Integrand size = 23, antiderivative size = 86 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {x}{a}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a (a+b)^{5/2} f}+\frac {(a+2 b) \cot (e+f x)}{(a+b)^2 f}-\frac {\cot ^3(e+f x)}{3 (a+b) f} \]
x/a-b^(5/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a/(a+b)^(5/2)/f+(a+2*b) *cot(f*x+e)/(a+b)^2/f-1/3*cot(f*x+e)^3/(a+b)/f
Result contains complex when optimal does not.
Time = 4.69 (sec) , antiderivative size = 390, normalized size of antiderivative = 4.53 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (3 b^3 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\frac {1}{8} \sqrt {a+b} \csc (e) \csc ^3(e+f x) \sqrt {b (\cos (e)-i \sin (e))^4} \left (9 (a+b)^2 f x \cos (f x)-9 (a+b)^2 f x \cos (2 e+f x)-3 a^2 f x \cos (2 e+3 f x)-6 a b f x \cos (2 e+3 f x)-3 b^2 f x \cos (2 e+3 f x)+3 a^2 f x \cos (4 e+3 f x)+6 a b f x \cos (4 e+3 f x)+3 b^2 f x \cos (4 e+3 f x)-12 a^2 \sin (f x)-24 a b \sin (f x)-12 a^2 \sin (2 e+f x)-18 a b \sin (2 e+f x)+8 a^2 \sin (2 e+3 f x)+14 a b \sin (2 e+3 f x)\right )\right )}{6 a (a+b)^{5/2} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(3*b^3*ArcTan[(Sec[f*x]*(Co s[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + (Sqrt[a + b]*Csc[e]*Csc[e + f*x]^3*Sqrt[b*(Cos[e] - I*Sin[e])^4]*(9*(a + b)^2*f*x*Co s[f*x] - 9*(a + b)^2*f*x*Cos[2*e + f*x] - 3*a^2*f*x*Cos[2*e + 3*f*x] - 6*a *b*f*x*Cos[2*e + 3*f*x] - 3*b^2*f*x*Cos[2*e + 3*f*x] + 3*a^2*f*x*Cos[4*e + 3*f*x] + 6*a*b*f*x*Cos[4*e + 3*f*x] + 3*b^2*f*x*Cos[4*e + 3*f*x] - 12*a^2 *Sin[f*x] - 24*a*b*Sin[f*x] - 12*a^2*Sin[2*e + f*x] - 18*a*b*Sin[2*e + f*x ] + 8*a^2*Sin[2*e + 3*f*x] + 14*a*b*Sin[2*e + 3*f*x]))/8))/(6*a*(a + b)^(5 /2)*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])
Time = 0.38 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4629, 2075, 382, 27, 445, 397, 216, 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.
| \(\displaystyle \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^4 \left (a+b \sec (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4629 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 2075 |
| \(\displaystyle \frac {\int \frac {\cot ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 382 |
| \(\displaystyle \frac {\frac {\int -\frac {3 \cot ^2(e+f x) \left (b \tan ^2(e+f x)+a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{3 (a+b)}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\cot ^2(e+f x) \left (b \tan ^2(e+f x)+a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle \frac {-\frac {-\frac {\int \frac {a^2+3 b a+3 b^2+b (a+2 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a+b}-\frac {(a+2 b) \cot (e+f x)}{a+b}}{a+b}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {(a+b)^2 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {(a+2 b) \cot (e+f x)}{a+b}}{a+b}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {(a+b)^2 \arctan (\tan (e+f x))}{a}-\frac {b^3 \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a+b}-\frac {(a+2 b) \cot (e+f x)}{a+b}}{a+b}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {-\frac {\frac {(a+b)^2 \arctan (\tan (e+f x))}{a}-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a+b}-\frac {(a+2 b) \cot (e+f x)}{a+b}}{a+b}-\frac {\cot ^3(e+f x)}{3 (a+b)}}{f}\) |
(-1/3*Cot[e + f*x]^3/(a + b) - (-((((a + b)^2*ArcTan[Tan[e + f*x]])/a - (b ^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(a + b )) - ((a + 2*b)*Cot[e + f*x])/(a + b))/(a + b))/f
3.4.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ (a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b* x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m + 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
Time = 1.81 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}-\frac {1}{3 \left (a +b \right ) \tan \left (f x +e \right )^{3}}-\frac {-a -2 b}{\left (a +b \right )^{2} \tan \left (f x +e \right )}-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{2} a \sqrt {\left (a +b \right ) b}}}{f}\) | \(90\) |
| default | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a}-\frac {1}{3 \left (a +b \right ) \tan \left (f x +e \right )^{3}}-\frac {-a -2 b}{\left (a +b \right )^{2} \tan \left (f x +e \right )}-\frac {b^{3} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{2} a \sqrt {\left (a +b \right ) b}}}{f}\) | \(90\) |
| risch | \(\frac {x}{a}+\frac {2 i \left (6 a \,{\mathrm e}^{4 i \left (f x +e \right )}+9 b \,{\mathrm e}^{4 i \left (f x +e \right )}-6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-12 b \,{\mathrm e}^{2 i \left (f x +e \right )}+4 a +7 b \right )}{3 f \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3}}+\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{3} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{3} f a}\) | \(199\) |
1/f*(1/a*arctan(tan(f*x+e))-1/3/(a+b)/tan(f*x+e)^3-(-a-2*b)/(a+b)^2/tan(f* x+e)-1/(a+b)^2*b^3/a/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (76) = 152\).
