Integrand size = 21, antiderivative size = 138 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{7/2} f}-\frac {15 \cos (e+f x)}{8 (a-b)^3 f}+\frac {\cos (e+f x)}{4 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {5 \cos (e+f x)}{8 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )} \]
-15/8*cos(f*x+e)/(a-b)^3/f+1/4*cos(f*x+e)/(a-b)/f/(a-b+b*sec(f*x+e)^2)^2+5 /8*cos(f*x+e)/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)-15/8*arctan(sec(f*x+e)*b^(1/2 )/(a-b)^(1/2))*b^(1/2)/(a-b)^(7/2)/f
Time = 2.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.23 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}+\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}+\frac {2 \cos (e+f x) \left (-4+\frac {4 b^2}{(a+b+(a-b) \cos (2 (e+f x)))^2}-\frac {9 b}{a+b+(a-b) \cos (2 (e+f x))}\right )}{(a-b)^3}}{8 f} \]
((15*Sqrt[b]*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) + (15*Sqrt[b]*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/S qrt[b]])/(a - b)^(7/2) + (2*Cos[e + f*x]*(-4 + (4*b^2)/(a + b + (a - b)*Co s[2*(e + f*x)])^2 - (9*b)/(a + b + (a - b)*Cos[2*(e + f*x)])))/(a - b)^3)/ (8*f)
Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4147, 253, 253, 264, 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.
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {5 \int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{4 (a-b)}+\frac {\cos (e+f x)}{4 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}\right )}{4 (a-b)}+\frac {\cos (e+f x)}{4 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {b \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a-b}-\frac {\cos (e+f x)}{a-b}\right )}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}\right )}{4 (a-b)}+\frac {\cos (e+f x)}{4 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}-\frac {\cos (e+f x)}{a-b}\right )}{2 (a-b)}+\frac {\cos (e+f x)}{2 (a-b) \left (a+b \sec ^2(e+f x)-b\right )}\right )}{4 (a-b)}+\frac {\cos (e+f x)}{4 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
(Cos[e + f*x]/(4*(a - b)*(a - b + b*Sec[e + f*x]^2)^2) + (5*((3*(-((Sqrt[b ]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a - b)^(3/2)) - Cos[e + f*x ]/(a - b)))/(2*(a - b)) + Cos[e + f*x]/(2*(a - b)*(a - b + b*Sec[e + f*x]^ 2))))/(4*(a - b)))/f
3.1.82.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
Time = 5.46 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96
| method | result | size |
| derivativedivides | \(\frac {-\frac {\cos \left (f x +e \right )}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {b \left (\frac {\left (-\frac {9 a}{8}+\frac {9 b}{8}\right ) \cos \left (f x +e \right )^{3}-\frac {7 b \cos \left (f x +e \right )}{8}}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\) | \(132\) |
| default | \(\frac {-\frac {\cos \left (f x +e \right )}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {b \left (\frac {\left (-\frac {9 a}{8}+\frac {9 b}{8}\right ) \cos \left (f x +e \right )^{3}-\frac {7 b \cos \left (f x +e \right )}{8}}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\) | \(132\) |
| risch | \(-\frac {{\mathrm e}^{i \left (f x +e \right )}}{2 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {{\mathrm e}^{-i \left (f x +e \right )}}{2 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {b \left (-9 a \,{\mathrm e}^{7 i \left (f x +e \right )}+9 b \,{\mathrm e}^{7 i \left (f x +e \right )}-27 a \,{\mathrm e}^{5 i \left (f x +e \right )}-b \,{\mathrm e}^{5 i \left (f x +e \right )}-27 a \,{\mathrm e}^{3 i \left (f x +e \right )}-b \,{\mathrm e}^{3 i \left (f x +e \right )}-9 a \,{\mathrm e}^{i \left (f x +e \right )}+9 b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 \left (-a +b \right )^{3} \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )^{2} f}-\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{16 \left (a -b \right )^{4} f}+\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{16 \left (a -b \right )^{4} f}\) | \(364\) |
1/f*(-1/(a^3-3*a^2*b+3*a*b^2-b^3)*cos(f*x+e)+b/(a-b)^3*(((-9/8*a+9/8*b)*co s(f*x+e)^3-7/8*b*cos(f*x+e))/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)^2+15/8/(b*( a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (122) = 244\).
