Integrand size = 25, antiderivative size = 83 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=-\frac {9 (c-2 d) x}{d^2}+\frac {18 (c-d)^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^2 \sqrt {c^2-d^2} f}-\frac {9 \cos (e+f x)}{d f} \]
-a^2*(c-2*d)*x/d^2-a^2*cos(f*x+e)/d/f+2*a^2*(c-d)^2*arctan((d+c*tan(1/2*f* x+1/2*e))/(c^2-d^2)^(1/2))/d^2/f/(c^2-d^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=-\frac {9 \left ((c-2 d) (e+f x)-\frac {2 (c-d)^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+d \cos (e+f x)\right )}{d^2 f} \]
(-9*((c - 2*d)*(e + f*x) - (2*(c - d)^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sq rt[c^2 - d^2]])/Sqrt[c^2 - d^2] + d*Cos[e + f*x]))/(d^2*f)
Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3225, 3042, 3214, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{c+d \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{c+d \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3225 |
| \(\displaystyle \frac {\int \frac {a^2 d-a^2 (c-2 d) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2 d-a^2 (c-2 d) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {a^2 (c-d)^2 \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {a^2 x (c-2 d)}{d}}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (c-d)^2 \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {a^2 x (c-2 d)}{d}}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {2 a^2 (c-d)^2 \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}-\frac {a^2 x (c-2 d)}{d}}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {-\frac {4 a^2 (c-d)^2 \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}-\frac {a^2 x (c-2 d)}{d}}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {2 a^2 (c-d)^2 \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}-\frac {a^2 x (c-2 d)}{d}}{d}-\frac {a^2 \cos (e+f x)}{d f}\) |
(-((a^2*(c - 2*d)*x)/d) + (2*a^2*(c - d)^2*ArcTan[(2*d + 2*c*Tan[(e + f*x) /2])/(2*Sqrt[c^2 - d^2])])/(d*Sqrt[c^2 - d^2]*f))/d - (a^2*Cos[e + f*x])/( d*f)
3.5.39.3.1 Defintions of rubi rules used
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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.85 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {\frac {d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (c -2 d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}+\frac {\left (c^{2}-2 c d +d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{2} \sqrt {c^{2}-d^{2}}}\right )}{f}\) | \(105\) |
| default | \(\frac {2 a^{2} \left (-\frac {\frac {d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (c -2 d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}+\frac {\left (c^{2}-2 c d +d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{2} \sqrt {c^{2}-d^{2}}}\right )}{f}\) | \(105\) |
| risch | \(-\frac {a^{2} x c}{d^{2}}+\frac {2 a^{2} x}{d}-\frac {a^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 d f}-\frac {a^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 d f}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right ) f \,d^{2}}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {-i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) c}{\left (c +d \right ) f \,d^{2}}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {-i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}\) | \(303\) |
2/f*a^2*(-1/d^2*(d/(1+tan(1/2*f*x+1/2*e)^2)+(c-2*d)*arctan(tan(1/2*f*x+1/2 *e)))+(c^2-2*c*d+d^2)/d^2/(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)))
Time = 0.30 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.57 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=\left [-\frac {2 \, a^{2} d \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c - 2 \, a^{2} d\right )} f x + {\left (a^{2} c - a^{2} d\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right )}{2 \, d^{2} f}, -\frac {a^{2} d \cos \left (f x + e\right ) + {\left (a^{2} c - 2 \, a^{2} d\right )} f x + {\left (a^{2} c - a^{2} d\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right )}{d^{2} f}\right ] \]
[-1/2*(2*a^2*d*cos(f*x + e) + 2*(a^2*c - 2*a^2*d)*f*x + (a^2*c - a^2*d)*sq rt(-(c - d)/(c + d))*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e ) - c^2 - d^2 + 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos (f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e ) - c^2 - d^2)))/(d^2*f), -(a^2*d*cos(f*x + e) + (a^2*c - 2*a^2*d)*f*x + ( a^2*c - a^2*d)*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))))/(d^2*f)]
Leaf count of result is larger than twice the leaf count of optimal. 3709 vs. \(2 (76) = 152\).
Time = 124.12 (sec) , antiderivative size = 3709, normalized size of antiderivative = 44.69 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*x*(a*sin(e) + a)**2/sin(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)) , ((2*a**2*f*x*tan(e/2 + f*x/2)**2/(f*tan(e/2 + f*x/2)**2 + f) + 2*a**2*f* x/(f*tan(e/2 + f*x/2)**2 + f) + a**2*log(tan(e/2 + f*x/2))*tan(e/2 + f*x/2 )**2/(f*tan(e/2 + f*x/2)**2 + f) + a**2*log(tan(e/2 + f*x/2))/(f*tan(e/2 + f*x/2)**2 + f) - 2*a**2/(f*tan(e/2 + f*x/2)**2 + f))/d, Eq(c, 0)), (2*a** 2*d**2*f*x*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/ 2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - a**2 *d**2*f*x*tan(e/2 + f*x/2)**2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 2*a** 2*d**2*f*x*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - a**2*d* *2*f*x/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**( 3/2)*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 4*a**2*d**2*tan(e/2 + f*x/2) **2/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2 )*tan(e/2 + f*x/2)**2 - f*(d**2)**(3/2)) - 2*a**2*d**2*tan(e/2 + f*x/2)/(d **3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan( e/2 + f*x/2)**2 - f*(d**2)**(3/2)) + 4*a**2*d**2/(d**3*f*tan(e/2 + f*x/2)* *3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)**2 - f*(d* *2)**(3/2)) + a**2*d*f*x*sqrt(d**2)*tan(e/2 + f*x/2)**3/(d**3*f*tan(e/2 + f*x/2)**3 + d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)*tan(e/2 + f*x/2)*...
