Integrand size = 21, antiderivative size = 100 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 a b \text {arctanh}(\sin (c+d x))}{d}-\frac {\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac {\left (a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac {2 a b \csc (c+d x)}{d}-\frac {2 a b \csc ^3(c+d x)}{3 d}+\frac {b^2 \tan (c+d x)}{d} \]
2*a*b*arctanh(sin(d*x+c))/d-(a^2+2*b^2)*cot(d*x+c)/d-1/3*(a^2+b^2)*cot(d*x +c)^3/d-2*a*b*csc(d*x+c)/d-2/3*a*b*csc(d*x+c)^3/d+b^2*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(259\) vs. \(2(100)=200\).
Time = 1.28 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.59 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {\csc ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-3 a^2-14 a b \cos (c+d x)-2 \left (a^2+4 b^2\right ) \cos (2 (c+d x))+6 a b \cos (3 (c+d x))+a^2 \cos (4 (c+d x))+4 b^2 \cos (4 (c+d x))-6 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+6 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (2 (c+d x))+3 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))-3 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 (c+d x))\right )}{96 d \left (-1+\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]
(Csc[(c + d*x)/2]^5*Sec[(c + d*x)/2]^3*(-3*a^2 - 14*a*b*Cos[c + d*x] - 2*( a^2 + 4*b^2)*Cos[2*(c + d*x)] + 6*a*b*Cos[3*(c + d*x)] + a^2*Cos[4*(c + d* x)] + 4*b^2*Cos[4*(c + d*x)] - 6*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/ 2]]*Sin[2*(c + d*x)] + 6*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[ 2*(c + d*x)] + 3*a*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[4*(c + d *x)] - 3*a*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*(c + d*x)]))/( 96*d*(-1 + Cot[(c + d*x)/2]^2))
Time = 0.69 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4360, 3042, 3390, 25, 3042, 3101, 25, 254, 2009, 4889, 355, 2009}
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.
| \(\displaystyle \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \csc ^4(c+d x) \sec ^2(c+d x) (-a \cos (c+d x)-b)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-b\right )^2}{\sin \left (c+d x-\frac {\pi }{2}\right )^2 \cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3390 |
| \(\displaystyle \int \left (b^2+a^2 \cos ^2(c+d x)\right ) \csc ^4(c+d x) \sec ^2(c+d x)dx-2 a b \int -\csc ^4(c+d x) \sec (c+d x)dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \left (b^2+a^2 \cos ^2(c+d x)\right ) \csc ^4(c+d x) \sec ^2(c+d x)dx+2 a b \int \csc ^4(c+d x) \sec (c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {b^2+a^2 \cos (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^4}dx+2 a b \int \csc (c+d x)^4 \sec (c+d x)dx\) |
| \(\Big \downarrow \) 3101 |
| \(\displaystyle \int \frac {b^2+a^2 \cos (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^4}dx-\frac {2 a b \int -\frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {b^2+a^2 \cos (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^4}dx+\frac {2 a b \int \frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \int \frac {b^2+a^2 \cos (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^4}dx+\frac {2 a b \int \left (-\csc ^2(c+d x)+\frac {1}{1-\csc ^2(c+d x)}-1\right )d\csc (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \frac {b^2+a^2 \cos (c+d x)^2}{\cos (c+d x)^2 \sin (c+d x)^4}dx-\frac {2 a b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \frac {\int \cot ^4(c+d x) \left (\tan ^2(c+d x)+1\right ) \left (a^2+b^2+b^2 \tan ^2(c+d x)\right )d\tan (c+d x)}{d}-\frac {2 a b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 355 |
| \(\displaystyle \frac {\int \left (\left (a^2+b^2\right ) \cot ^4(c+d x)+\left (a^2+2 b^2\right ) \cot ^2(c+d x)+b^2\right )d\tan (c+d x)}{d}-\frac {2 a b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{3} \left (a^2+b^2\right ) \cot ^3(c+d x)-\left (a^2+2 b^2\right ) \cot (c+d x)+b^2 \tan (c+d x)}{d}-\frac {2 a b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
(-2*a*b*(-ArcTanh[Csc[c + d*x]] + Csc[c + d*x] + Csc[c + d*x]^3/3))/d + (- ((a^2 + 2*b^2)*Cot[c + d*x]) - ((a^2 + b^2)*Cot[c + d*x]^3)/3 + b^2*Tan[c + d*x])/d
3.2.83.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Time = 1.67 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+2 a b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) | \(116\) |
| default | \(\frac {a^{2} \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+2 a b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {1}{3 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )}+\frac {4}{3 \sin \left (d x +c \right ) \cos \left (d x +c \right )}-\frac {8 \cot \left (d x +c \right )}{3}\right )}{d}\) | \(116\) |
