\(\int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1010]

Optimal result

Integrand size = 35, antiderivative size = 153 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {6 i c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {2 i c (c-i c \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}} \] Output:

6*I*c^(5/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x 
+e))^(1/2))/a^(1/2)/f+3*I*c^2*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^ 
(1/2)/a/f+2*I*c*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.83 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.56 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {4 i c^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) \sqrt {2-2 i \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

((4*I)*c^3*Hypergeometric2F1[-3/2, -1/2, 1/2, (1 + I*Tan[e + f*x])/2]*Sqrt 
[2 - (2*I)*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e 
 + f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 57, 60, 45, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(a*c*(((2*I)*(c - I*c*Tan[e + f*x])^(3/2))/(a*Sqrt[a + I*a*Tan[e + f*x]]) 
- (3*c*(((-2*I)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[ 
a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[a] - (I*Sqrt[a + I*a*Tan[e + f*x]]*S 
qrt[c - I*c*Tan[e + f*x]])/a))/a))/f
 

Defintions of rubi rules used

rule 45

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 4006

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (124 ) = 248\).

Time = 0.35 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.96

method	result	size
derivativedivides	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -6 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+6 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-5 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f a \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (-\tan \left (f x +e \right )+i\right )^{2}}\)	\(300\)
default	\(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -6 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+6 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-5 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f a \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (-\tan \left (f x +e \right )+i\right )^{2}}\)	\(300\)

Input:

int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a*(3*I*ln(( 
a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c* 
tan(f*x+e)^2-3*I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/ 
2))/(a*c)^(1/2))*a*c-6*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+ 
e)+6*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1 
/2))*a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2- 
5*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/( 
a*c)^(1/2)/(-tan(f*x+e)+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. 360 vs. \(2 (115) = 230\).

Time = 0.10 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.35 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {c^{5}}{a f^{2}}} a f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {c^{5}}{a f^{2}}}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 3 \, \sqrt {\frac {c^{5}}{a f^{2}}} a f e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {c^{5}}{a f^{2}}}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) + 4 \, {\left (-3 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f} \] Input:

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm=" 
fricas")
 

Output:

-1/2*(3*sqrt(c^5/(a*f^2))*a*f*e^(I*f*x + I*e)*log(4*(2*(c^2*e^(3*I*f*x + 3 
*I*e) + c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^( 
2*I*f*x + 2*I*e) + 1)) - (I*a*f*e^(2*I*f*x + 2*I*e) - I*a*f)*sqrt(c^5/(a*f 
^2)))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - 3*sqrt(c^5/(a*f^2))*a*f*e^(I*f*x 
+ I*e)*log(4*(2*(c^2*e^(3*I*f*x + 3*I*e) + c^2*e^(I*f*x + I*e))*sqrt(a/(e^ 
(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (-I*a*f*e^(2*I 
*f*x + 2*I*e) + I*a*f)*sqrt(c^5/(a*f^2)))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) 
 + 4*(-3*I*c^2*e^(2*I*f*x + 2*I*e) - 2*I*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) 
+ 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)
 

Sympy [F]

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

integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.21 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {{\left (6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) + 6 i \, c^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) \cos \left (f x + e\right ) - 8 i \, c^{2} \cos \left (f x + e\right )^{2} + 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 \, c^{2} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 8 \, c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 i \, c^{2}\right )} \sqrt {c}}{2 \, \sqrt {a} f \cos \left (f x + e\right )} \] Input:

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm=" 
maxima")
 

Output:

-1/2*(6*I*c^2*arctan2(cos(f*x + e), sin(f*x + e) + 1)*cos(f*x + e) + 6*I*c 
^2*arctan2(cos(f*x + e), -sin(f*x + e) + 1)*cos(f*x + e) - 8*I*c^2*cos(f*x 
 + e)^2 + 3*c^2*cos(f*x + e)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f 
*x + e) + 1) - 3*c^2*cos(f*x + e)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2* 
sin(f*x + e) + 1) - 8*c^2*cos(f*x + e)*sin(f*x + e) - 2*I*c^2)*sqrt(c)/(sq 
rt(a)*f*cos(f*x + e))
 

Giac [F]

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

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{2} \left (\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2} i +\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )+5 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, i -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f \right )}{a f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:

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

Output:

(sqrt(c)*sqrt(a)*c**2*(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1 
)*tan(e + f*x)**2*i + sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1) 
*tan(e + f*x) + 5*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*i - 
 3*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)* 
*2)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f - 3*int((sqrt(tan(e + f*x)* 
i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1), 
x)*f))/(a*f*(tan(e + f*x)**2 + 1))