\(\int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\) [645]

Optimal result

Integrand size = 25, antiderivative size = 687 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {5 a b^{2/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {5}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {5}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )} \] Output:

-5/6*a*b^(2/3)*arctan(-1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)*3^(1/2 
)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(1/3)*3^(1/2)/(a^2+b^2)^(11/6)/f/(sec(f* 
x+e)^2)^(1/6)-5/6*a*b^(2/3)*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^ 
(1/6)*3^(1/2)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(1/3)*3^(1/2)/(a^2+b^2)^(11/ 
6)/f/(sec(f*x+e)^2)^(1/6)-5/3*a*b^(2/3)*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/ 
6)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2) 
^(1/6)+5/12*a*b^(2/3)*ln((a^2+b^2)^(1/3)-b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+ 
e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^( 
11/6)/f/(sec(f*x+e)^2)^(1/6)-5/12*a*b^(2/3)*ln((a^2+b^2)^(1/3)+b^(1/3)*(a^ 
2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(d*sec(f*x 
+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1/6)+AppellF1(1/2,2,5/6,3/2, 
b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(1/3)*tan(f*x+e)/a^2/f/ 
(sec(f*x+e)^2)^(1/6)+1/3*b^2*AppellF1(3/2,2,5/6,5/2,b^2*tan(f*x+e)^2/a^2,- 
tan(f*x+e)^2)*(d*sec(f*x+e))^(1/3)*tan(f*x+e)^3/a^4/f/(sec(f*x+e)^2)^(1/6) 
-a*b*(d*sec(f*x+e))^(1/3)/(a^2+b^2)/f/(a^2-b^2*tan(f*x+e)^2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4560\) vs. \(2(687)=1374\).

Time = 70.80 (sec) , antiderivative size = 4560, normalized size of antiderivative = 6.64 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*Sec[e + f*x])^(1/3)/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Sec[e + f*x])^(1/3)*((-2*b^2*AppellF1[7/6, 1/2, 1, 13/6, Sec[e + f*x]^ 
2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x] 
^(10/3))/(21*(a^2 + b^2)^2*Sqrt[1 - Sec[e + f*x]^2]) - (7*(3*a^2 - 2*b^2)* 
AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2 
)]*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(4/3))/(3*(-1 + Sec[e + f*x]^2)*( 
7*(a^2 + b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x] 
^2)/(a^2 + b^2)] + 3*(2*b^2*AppellF1[7/6, 1/2, 2, 13/6, Sec[e + f*x]^2, (b 
^2*Sec[e + f*x]^2)/(a^2 + b^2)] + (a^2 + b^2)*AppellF1[7/6, 3/2, 1, 13/6, 
Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])*Sec[e + f*x]^2)*(-a^2 + 
 b^2*(-1 + Sec[e + f*x]^2))) + (b^(2/3)*((-10*(-1)^(5/6)*a*ArcTan[Sqrt[3] 
- (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)])/(a^2 + b^2 
)^(11/6) + (10*(-1)^(5/6)*a*ArcTan[Sqrt[3] + (2*(-1)^(1/6)*b^(1/3)*Sec[e + 
 f*x]^(1/3))/(a^2 + b^2)^(1/6)])/(a^2 + b^2)^(11/6) + (20*(-1)^(5/6)*a*Arc 
Tan[((-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)])/(a^2 + b^2 
)^(11/6) - (5*(-1)^(5/6)*Sqrt[3]*a*Log[(a^2 + b^2)^(1/3) - (-1)^(1/6)*Sqrt 
[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[ 
e + f*x]^(2/3)])/(a^2 + b^2)^(11/6) + (5*(-1)^(5/6)*Sqrt[3]*a*Log[(a^2 + b 
^2)^(1/3) + (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3 
) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)])/(a^2 + b^2)^(11/6) - (12*a*b^( 
1/3)*Sec[e + f*x]^(1/3))/((a^2 + b^2)*(a^2 + b^2 - b^2*Sec[e + f*x]^2))...
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 545, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3994, 505, 2009}

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.