Exception generated. \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.59 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=-\frac {\frac {{\left (a^{2} c - 2 \, a^{2} d\right )} {\left (f x + e\right )}}{d^{2}} + \frac {2 \, a^{2}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )} d} - \frac {2 \, {\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{2}}}{f} \]
-((a^2*c - 2*a^2*d)*(f*x + e)/d^2 + 2*a^2/((tan(1/2*f*x + 1/2*e)^2 + 1)*d) - 2*(a^2*c^2 - 2*a^2*c*d + a^2*d^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)))/(sqrt(c^2 - d^ 2)*d^2))/f
Time = 8.21 (sec) , antiderivative size = 940, normalized size of antiderivative = 11.33 \[ \int \frac {(3+3 \sin (e+f x))^2}{c+d \sin (e+f x)} \, dx=\frac {4\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,\left (c+d\right )}-\frac {a^2\,\cos \left (e+f\,x\right )}{f\,\left (c+d\right )}+\frac {2\,a^2\,c\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{d\,f\,\left (c+d\right )}-\frac {2\,a^2\,c^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{d^2\,f\,\left (c+d\right )}-\frac {a^2\,c\,\cos \left (e+f\,x\right )}{d\,f\,\left (c+d\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {\left (3\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (-c^4+2\,c^3\,d-2\,c\,d^3+d^4\right )}^{3/2}-2\,c^6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}-2\,c^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (-c^4+2\,c^3\,d-2\,c\,d^3+d^4\right )}^{3/2}+7\,d^6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+10\,c\,d^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+4\,c^5\,d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+4\,c^2\,d^4\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}-3\,c^3\,d^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}-2\,c^4\,d^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}-9\,c^2\,d^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}-12\,c^3\,d^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+6\,c^4\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+c\,d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (-c^4+2\,c^3\,d-2\,c\,d^3+d^4\right )}^{3/2}+4\,c\,d^5\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}+c^5\,d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {-c^4+2\,c^3\,d-2\,c\,d^3+d^4}\right )\,1{}\mathrm {i}}{d\,\left (c+d\right )\,\left (3\,c^5\,d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-5\,c\,d^5\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-10\,d^6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+16\,c\,d^5\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+8\,c^2\,d^4\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,c^3\,d^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-8\,c^4\,d^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+4\,c^2\,d^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-16\,c^3\,d^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+6\,c^4\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}\right )\,\sqrt {-\left (c+d\right )\,{\left (c-d\right )}^3}\,2{}\mathrm {i}}{d^2\,f\,\left (c+d\right )} \]
(4*a^2*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(f*(c + d)) - (a^2*cos (e + f*x))/(f*(c + d)) + (2*a^2*c*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/ 2)))/(d*f*(c + d)) - (2*a^2*c^2*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2) ))/(d^2*f*(c + d)) + (a^2*atan(((3*d^2*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d ^3 - c^4 + d^4)^(3/2) - 2*c^6*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) - 2*c^2*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(3 /2) + 7*d^6*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) + 10* c*d^5*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) + 4*c^5*d*s in(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) + 4*c^2*d^4*cos(e/ 2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) - 3*c^3*d^3*cos(e/2 + ( f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) - 2*c^4*d^2*cos(e/2 + (f*x)/ 2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) - 9*c^2*d^4*sin(e/2 + (f*x)/2)*(2 *c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) - 12*c^3*d^3*sin(e/2 + (f*x)/2)*(2*c^3 *d - 2*c*d^3 - c^4 + d^4)^(1/2) + 6*c^4*d^2*sin(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2) + c*d*cos(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c ^4 + d^4)^(3/2) + 4*c*d^5*cos(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^ 4)^(1/2) + c^5*d*cos(e/2 + (f*x)/2)*(2*c^3*d - 2*c*d^3 - c^4 + d^4)^(1/2)) *1i)/(d*(c + d)*(3*c^5*d*cos(e/2 + (f*x)/2) - 5*c*d^5*cos(e/2 + (f*x)/2) - 10*d^6*sin(e/2 + (f*x)/2) + 16*c*d^5*sin(e/2 + (f*x)/2) + 8*c^2*d^4*cos(e /2 + (f*x)/2) + 2*c^3*d^3*cos(e/2 + (f*x)/2) - 8*c^4*d^2*cos(e/2 + (f*x...