Integrand size = 31, antiderivative size = 99 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {(A b-a B) \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \]
-(A*b-B*a)*arctanh(sin(d*x+c))/a^2/d+2*b*(A*b-B*a)*arctan((a-b)^(1/2)*tan( 1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/d/(a-b)^(1/2)/(a+b)^(1/2)+A*tan(d*x+c)/a/d
Time = 0.74 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {2 b (A b-a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+(A b-a B) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+a A \tan (c+d x)}{a^2 d} \]
((-2*b*(A*b - a*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/S qrt[-a^2 + b^2] + (A*b - a*B)*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + a*A*Tan[c + d*x])/(a^2*d)
Time = 0.52 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3479, 25, 27, 3042, 3226, 3042, 3138, 218, 4257}
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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3479 |
| \(\displaystyle \frac {\int -\frac {(A b-a B) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {A \tan (c+d x)}{a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {\int \frac {(A b-a B) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \int \frac {\sec (c+d x)}{a+b \cos (c+d x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\) |
| \(\Big \downarrow \) 3226 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \left (\frac {\int \sec (c+d x)dx}{a}-\frac {b \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \left (\frac {\text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}\) |
-(((A*b - a*B)*((-2*b*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/ (a*Sqrt[a - b]*Sqrt[a + b]*d) + ArcTanh[Sin[c + d*x]]/(a*d)))/a) + (A*Tan[ c + d*x])/(a*d)
3.3.55.3.1 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/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[d/(b*c - a*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_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.64 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.45
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(144\) |
| default | \(\frac {\frac {2 b \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {A}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(144\) |
| risch | \(\frac {2 i A}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}\) | \(407\) |
1/d*(2*b*(A*b-B*a)/a^2/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c) /((a-b)*(a+b))^(1/2))-A/a/(tan(1/2*d*x+1/2*c)-1)+(A*b-B*a)/a^2*ln(tan(1/2* d*x+1/2*c)-1)-A/a/(tan(1/2*d*x+1/2*c)+1)+1/a^2*(-A*b+B*a)*ln(tan(1/2*d*x+1 /2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (89) = 178\).
Time = 0.39 (sec) , antiderivative size = 460, normalized size of antiderivative = 4.65 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (B a b - A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, -\frac {2 \, {\left (B a b - A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \]
[1/2*((B*a*b - A*b^2)*sqrt(-a^2 + b^2)*cos(d*x + c)*log((2*a*b*cos(d*x + c ) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b) *sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^ 2)) + (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(A*a^3 - A*a*b^2)*sin(d*x + c))/((a^4 - a^2*b^2)*d*cos(d*x + c)), - 1/2*(2*(B*a*b - A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt( a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (B*a^3 - A*a^2*b - B*a*b^2 + A*b^ 3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (B*a^3 - A*a^2*b - B*a*b^2 + A*b^3 )*cos(d*x + c)*log(-sin(d*x + c) + 1) - 2*(A*a^3 - A*a*b^2)*sin(d*x + c))/ ((a^4 - a^2*b^2)*d*cos(d*x + c))]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.77 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a} + \frac {2 \, {\left (B a b - A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2}}}{d} \]
((B*a - A*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - (B*a - A*b)*log(abs( tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a) + 2*(B*a*b - A*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn (-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/s qrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^2))/d
Time = 2.30 (sec) , antiderivative size = 675, normalized size of antiderivative = 6.82 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {A\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d\,\left (a^2-b^2\right )}-\frac {B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d\,\left (a^2-b^2\right )}+\frac {A\,a\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}-\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{a^2\,d\,\left (a^2-b^2\right )}+\frac {B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{a\,d\,\left (a^2-b^2\right )}-\frac {A\,b^2\,\mathrm {tan}\left (c+d\,x\right )}{a\,d\,\left (a^2-b^2\right )}-\frac {B\,b\,\mathrm {atan}\left (\frac {\left (a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a\,b^2-a^3\right )}^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,2{}\mathrm {i}}{a\,d\,\left (a^2-b^2\right )}+\frac {A\,b^2\,\mathrm {atan}\left (\frac {\left (a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (b^2-a^2\right )}^{3/2}-2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}+3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}-a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a\,b^2-a^3\right )}^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,2{}\mathrm {i}}{a^2\,d\,\left (a^2-b^2\right )} \]
(A*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(d*(a^2 - b^2)) - (B*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(d*(a^2 - b^2) ) + (A*a*tan(c + d*x))/(d*(a^2 - b^2)) - (A*b^3*atan((sin(c/2 + (d*x)/2)*1 i)/cos(c/2 + (d*x)/2))*2i)/(a^2*d*(a^2 - b^2)) + (B*b^2*atan((sin(c/2 + (d *x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/(a*d*(a^2 - b^2)) - (A*b^2*tan(c + d*x) )/(a*d*(a^2 - b^2)) - (B*b*atan(((a^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*b^5*sin(c/2 + (d*x)/2)*( b^2 - a^2)^(1/2) + 3*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^3*b^ 2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^4*b*sin(c/2 + (d*x)/2)*(b^2 - a ^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)^2))*(-(a + b)*(a - b))^(1 /2)*2i)/(a*d*(a^2 - b^2)) + (A*b^2*atan(((a^5*sin(c/2 + (d*x)/2)*(b^2 - a^ 2)^(1/2) + 2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 2*b^5*sin(c/2 + (d *x)/2)*(b^2 - a^2)^(1/2) + 3*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^3*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - a^4*b*sin(c/2 + (d*x)/2)* (b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)^2))*(-(a + b)*(a - b))^(1/2)*2i)/(a^2*d*(a^2 - b^2))