\(\int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [120]

Optimal result

Integrand size = 25, antiderivative size = 91 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 a f} \] Output:

-1/2*(a-b)*arctanh(a^(1/2)*sec(f*x+e)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(3/2)/ 
f-1/2*cot(f*x+e)*csc(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/a/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(367\) vs. \(2(91)=182\).

Time = 2.36 (sec) , antiderivative size = 367, normalized size of antiderivative = 4.03 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\cot ^2(e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (2 (-a+b) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )-\left (-2 (a-b) \log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )+\sqrt {2} \sqrt {a} \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}\right ) \sec (e+f x)+8 (a-b) \text {arctanh}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right ) \sec (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )}{4 a^{3/2} f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )}} \] Input:

Integrate[Csc[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

(Cot[e + f*x]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(2 
*(-a + b)*Log[a - 2*b - a*Tan[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f 
*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] - (-2*(a - b)*Log[a - 2*b - a*T 
an[(e + f*x)/2]^2 + Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + 
 f*x)/2]^2)^2]] + Sqrt[2]*Sqrt[a]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])* 
Sec[(e + f*x)/2]^4])*Sec[e + f*x] + 8*(a - b)*ArcTanh[Tan[(e + f*x)/2]^2 - 
 Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]/Sqrt[a]]*Sec 
[e + f*x]*Sin[(e + f*x)/2]^2))/(4*a^(3/2)*f*Sqrt[(a + b + (a - b)*Cos[2*(e 
 + f*x)])*Sec[(e + f*x)/2]^4])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4147, 373, 27, 291, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^3 \sqrt {a+b \tan (e+f x)^2}}dx\)

\(\Big \downarrow \) 4147

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 a \left (1-\sec ^2(e+f x)\right )}-\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{3/2}}}{f}\)

Input:

Int[Csc[e + f*x]^3/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

(-1/2*((a - b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^ 
2]])/a^(3/2) + (Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(2*a*(1 - Sec 
[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 291

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

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 3042

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

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ 
m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 
)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( 
m - 1)/2]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(646\) vs. \(2(79)=158\).

Time = 3.33 (sec) , antiderivative size = 647, normalized size of antiderivative = 7.11

method	result	size
default	\(\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\ln \left (\frac {2 \sqrt {a}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}-2 a \cos \left (f x +e \right )+2 \cos \left (f x +e \right ) b +2 b}{\sqrt {a}\, \left (\cos \left (f x +e \right )+1\right )}\right ) a^{2} \left (1-\cos \left (f x +e \right )\right )^{2}+\ln \left (\frac {-2 a \left (1-\cos \left (f x +e \right )\right )^{2}+4 \left (1-\cos \left (f x +e \right )\right )^{2} b +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}\, \sin \left (f x +e \right )^{2}+2 a \sin \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2}}\right ) a^{2} \left (1-\cos \left (f x +e \right )\right )^{2}-\ln \left (\frac {2 \sqrt {a}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}-2 a \cos \left (f x +e \right )+2 \cos \left (f x +e \right ) b +2 b}{\sqrt {a}\, \left (\cos \left (f x +e \right )+1\right )}\right ) a \left (1-\cos \left (f x +e \right )\right )^{2} b -\ln \left (\frac {-2 a \left (1-\cos \left (f x +e \right )\right )^{2}+4 \left (1-\cos \left (f x +e \right )\right )^{2} b +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {a}\, \sin \left (f x +e \right )^{2}+2 a \sin \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2}}\right ) a \left (1-\cos \left (f x +e \right )\right )^{2} b +\left (2-2 \cos \left (f x +e \right )\right ) a^{\frac {3}{2}} \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{2 f \,a^{\frac {5}{2}} \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (1-\cos \left (f x +e \right )\right )^{2}}\)	\(647\)

Input:

int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/f/a^(5/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)/(a+ 
b*tan(f*x+e)^2)^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)/(1-cos(f*x+e))^2*( 
ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^( 
1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a 
^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/(cos(f*x+e)+1))*a^2*(1-cos(f*x+e))^2+l 
n(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2+2*(1-cos(f*x+e))^2*b+2*((a*cos(f 
*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin 
(f*x+e)^2))*a^2*(1-cos(f*x+e))^2-ln(2/a^(1/2)*(a^(1/2)*((a*cos(f*x+e)^2+b* 
sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+((a*cos(f*x+e)^2+b*sin(f* 
x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+cos(f*x+e)*b+b)/(cos( 
f*x+e)+1))*a*(1-cos(f*x+e))^2*b-ln(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2 
+2*(1-cos(f*x+e))^2*b+2*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2) 
^(1/2)*a^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2))*a*(1-cos(f*x+e))^2*b+(2-2*cos 
(f*x+e))*a^(3/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.31 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {2 \, a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, -\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - a + b\right )} \sqrt {-a} \arctan \left (-\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - a \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \] Input:

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

Output:

[1/4*(2*a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - 
 ((a - b)*cos(f*x + e)^2 - a + b)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 + 
 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) 
+ a + b)/(cos(f*x + e)^2 - 1)))/(a^2*f*cos(f*x + e)^2 - a^2*f), -1/2*(((a 
- b)*cos(f*x + e)^2 - a + b)*sqrt(-a)*arctan(-sqrt(-a)*sqrt(((a - b)*cos(f 
*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/((a - b)*cos(f*x + e)^2 + b)) 
- a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(a^2*f 
*cos(f*x + e)^2 - a^2*f)]
 

Sympy [F]

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

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

Output:

Integral(csc(e + f*x)**3/sqrt(a + b*tan(e + f*x)**2), x)
 

Maxima [F]

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

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

Output:

integrate(csc(f*x + e)^3/sqrt(b*tan(f*x + e)^2 + a), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (79) = 158\).

Time = 1.13 (sec) , antiderivative size = 464, normalized size of antiderivative = 5.10 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/8*(4*(a - b)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f 
*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + 
a))/sqrt(-a))/(sqrt(-a)*a) + 2*(a^(3/2) - sqrt(a)*b)*log(abs(-(sqrt(a)*tan 
(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/ 
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a + a^(3/2) - 2*sqrt(a)*b))/a^2 
- sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2 
*f*x + 1/2*e)^2 + a)/a - 2*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1 
/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^ 
2 + a))*a - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e 
)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b - a^ 
(3/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 
 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)*a))/ 
(f*sgn(tan(1/2*f*x + 1/2*e)^2 - 1))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(tan(e + f*x)**2*b + a)*csc(e + f*x)**3)/(tan(e + f*x)**2*b + a), 
x)