| parallelrisch | \(\frac {-96 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+96 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\left (a^{2}+4 b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {a^{2}}{2}-2 b^{2}\right ) \cos \left (4 d x +4 c \right )+\frac {3 a \left (\frac {14 \cos \left (d x +c \right ) b}{3}-2 b \cos \left (3 d x +3 c \right )+a \right )}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{48 d \cos \left (d x +c \right )}\) | \(150\) |
| risch | \(-\frac {4 i \left (3 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-7 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-7 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{i \left (d x +c \right )}+a^{2}+4 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(177\) |
| norman | \(\frac {\frac {a^{2}+2 a b +b^{2}}{24 d}-\frac {3 \left (a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}+\frac {\left (a^{2}-2 a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{24 d}+\frac {\left (2 a^{2}-7 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{6 d}+\frac {\left (2 a^{2}+7 a b +5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(198\) |
1/d*(a^2*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+2*a*b*(-1/3/sin(d*x+c)^3-1/sin (d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+b^2*(-1/3/sin(d*x+c)^3/cos(d*x+c)+4/3/s in(d*x+c)/cos(d*x+c)-8/3*cot(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.78 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {6 \, a b \cos \left (d x + c\right )^{3} + 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 8 \, a b \cos \left (d x + c\right ) - 3 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a b \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, b^{2}}{3 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \]
-1/3*(6*a*b*cos(d*x + c)^3 + 2*(a^2 + 4*b^2)*cos(d*x + c)^4 - 8*a*b*cos(d* x + c) - 3*(a^2 + 4*b^2)*cos(d*x + c)^2 - 3*(a*b*cos(d*x + c)^3 - a*b*cos( d*x + c))*log(sin(d*x + c) + 1)*sin(d*x + c) + 3*(a*b*cos(d*x + c)^3 - a*b *cos(d*x + c))*log(-sin(d*x + c) + 1)*sin(d*x + c) + 3*b^2)/((d*cos(d*x + c)^3 - d*cos(d*x + c))*sin(d*x + c))
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \csc ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + b^{2} {\left (\frac {6 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{3}} - 3 \, \tan \left (d x + c\right )\right )} + \frac {{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]
-1/3*(a*b*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + b^2*((6*tan(d*x + c)^2 + 1)/tan(d*x + c)^3 - 3*tan(d*x + c)) + (3*tan(d*x + c)^2 + 1)*a^2/tan(d*x + c)^3)/d
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (96) = 192\).
Time = 0.34 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.26 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {48 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} + 2 \, a b + b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]
1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 2*a*b*tan(1/2*d*x + 1/2*c)^3 + b^2*tan( 1/2*d*x + 1/2*c)^3 + 48*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 48*a*b*lo g(abs(tan(1/2*d*x + 1/2*c) - 1)) + 9*a^2*tan(1/2*d*x + 1/2*c) - 30*a*b*tan (1/2*d*x + 1/2*c) + 21*b^2*tan(1/2*d*x + 1/2*c) - 48*b^2*tan(1/2*d*x + 1/2 *c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (9*a^2*tan(1/2*d*x + 1/2*c)^2 + 30*a*b* tan(1/2*d*x + 1/2*c)^2 + 21*b^2*tan(1/2*d*x + 1/2*c)^2 + a^2 + 2*a*b + b^2 )/tan(1/2*d*x + 1/2*c)^3)/d
Time = 14.68 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.82 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\left (a-b\right )}^2}{24\,d}-\frac {\frac {2\,a\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^2+10\,a\,b+23\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {8\,a^2}{3}+\frac {28\,a\,b}{3}+\frac {20\,b^2}{3}\right )+\frac {a^2}{3}+\frac {b^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2}{8}-\frac {3\,a\,b}{4}+\frac {5\,b^2}{8}+\frac {{\left (a-b\right )}^2}{4}\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]
(tan(c/2 + (d*x)/2)^3*(a - b)^2)/(24*d) - ((2*a*b)/3 - tan(c/2 + (d*x)/2)^ 4*(10*a*b + 3*a^2 + 23*b^2) + tan(c/2 + (d*x)/2)^2*((28*a*b)/3 + (8*a^2)/3 + (20*b^2)/3) + a^2/3 + b^2/3)/(d*(8*tan(c/2 + (d*x)/2)^3 - 8*tan(c/2 + ( d*x)/2)^5)) + (tan(c/2 + (d*x)/2)*(a^2/8 - (3*a*b)/4 + (5*b^2)/8 + (a - b) ^2/4))/d + (4*a*b*atanh(tan(c/2 + (d*x)/2)))/d