\(\displaystyle \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \int \frac {1}{(a+b \tan (e+f x))^2 \left (\tan ^2(e+f x)+1\right )^{5/6}}d(b \tan (e+f x))}{b f \sqrt [6]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 505

\(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \int \left (\frac {a^2}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 b \tan (e+f x) a}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f \sqrt [6]{\sec ^2(e+f x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {5}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}+\frac {5 a b^{5/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt [6]{a^2+b^2}}\right )}{3 \left (a^2+b^2\right )^{11/6}}-\frac {a b^2 \sqrt [6]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {5 a b^{5/3} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{11/6}}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {5}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\)

Input:

Int[(d*Sec[e + f*x])^(1/3)/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Sec[e + f*x])^(1/3)*((5*a*b^(5/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(1 + T 
an[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(2*Sqrt[3]*(a^2 + b^2) 
^(11/6)) - (5*a*b^(5/3)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(1 + Tan[e + f*x]^2) 
^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(2*Sqrt[3]*(a^2 + b^2)^(11/6)) - (5* 
a*b^(5/3)*ArcTanh[(b^(1/3)*(1 + Tan[e + f*x]^2)^(1/6))/(a^2 + b^2)^(1/6)]) 
/(3*(a^2 + b^2)^(11/6)) + (5*a*b^(5/3)*Log[(a^2 + b^2)^(1/3) - b^(1/3)*(a^ 
2 + b^2)^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f*x]^2)^( 
1/3)])/(12*(a^2 + b^2)^(11/6)) - (5*a*b^(5/3)*Log[(a^2 + b^2)^(1/3) + b^(1 
/3)*(a^2 + b^2)^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f* 
x]^2)^(1/3)])/(12*(a^2 + b^2)^(11/6)) + (b*AppellF1[1/2, 2, 5/6, 3/2, (b^2 
*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a^2 + (b^3*AppellF1[3 
/2, 2, 5/6, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x]^3 
)/(3*a^4) - (a*b^2*(1 + Tan[e + f*x]^2)^(1/6))/((a^2 + b^2)*(a^2 - b^2*Tan 
[e + f*x]^2))))/(b*f*(Sec[e + f*x]^2)^(1/6))
 

Defintions of rubi rules used

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( 
-n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]))   Subst[Int[(a + x)^n*(1 + 
x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

Maple [F]

\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]

Input:

int((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x)
 

Output:

int((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\sqrt [3]{d \sec {\left (e + f x \right )}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate((d*sec(f*x+e))**(1/3)/(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((d*sec(e + f*x))**(1/3)/(a + b*tan(e + f*x))**2, x)
 

Maxima [F]

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

integrate((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^(1/3)/(b*tan(f*x + e) + a)^2, x)
 

Giac [F]

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

integrate((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(1/3)/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((d/cos(e + f*x))^(1/3)/(a + b*tan(e + f*x))^2,x)
 

Output:

int((d/cos(e + f*x))^(1/3)/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {d^{\frac {1}{3}} \left (3 \sec \left (f x +e \right )^{\frac {1}{3}} \tan \left (f x +e \right )+3 \left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f +3 \left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f -\left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) b^{2} f -\left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a b f -4 \left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \tan \left (f x +e \right ) a b f -4 \left (\int \frac {\sec \left (f x +e \right )^{\frac {1}{3}} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) a^{2} f \right )}{6 a f \left (\tan \left (f x +e \right ) b +a \right )} \] Input:

int((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x)
 

Output:

(d**(1/3)*(3*sec(e + f*x)**(1/3)*tan(e + f*x) + 3*int(sec(e + f*x)**(1/3)/ 
(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*a*b*f + 
 3*int(sec(e + f*x)**(1/3)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a* 
*2),x)*a**2*f - int((sec(e + f*x)**(1/3)*tan(e + f*x)**3)/(tan(e + f*x)**2 
*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*b**2*f - int((sec(e + f 
*x)**(1/3)*tan(e + f*x)**3)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a 
**2),x)*a*b*f - 4*int((sec(e + f*x)**(1/3)*tan(e + f*x)**2)/(tan(e + f*x)* 
*2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*a*b*f - 4*int((sec(e 
+ f*x)**(1/3)*tan(e + f*x)**2)/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b 
+ a**2),x)*a**2*f))/(6*a*f*(tan(e + f*x)*b + a))