Integrand size = 25, antiderivative size = 148 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=-\frac {2 d (2 c+d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{3 (c-d) \left (c^2-d^2\right )^{3/2} f}-\frac {d (c+2 d) \cos (e+f x)}{3 (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{(c-d) f (3+3 \sin (e+f x)) (c+d \sin (e+f x))} \]
-2*d*(2*c+d)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a/(c-d)/(c^2 -d^2)^(3/2)/f-d*(c+2*d)*cos(f*x+e)/a/(c-d)^2/(c+d)/f/(c+d*sin(f*x+e))-cos( f*x+e)/(c-d)/f/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))
Time = 0.42 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\frac {\cos (e+f x) \left (\frac {2 d (2 c+d) \text {arctanh}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right )}{\sqrt {-c-d} (c-d)^{3/2} \sqrt {\cos ^2(e+f x)}}+\frac {c+2 d}{(c-d) (1+\sin (e+f x))}-\frac {d}{(1+\sin (e+f x)) (c+d \sin (e+f x))}\right )}{3 (-c+d) (c+d) f} \]
(Cos[e + f*x]*((2*d*(2*c + d)*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]]) /(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/(Sqrt[-c - d]*(c - d)^(3/2)*Sqrt[ Cos[e + f*x]^2]) + (c + 2*d)/((c - d)*(1 + Sin[e + f*x])) - d/((1 + Sin[e + f*x])*(c + d*Sin[e + f*x]))))/(3*(-c + d)*(c + d)*f)
Time = 0.55 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3247, 25, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) (c+d \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a) (c+d \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3247 |
| \(\displaystyle \frac {d \int -\frac {2 a-a \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {d \int \frac {2 a-a \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {d \int \frac {2 a-a \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle -\frac {d \left (\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\int -\frac {a (2 c+d)}{c+d \sin (e+f x)}dx}{c^2-d^2}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {d \left (\frac {\int \frac {a (2 c+d)}{c+d \sin (e+f x)}dx}{c^2-d^2}+\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d \left (\frac {a (2 c+d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}+\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {d \left (\frac {a (2 c+d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}+\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {d \left (\frac {2 a (2 c+d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}+\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {d \left (\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {4 a (2 c+d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {d \left (\frac {2 a (2 c+d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}+\frac {a (c+2 d) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\right )}{a^2 (c-d)}-\frac {\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\) |
-(Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x]))) - (d *((2*a*(2*c + d)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])]) /((c^2 - d^2)^(3/2)*f) + (a*(c + 2*d)*Cos[e + f*x])/((c^2 - d^2)*f*(c + d* Sin[e + f*x]))))/(a^2*(c - d))
3.5.59.3.1 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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x]))), x] + Simp[d/(a*(b*c - a*d)) Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[ c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])
Time = 1.18 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {-\frac {2}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d \left (\frac {\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 c +d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}}{f a}\) | \(154\) |
| default | \(\frac {-\frac {2}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 d \left (\frac {\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 c +d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}}{f a}\) | \(154\) |
| risch | \(-\frac {2 \left (-2 i c d \,{\mathrm e}^{2 i \left (f x +e \right )}-i d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+i c d +2 i d^{2}+2 c^{2} {\mathrm e}^{i \left (f x +e \right )}+3 d \,{\mathrm e}^{i \left (f x +e \right )} c +d^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) \left (c +d \right ) \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right ) f \left (c -d \right )^{2} a}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}\) | \(482\) |
2/f/a*(-1/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)-1/(c-d)^2*d*((d^2/(c+d)/c*tan(1/2 *f*x+1/2*e)+d/(c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(2* c+d)/(c+d)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^ 2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (145) = 290\).
