Integrand size = 21, antiderivative size = 129 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=-\frac {25 a^4 \log (1-\sin (c+d x))}{d}-\frac {16 a^4 \sin (c+d x)}{d}-\frac {9 a^4 \sin ^2(c+d x)}{2 d}-\frac {4 a^4 \sin ^3(c+d x)}{3 d}-\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {a^6}{d (a-a \sin (c+d x))^2}-\frac {11 a^5}{d (a-a \sin (c+d x))} \] Output:
-25*a^4*ln(1-sin(d*x+c))/d-16*a^4*sin(d*x+c)/d-9/2*a^4*sin(d*x+c)^2/d-4/3* a^4*sin(d*x+c)^3/d-1/4*a^4*sin(d*x+c)^4/d+a^6/d/(a-a*sin(d*x+c))^2-11*a^5/ d/(a-a*sin(d*x+c))
Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.64 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=-\frac {a^4 \left (300 \log (1-\sin (c+d x))+\frac {120-132 \sin (c+d x)}{(-1+\sin (c+d x))^2}+192 \sin (c+d x)+54 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \] Input:
Integrate[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^5,x]
Output:
-1/12*(a^4*(300*Log[1 - Sin[c + d*x]] + (120 - 132*Sin[c + d*x])/(-1 + Sin [c + d*x])^2 + 192*Sin[c + d*x] + 54*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/d
Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 86, 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.
\(\displaystyle \int \tan ^5(c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^5 (a \sin (c+d x)+a)^4dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x) (\sin (c+d x) a+a)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {2 a^6}{(a-a \sin (c+d x))^3}-\frac {11 a^5}{(a-a \sin (c+d x))^2}+\frac {25 a^4}{a-a \sin (c+d x)}-\sin ^3(c+d x) a^3-4 \sin ^2(c+d x) a^3-9 \sin (c+d x) a^3-16 a^3\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^6}{(a-a \sin (c+d x))^2}-\frac {11 a^5}{a-a \sin (c+d x)}-\frac {1}{4} a^4 \sin ^4(c+d x)-\frac {4}{3} a^4 \sin ^3(c+d x)-\frac {9}{2} a^4 \sin ^2(c+d x)-16 a^4 \sin (c+d x)-25 a^4 \log (a-a \sin (c+d x))}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^5,x]
Output:
(-25*a^4*Log[a - a*Sin[c + d*x]] - 16*a^4*Sin[c + d*x] - (9*a^4*Sin[c + d* x]^2)/2 - (4*a^4*Sin[c + d*x]^3)/3 - (a^4*Sin[c + d*x]^4)/4 + a^6/(a - a*S in[c + d*x])^2 - (11*a^5)/(a - a*Sin[c + d*x]))/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 35.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(-\frac {a^{4} \left (25 \ln \left (\sin \left (d x +c \right )-1\right )-\frac {11}{\sin \left (d x +c \right )-1}-\frac {1}{\left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}+\frac {9 \sin \left (d x +c \right )^{2}}{2}+16 \sin \left (d x +c \right )\right )}{d}\) | \(83\) |
default | \(-\frac {a^{4} \left (25 \ln \left (\sin \left (d x +c \right )-1\right )-\frac {11}{\sin \left (d x +c \right )-1}-\frac {1}{\left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}+\frac {9 \sin \left (d x +c \right )^{2}}{2}+16 \sin \left (d x +c \right )\right )}{d}\) | \(83\) |
risch | \(25 i a^{4} x -\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}+\frac {19 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {17 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {17 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {19 a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {i a^{4} {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d}+\frac {50 i a^{4} c}{d}+\frac {2 i \left (-11 \,{\mathrm e}^{i \left (d x +c \right )} a^{4}-20 i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+11 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {50 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}\) | \(227\) |
parts | \(\frac {a^{4} \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{4} \left (\frac {\sin \left (d x +c \right )^{10}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{10}}{4 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{8}}{4}-\sin \left (d x +c \right )^{6}-\frac {3 \sin \left (d x +c \right )^{4}}{2}-3 \sin \left (d x +c \right )^{2}-6 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {4 a^{4} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {6 a^{4} \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {4 a^{4} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(408\) |
Input:
int((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