Time = 0.31 (sec) , antiderivative size = 533, normalized size of antiderivative = 6.20 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {4 \, {\left (4 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 12 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right ) + 12 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f x \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f x\right )} \sin \left (f x + e\right )}{12 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac {2 \, {\left (4 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 6 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right ) + 6 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f x \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f x\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/12*(4*(4*a^2 + 7*a*b)*cos(f*x + e)^3 + 3*(b^2*cos(f*x + e)^2 - b^2)*sqr t(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2 )*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*c os(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2* a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) - 12*(a^2 + 2*a*b)*cos(f*x + e) + 12*((a^2 + 2*a*b + b^2)*f*x*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f*x)*sin( f*x + e))/(((a^3 + 2*a^2*b + a*b^2)*f*cos(f*x + e)^2 - (a^3 + 2*a^2*b + a* b^2)*f)*sin(f*x + e)), 1/6*(2*(4*a^2 + 7*a*b)*cos(f*x + e)^3 + 3*(b^2*cos( f*x + e)^2 - b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b )*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 6*(a^2 + 2 *a*b)*cos(f*x + e) + 6*((a^2 + 2*a*b + b^2)*f*x*cos(f*x + e)^2 - (a^2 + 2* a*b + b^2)*f*x)*sin(f*x + e))/(((a^3 + 2*a^2*b + a*b^2)*f*cos(f*x + e)^2 - (a^3 + 2*a^2*b + a*b^2)*f)*sin(f*x + e))]
\[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, b^{3} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} b}} - \frac {3 \, {\left (f x + e\right )}}{a} - \frac {3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(3*b^3*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 2*a^2*b + a*b^2 )*sqrt((a + b)*b)) - 3*(f*x + e)/a - (3*(a + 2*b)*tan(f*x + e)^2 - a - b)/ ((a^2 + 2*a*b + b^2)*tan(f*x + e)^3))/f
Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b + b^{2}}} - \frac {3 \, {\left (f x + e\right )}}{a} - \frac {3 \, a \tan \left (f x + e\right )^{2} + 6 \, b \tan \left (f x + e\right )^{2} - a - b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt( a*b + b^2)))*b^3/((a^3 + 2*a^2*b + a*b^2)*sqrt(a*b + b^2)) - 3*(f*x + e)/a - (3*a*tan(f*x + e)^2 + 6*b*tan(f*x + e)^2 - a - b)/((a^2 + 2*a*b + b^2)* tan(f*x + e)^3))/f
Time = 24.73 (sec) , antiderivative size = 2644, normalized size of antiderivative = 30.74 \[ \int \frac {\cot ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Too large to display} \]
atan((10*b^12*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3* b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (80*a*b^11*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (290*a^2*b^10*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^ 10*b^2) + (630*a^3*b^9*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110 *a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (912*a^4*b^8*tan(e + f*x))/(80*a*b^1 1 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660 *a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (922*a^5 *b^7*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912 *a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9* b^3 + 2*a^10*b^2) + (660*a^6*b^6*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290* a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5*b^7 + 660*a^6*b^6 + 330*a^7 *b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (330*a^7*b^5*tan(e + f*x)) /(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*a^3*b^9 + 912*a^4*b^8 + 922*a^5 *b^7 + 660*a^6*b^6 + 330*a^7*b^5 + 110*a^8*b^4 + 22*a^9*b^3 + 2*a^10*b^2) + (110*a^8*b^4*tan(e + f*x))/(80*a*b^11 + 10*b^12 + 290*a^2*b^10 + 630*...