3.4.9 \(\int \cot (e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [309]

3.4.9.1 Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {(a-b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {b \sqrt {a+b \tan ^2(e+f x)}}{f} \]

output

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

3.4.9.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {-a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )+(a-b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+b \sqrt {a+b \tan ^2(e+f x)}}{f} \]

input

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

output

(-(a^(3/2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]]) + (a - b)^(3/2)*Ar 
cTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + b*Sqrt[a + b*Tan[e + f*x]^ 
2])/f
 

3.4.9.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4153, 354, 95, 174, 73, 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.

\(\displaystyle \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \tan (e+f x)^2\right )^{3/2}}{\tan (e+f x)}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {\int \frac {\cot (e+f x) \left (b \tan ^2(e+f x)+a\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\cot (e+f x) \left (b \tan ^2(e+f x)+a\right )^{3/2}}{\tan ^2(e+f x)+1}d\tan ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 95

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {2 a^2 \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \tan ^2(e+f x)+a}}{b}-\frac {2 (a-b)^2 \int \frac {1}{\frac {\tan ^4(e+f x)}{b}-\frac {a}{b}+1}d\sqrt {b \tan ^2(e+f x)+a}}{b}+2 b \sqrt {a+b \tan ^2(e+f x)}}{2 f}\)

\(\Big \downarrow \) 221

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

input

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

output

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

3.4.9.3.1 Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 95

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d)   Int[(b 
*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + 
b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

3.4.9.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(884\) vs. \(2(81)=162\).

Time = 2.25 (sec) , antiderivative size = 885, normalized size of antiderivative = 9.32

method	result	size
default	\(\text {Expression too large to display}\)	\(885\)

input

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

output

1/2/f/a^(3/2)/(a-b)^(1/2)*(a+b*tan(f*x+e)^2)^(3/2)/(cos(f*x+e)+1)/(a*cos(f 
*x+e)^2-b*cos(f*x+e)^2+b)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1 
)^2)^(1/2)*(2*cos(f*x+e)^3*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b* 
cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a 
-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(7 
/2)-4*cos(f*x+e)^3*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+ 
e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2 
)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(5/2)*b+2* 
cos(f*x+e)^3*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b 
)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a* 
cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(3/2)*b^2+2*cos( 
f*x+e)^3*a^(3/2)*(a-b)^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e 
)+1)^2)^(1/2)*b+cos(f*x+e)^3*ln(2/a^(1/2)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+ 
e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+((a*cos(f*x+e)^2-b*cos(f*x+ 
e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+b)/(cos( 
f*x+e)+1))*(a-b)^(1/2)*a^3-cos(f*x+e)^3*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos( 
f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*a-b*cos(f*x+ 
e)+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/( 
cos(f*x+e)-1))*(a-b)^(1/2)*a^3+2*cos(f*x+e)^2*a^(3/2)*(a-b)^(1/2)*((a*cos( 
f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b)
 

3.4.9.5 Fricas [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 591, normalized size of antiderivative = 6.22 \[ \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {{\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, a^{\frac {3}{2}} \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) - 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{4 \, f}, \frac {4 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) - {\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{4 \, f}, \frac {{\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + a^{\frac {3}{2}} \log \left (\frac {b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{2 \, f}, \frac {2 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) + {\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{2 \, f}\right ] \]

input

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

output

[-1/4*((a - b)^(3/2)*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x 
+ e)^2 - 4*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a 
- b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) - 2*a 
^(3/2)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a) 
/tan(f*x + e)^2) - 4*sqrt(b*tan(f*x + e)^2 + a)*b)/f, 1/4*(4*sqrt(-a)*a*ar 
ctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) - (a - b)^(3/2)*log(-(b^2*tan( 
f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2*a 
- b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f* 
x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*sqrt(b*tan(f*x + e)^2 + a)*b)/f, 1/2 
*((-a + b)^(3/2)*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f 
*x + e)^2 + 2*a - b)) + a^(3/2)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + 
 e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) + 2*sqrt(b*tan(f*x + e)^2 + a)*b 
)/f, 1/2*(2*sqrt(-a)*a*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + (-a 
 + b)^(3/2)*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + 
e)^2 + 2*a - b)) + 2*sqrt(b*tan(f*x + e)^2 + a)*b)/f]
 

3.4.9.6 Sympy [F]

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

input

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

output

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

3.4.9.7 Maxima [F]

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

input

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

output

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

3.4.9.8 Giac [F(-1)]

Timed out. \[ \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.9.9 Mupad [B] (verification not implemented)

Time = 11.87 (sec) , antiderivative size = 546, normalized size of antiderivative = 5.75 \[ \int \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx=\frac {b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{f}+\frac {\mathrm {atanh}\left (\frac {6\,a^3\,b^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}-\frac {6\,a^2\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}+\frac {2\,a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}\right )\,\sqrt {{\left (a-b\right )}^3}}{f}-\frac {\mathrm {atanh}\left (\frac {2\,b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {8\,a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}+\frac {12\,a^2\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {6\,a^3\,b^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}\right )\,\sqrt {a^3}}{f} \]

input

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

output

(b*(a + b*tan(e + f*x)^2)^(1/2))/f + (atanh((6*a^3*b^3*(a + b*tan(e + f*x) 
^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 2 
0*a^3*b^5 - 18*a^4*b^4 + 6*a^5*b^3) - (6*a^2*b^4*(a + b*tan(e + f*x)^2)^(1 
/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3* 
b^5 - 18*a^4*b^4 + 6*a^5*b^3) + (2*a*b^5*(a + b*tan(e + f*x)^2)^(1/2)*(3*a 
*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5 - 18 
*a^4*b^4 + 6*a^5*b^3))*((a - b)^3)^(1/2))/f - (atanh((2*b^6*(a + b*tan(e + 
 f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^ 
3) - (8*a*b^5*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 - 8*a^3 
*b^5 + 12*a^4*b^4 - 6*a^5*b^3) + (12*a^2*b^4*(a + b*tan(e + f*x)^2)^(1/2)* 
(a^3)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) - (6*a^3*b^3 
*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5 + 12*a^4 
*b^4 - 6*a^5*b^3))*(a^3)^(1/2))/f