Time = 0.36 (sec) , antiderivative size = 556, normalized size of antiderivative = 4.03 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\left [-\frac {16 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) + 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right )}{16 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f\right )}}, -\frac {8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) + 15 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right )}{8 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f\right )}}\right ] \]
[-1/16*(16*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 + 50*(a*b - b^2)*cos(f*x + e )^3 + 30*b^2*cos(f*x + e) + 15*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a* b - b^2)*cos(f*x + e)^2 + b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e) ^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f* x + e)^4 + 2*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f*cos(f*x + e )^2 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f), -1/8*(8*(a^2 - 2*a*b + b^2 )*cos(f*x + e)^5 + 25*(a*b - b^2)*cos(f*x + e)^3 + 15*b^2*cos(f*x + e) + 1 5*((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2 )*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*f*cos(f*x + e)^4 + 2*( a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*f)]
Timed out. \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Time = 0.99 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.56 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {f^{5} \cos \left (f x + e\right )}{a^{3} f^{6} - 3 \, a^{2} b f^{6} + 3 \, a b^{2} f^{6} - b^{3} f^{6}} + \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}} f} - \frac {\frac {9 \, a b \cos \left (f x + e\right )^{3}}{f} - \frac {9 \, b^{2} \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, b^{2} \cos \left (f x + e\right )}{f}}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \]
-f^5*cos(f*x + e)/(a^3*f^6 - 3*a^2*b*f^6 + 3*a*b^2*f^6 - b^3*f^6) + 15/8*b *arctan((a*cos(f*x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)*f) - 1/8*(9*a*b*cos(f*x + e)^3/f - 9*b^2 *cos(f*x + e)^3/f + 7*b^2*cos(f*x + e)/f)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3) *(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)^2)
Time = 14.01 (sec) , antiderivative size = 780, normalized size of antiderivative = 5.65 \[ \int \frac {\sin (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {8\,a^2+9\,a\,b-2\,b^2}{4\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (16\,a^4-41\,a^3\,b+27\,a^2\,b^2-40\,a\,b^3+8\,b^4\right )}{2\,a^2\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (24\,a^4-64\,a^3\,b+53\,a^2\,b^2+40\,a\,b^3-8\,b^4\right )}{2\,a^2\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (8\,a^3-9\,a^2\,b+24\,a\,b^2-8\,b^3\right )}{4\,a\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-16\,a^3+23\,a^2\,b+27\,a\,b^2-4\,b^3\right )}{2\,a\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,a\,b-3\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (8\,a\,b-3\,a^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,a^2-8\,a\,b+16\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,a^2-8\,a\,b+16\,b^2\right )+a^2\right )}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {{\left (a-b\right )}^7\,\left (2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (225\,a^8\,b-1350\,a^7\,b^2+3375\,a^6\,b^3-4500\,a^5\,b^4+3375\,a^4\,b^5-1350\,a^3\,b^6+225\,a^2\,b^7\right )}{a\,{\left (a-b\right )}^{13/2}}+\frac {225\,\sqrt {b}\,\left (a-2\,b\right )\,\left (128\,a^{12}-1408\,a^{11}\,b+6912\,a^{10}\,b^2-19968\,a^9\,b^3+37632\,a^8\,b^4-48384\,a^7\,b^5+43008\,a^6\,b^6-26112\,a^5\,b^7+10368\,a^4\,b^8-2432\,a^3\,b^9+256\,a^2\,b^{10}\right )}{512\,a\,{\left (a-b\right )}^{21/2}}\right )+\frac {225\,\sqrt {b}\,\left (a-2\,b\right )\,\left (-128\,a^{12}+1152\,a^{11}\,b-4608\,a^{10}\,b^2+10752\,a^9\,b^3-16128\,a^8\,b^4+16128\,a^7\,b^5-10752\,a^6\,b^6+4608\,a^5\,b^7-1152\,a^4\,b^8+128\,a^3\,b^9\right )}{256\,a\,{\left (a-b\right )}^{21/2}}\right )}{225\,a^8\,b-1350\,a^7\,b^2+3375\,a^6\,b^3-4500\,a^5\,b^4+3375\,a^4\,b^5-1350\,a^3\,b^6+225\,a^2\,b^7}\right )}{8\,f\,{\left (a-b\right )}^{7/2}} \]
- ((9*a*b + 8*a^2 - 2*b^2)/(4*(a - b)*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f *x)/2)^6*(16*a^4 - 41*a^3*b - 40*a*b^3 + 8*b^4 + 27*a^2*b^2))/(2*a^2*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^4*(40*a*b^3 - 64*a^3*b + 24* a^4 - 8*b^4 + 53*a^2*b^2))/(2*a^2*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^8*(24*a*b^2 - 9*a^2*b + 8*a^3 - 8*b^3))/(4*a*(a - b)*(a^2 - 2*a *b + b^2)) + (tan(e/2 + (f*x)/2)^2*(27*a*b^2 + 23*a^2*b - 16*a^3 - 4*b^3)) /(2*a*(a - b)*(a^2 - 2*a*b + b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(8*a*b - 3*a^ 2) + tan(e/2 + (f*x)/2)^8*(8*a*b - 3*a^2) + a^2*tan(e/2 + (f*x)/2)^10 + ta n(e/2 + (f*x)/2)^4*(2*a^2 - 8*a*b + 16*b^2) + tan(e/2 + (f*x)/2)^6*(2*a^2 - 8*a*b + 16*b^2) + a^2)) - (15*b^(1/2)*atan(((a - b)^7*(2*tan(e/2 + (f*x) /2)^2*((b^(1/2)*(225*a^8*b + 225*a^2*b^7 - 1350*a^3*b^6 + 3375*a^4*b^5 - 4 500*a^5*b^4 + 3375*a^6*b^3 - 1350*a^7*b^2))/(a*(a - b)^(13/2)) + (225*b^(1 /2)*(a - 2*b)*(128*a^12 - 1408*a^11*b + 256*a^2*b^10 - 2432*a^3*b^9 + 1036 8*a^4*b^8 - 26112*a^5*b^7 + 43008*a^6*b^6 - 48384*a^7*b^5 + 37632*a^8*b^4 - 19968*a^9*b^3 + 6912*a^10*b^2))/(512*a*(a - b)^(21/2))) + (225*b^(1/2)*( a - 2*b)*(1152*a^11*b - 128*a^12 + 128*a^3*b^9 - 1152*a^4*b^8 + 4608*a^5*b ^7 - 10752*a^6*b^6 + 16128*a^7*b^5 - 16128*a^8*b^4 + 10752*a^9*b^3 - 4608* a^10*b^2))/(256*a*(a - b)^(21/2))))/(225*a^8*b + 225*a^2*b^7 - 1350*a^3*b^ 6 + 3375*a^4*b^5 - 4500*a^5*b^4 + 3375*a^6*b^3 - 1350*a^7*b^2)))/(8*f*(a - b)^(7/2))