4*tan(f*x+e)^3+60 
*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...
 

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. 655 vs. \(2 (169) = 338\).

Time = 0.12 (sec) , antiderivative size = 655, normalized size of antiderivative = 2.91 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fr 
icas")
 

Output:

-1/120*(15*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2* 
d^3 + 5*c*d^4 - I*d^5)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(2*I*sqrt(1/2)* 
a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5 
)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c^2 - 2*I*c 
*d - d^2)*e^(2*I*f*x + 2*I*e))*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)))/(c^2 - 2 
*I*c*d - d^2)) - 15*sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 
10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(-2*I* 
sqrt(1/2)*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d 
^4 - I*d^5)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*(c^2 - 2*I*c*d - d^2 + (c 
^2 - 2*I*c*d - d^2)*e^(2*I*f*x + 2*I*e))*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)) 
)/(c^2 - 2*I*c*d - d^2)) - sqrt(2)*(3*I*c^2 - 6*c*d - 3*I*d^2 - 23*(-I*c^2 
 - 2*c*d + I*d^2)*e^(6*I*f*x + 6*I*e) - 2*(-17*I*c^2 - 23*c*d + 6*I*d^2)*e 
^(4*I*f*x + 4*I*e) - 2*(-7*I*c^2 + 3*c*d - 4*I*d^2)*e^(2*I*f*x + 2*I*e))*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^(-5*I*f*x - 5*I*e)/(a^3*f)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((c+d*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(5/2),x)
 

Output:

Integral((c + d*tan(e + f*x))**(5/2)/(I*a*(tan(e + f*x) - I))**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="ma 
xima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="gi 
ac")
 

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 [F(-1)]

Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^(5/2),x)
 

Output:

int((c + d*tan(e + f*x))^(5/2)/(a + a*tan(e + f*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {-\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) c^{2}-\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) d^{2}-2 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) c d}{\sqrt {a}\, a^{2}} \] Input:

int((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x)
 

Output:

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