Time = 0.33 (sec) , antiderivative size = 1120, normalized size of antiderivative = 7.57 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]
[1/2*(2*c^4 - 4*c^2*d^2 + 2*d^4 + 2*(c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*co s(f*x + e)^2 + (2*c^2*d + 3*c*d^2 + d^3 - (2*c*d^2 + d^3)*cos(f*x + e)^2 + (2*c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 + d^ 3)*cos(f*x + e))*sin(f*x + e))*sqrt(-c^2 + d^2)*log(-((2*c^2 - d^2)*cos(f* x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 - 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*(c^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e) - 2*(c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e))*sin(f* x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^ 6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2 *d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^ 6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6 )*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e)), (c^4 - 2*c^2*d^2 + d^4 + (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f* x + e)^2 - (2*c^2*d + 3*c*d^2 + d^3 - (2*c*d^2 + d^3)*cos(f*x + e)^2 + (2* c^2*d + c*d^2)*cos(f*x + e) + (2*c^2*d + 3*c*d^2 + d^3 + (2*c*d^2 + d^3)*c os(f*x + e))*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(s qrt(c^2 - d^2)*cos(f*x + e))) + (c^4 + c^3*d - c*d^3 - d^4)*cos(f*x + e) - (c^4 - 2*c^2*d^2 + d^4 - (c^3*d + 2*c^2*d^2 - c*d^3 - 2*d^4)*cos(f*x + e) )*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c...
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (145) = 290\).
Time = 0.32 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.01 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (2 \, c d + d^{2}\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c^{3} + c^{2} d + c d^{2}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \]
-2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2 *e) + d)/sqrt(c^2 - d^2)))*(2*c*d + d^2)/((a*c^3 - a*c^2*d - a*c*d^2 + a*d ^3)*sqrt(c^2 - d^2)) + (c^3*tan(1/2*f*x + 1/2*e)^2 + c^2*d*tan(1/2*f*x + 1 /2*e)^2 + d^3*tan(1/2*f*x + 1/2*e)^2 + 2*c^2*d*tan(1/2*f*x + 1/2*e) + 3*c* d^2*tan(1/2*f*x + 1/2*e) + d^3*tan(1/2*f*x + 1/2*e) + c^3 + c^2*d + c*d^2) /((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2*e)^3 + c*ta n(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 1/2*e) + 2*d*tan(1/2*f*x + 1/2*e) + c)))/f
Time = 8.93 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.09 \[ \int \frac {1}{(3+3 \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=-\frac {\frac {2\,\left (c^2+c\,d+d^2\right )}{\left (c+d\right )\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (d^2+2\,c\,d\right )}{c\,{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^3+c^2\,d+d^3\right )}{c\,\left (c+d\right )\,{\left (c-d\right )}^2}}{f\,\left (a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (a\,c+2\,a\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,c\right )}-\frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (2\,c+d\right )\,\left (2\,a\,c^3\,d-2\,a\,c^2\,d^2-2\,a\,c\,d^3+2\,a\,d^4\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c+d\right )\,\left (a\,c^3-a\,c^2\,d-a\,c\,d^2+a\,d^3\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}}{2\,d^2+4\,c\,d}\right )\,\left (2\,c+d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}} \]
- ((2*(c*d + c^2 + d^2))/((c + d)*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(2*c* d + d^2))/(c*(c - d)^2) + (2*tan(e/2 + (f*x)/2)^2*(c^2*d + c^3 + d^3))/(c* (c + d)*(c - d)^2))/(f*(a*c + tan(e/2 + (f*x)/2)^2*(a*c + 2*a*d) + tan(e/2 + (f*x)/2)*(a*c + 2*a*d) + a*c*tan(e/2 + (f*x)/2)^3)) - (2*d*atan(((d*(2* c + d)*(2*a*d^4 - 2*a*c^2*d^2 - 2*a*c*d^3 + 2*a*c^3*d))/(a*(c + d)^(3/2)*( c - d)^(5/2)) + (2*c*d*tan(e/2 + (f*x)/2)*(2*c + d)*(a*c^3 + a*d^3 - a*c*d ^2 - a*c^2*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)))/(4*c*d + 2*d^2))*(2*c + d) )/(a*f*(c + d)^(3/2)*(c - d)^(5/2))