Integrand size = 30, antiderivative size = 98 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {3 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}+\frac {8 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \]
-A*arctanh(cos(d*x+c))/a^3/d+2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3+3/5*A *cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2+8/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(313\) vs. \(2(98)=196\).
Time = 6.55 (sec) , antiderivative size = 313, normalized size of antiderivative = 3.19 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (2 \cos \left (\frac {c}{2}\right )-2 \sin \left (\frac {c}{2}\right )+3 \cos \left (\frac {c}{2}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-3 \sin \left (\frac {c}{2}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2-5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4\right )+2 \sin \left (\frac {d x}{2}\right ) (-17+4 \cos (2 (c+d x))-19 \sin (c+d x))\right ) (A-A \sin (c+d x))}{5 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]
(((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(2*Cos[c/2] - 2*Sin[c/2] + 3*Cos[c /2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 3*Sin[c/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 5*Log[Cos[(c + d*x)/2]]*(Cos[c/2] + Sin[c/2])*(Co s[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 5*Log[Sin[(c + d*x)/2]]*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + 2*Sin[(d*x)/2]*(-17 + 4*Cos[2*(c + d*x)] - 19*Sin[c + d*x]))*(A - A*Sin[c + d*x]))/(5*a^3*d*(C os[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2*(Cos[(c + d*x) /2] + Sin[(c + d*x)/2])^5)
Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3445, 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.
| \(\displaystyle \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A-A \sin (c+d x)}{\sin (c+d x) (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3445 |
| \(\displaystyle \int \left (-\frac {A}{a^3 (\sin (c+d x)+1)}-\frac {A}{a^3 (\sin (c+d x)+1)^2}-\frac {2 A}{a^3 (\sin (c+d x)+1)^3}+\frac {A \csc (c+d x)}{a^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {A \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {8 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {3 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}+\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}\) |
-((A*ArcTanh[Cos[c + d*x]])/(a^3*d)) + (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Si n[c + d*x])^3) + (3*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^2) + (8*A* Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))
3.3.40.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[si n[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; FreeQ[{ a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ [m] && IntegerQ[n]
Time = 0.93 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {A \left (\frac {16}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {12}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {10}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{d \,a^{3}}\) | \(95\) |
| default | \(\frac {A \left (\frac {16}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {12}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {10}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{d \,a^{3}}\) | \(95\) |
| risch | \(\frac {2 A \left (25 i {\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{4 i \left (d x +c \right )}-35 i {\mathrm e}^{i \left (d x +c \right )}-55 \,{\mathrm e}^{2 i \left (d x +c \right )}+8\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(114\) |
| norman | \(\frac {\frac {8 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {18 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {38 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {26 A}{5 a d}+\frac {22 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {40 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {176 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\) | \(180\) |
| parallelrisch | \(\frac {5 \left (\left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-\frac {\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )}{5}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-\frac {\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )}{5}+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {19 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )}{5}-\frac {\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )}{5}+\frac {32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {7 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )}{5}-\frac {21 \cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )}{25}+4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3} \left (-\sin \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+5 \sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-5 \cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+10 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(222\) |
1/d*A/a^3*(16/5/(tan(1/2*d*x+1/2*c)+1)^5-8/(tan(1/2*d*x+1/2*c)+1)^4+12/(ta n(1/2*d*x+1/2*c)+1)^3-10/(tan(1/2*d*x+1/2*c)+1)^2+8/(tan(1/2*d*x+1/2*c)+1) +ln(tan(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (92) = 184\).
Time = 0.26 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.16 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \, A \cos \left (d x + c\right )^{3} - 22 \, A \cos \left (d x + c\right )^{2} - 42 \, A \cos \left (d x + c\right ) - 5 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (8 \, A \cos \left (d x + c\right )^{2} + 19 \, A \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 4 \, A}{10 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]
1/10*(16*A*cos(d*x + c)^3 - 22*A*cos(d*x + c)^2 - 42*A*cos(d*x + c) - 5*(A *cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + (A*cos(d*x + c)^ 2 - 2*A*cos(d*x + c) - 4*A)*sin(d*x + c) - 4*A)*log(1/2*cos(d*x + c) + 1/2 ) + 5*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + (A*cos(d *x + c)^2 - 2*A*cos(d*x + c) - 4*A)*sin(d*x + c) - 4*A)*log(-1/2*cos(d*x + c) + 1/2) - 2*(8*A*cos(d*x + c)^2 + 19*A*cos(d*x + c) - 2*A)*sin(d*x + c) - 4*A)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*si n(d*x + c))
\[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=- \frac {A \left (\int \left (- \frac {\csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\right )\, dx + \int \frac {\sin {\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx\right )}{a^{3}} \]
-A*(Integral(-csc(c + d*x)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x) + Integral(sin(c + d*x)*csc(c + d*x)/(sin(c + d*x)**3 + 3* sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (92) = 184\).
Time = 0.22 (sec) , antiderivative size = 433, normalized size of antiderivative = 4.42 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A {\left (\frac {2 \, {\left (\frac {115 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {185 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {135 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 32\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, A {\left (\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}}{15 \, d} \]
1/15*(A*(2*(115*sin(d*x + c)/(cos(d*x + c) + 1) + 185*sin(d*x + c)^2/(cos( d*x + c) + 1)^2 + 135*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 45*sin(d*x + c )^4/(cos(d*x + c) + 1)^4 + 32)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1 ) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(co s(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d* x + c)^5/(cos(d*x + c) + 1)^5) + 15*log(sin(d*x + c)/(cos(d*x + c) + 1))/a ^3) + 2*A*(20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/ (cos(d*x + c) + 1)^4 + 7)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 1 0*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c )^5/(cos(d*x + c) + 1)^5))/d
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {5 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {2 \, {\left (20 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 55 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 45 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, d} \]
1/5*(5*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 2*(20*A*tan(1/2*d*x + 1/2*c) ^4 + 55*A*tan(1/2*d*x + 1/2*c)^3 + 75*A*tan(1/2*d*x + 1/2*c)^2 + 45*A*tan( 1/2*d*x + 1/2*c) + 13*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 14.77 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.03 \[ \int \frac {\csc (c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A\,\left (5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+90\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+150\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+110\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+25\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+50\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+50\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+25\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+26\right )}{5\,a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]
(A*(5*log(tan(c/2 + (d*x)/2)) + 90*tan(c/2 + (d*x)/2) + 150*tan(c/2 + (d*x )/2)^2 + 110*tan(c/2 + (d*x)/2)^3 + 40*tan(c/2 + (d*x)/2)^4 + 25*log(tan(c /2 + (d*x)/2))*tan(c/2 + (d*x)/2) + 50*log(tan(c/2 + (d*x)/2))*tan(c/2 + ( d*x)/2)^2 + 50*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^3 + 25*log(tan(c /2 + (d*x)/2))*tan(c/2 + (d*x)/2)^4 + 5*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^5 + 26))/(5*a^3*d*(tan(c/2 + (d*x)/2) + 1)^5)