-a^4/d*(25*ln(sin(d*x+c)-1)-11/(sin(d*x+c)-1)-1/(sin(d*x+c)-1)^2+1/4*sin(d *x+c)^4+4/3*sin(d*x+c)^3+9/2*sin(d*x+c)^2+16*sin(d*x+c))
Time = 0.13 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.19 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=-\frac {24 \, a^{4} \cos \left (d x + c\right )^{6} - 272 \, a^{4} \cos \left (d x + c\right )^{4} - 2393 \, a^{4} \cos \left (d x + c\right )^{2} + 1906 \, a^{4} + 2400 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 2 \, a^{4} \sin \left (d x + c\right ) - 2 \, a^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (8 \, a^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{4} \cos \left (d x + c\right )^{2} + 181 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x, algorithm="fricas")
Output:
-1/96*(24*a^4*cos(d*x + c)^6 - 272*a^4*cos(d*x + c)^4 - 2393*a^4*cos(d*x + c)^2 + 1906*a^4 + 2400*(a^4*cos(d*x + c)^2 + 2*a^4*sin(d*x + c) - 2*a^4)* log(-sin(d*x + c) + 1) - 10*(8*a^4*cos(d*x + c)^4 - 96*a^4*cos(d*x + c)^2 + 181*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
\[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=a^{4} \left (\int 4 \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))**4*tan(d*x+c)**5,x)
Output:
a**4*(Integral(4*sin(c + d*x)*tan(c + d*x)**5, x) + Integral(6*sin(c + d*x )**2*tan(c + d*x)**5, x) + Integral(4*sin(c + d*x)**3*tan(c + d*x)**5, x) + Integral(sin(c + d*x)**4*tan(c + d*x)**5, x) + Integral(tan(c + d*x)**5, x))
Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=-\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 54 \, a^{4} \sin \left (d x + c\right )^{2} + 300 \, a^{4} \log \left (\sin \left (d x + c\right ) - 1\right ) + 192 \, a^{4} \sin \left (d x + c\right ) - \frac {12 \, {\left (11 \, a^{4} \sin \left (d x + c\right ) - 10 \, a^{4}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{12 \, d} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x, algorithm="maxima")
Output:
-1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 54*a^4*sin(d*x + c)^ 2 + 300*a^4*log(sin(d*x + c) - 1) + 192*a^4*sin(d*x + c) - 12*(11*a^4*sin( d*x + c) - 10*a^4)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=-\frac {25 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {11 \, a^{4} \sin \left (d x + c\right ) - 10 \, a^{4}}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {3 \, a^{4} d^{3} \sin \left (d x + c\right )^{4} + 16 \, a^{4} d^{3} \sin \left (d x + c\right )^{3} + 54 \, a^{4} d^{3} \sin \left (d x + c\right )^{2} + 192 \, a^{4} d^{3} \sin \left (d x + c\right )}{12 \, d^{4}} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x, algorithm="giac")
Output:
-25*a^4*log(abs(sin(d*x + c) - 1))/d + (11*a^4*sin(d*x + c) - 10*a^4)/(d*( sin(d*x + c) - 1)^2) - 1/12*(3*a^4*d^3*sin(d*x + c)^4 + 16*a^4*d^3*sin(d*x + c)^3 + 54*a^4*d^3*sin(d*x + c)^2 + 192*a^4*d^3*sin(d*x + c))/d^4
Time = 18.64 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.94 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=\frac {25\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {50\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}-\frac {50\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-150\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {950\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}-\frac {1700\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {2180\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {2452\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {2180\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {1700\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {950\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-150\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+50\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \] Input:
int(tan(c + d*x)^5*(a + a*sin(c + d*x))^4,x)
Output:
(25*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (50*a^4*log(tan(c/2 + (d*x)/2) - 1))/d - ((950*a^4*tan(c/2 + (d*x)/2)^3)/3 - 150*a^4*tan(c/2 + (d*x)/2)^2 - (1700*a^4*tan(c/2 + (d*x)/2)^4)/3 + (2180*a^4*tan(c/2 + (d*x)/2)^5)/3 - (2452*a^4*tan(c/2 + (d*x)/2)^6)/3 + (2180*a^4*tan(c/2 + (d*x)/2)^7)/3 - ( 1700*a^4*tan(c/2 + (d*x)/2)^8)/3 + (950*a^4*tan(c/2 + (d*x)/2)^9)/3 - 150* a^4*tan(c/2 + (d*x)/2)^10 + 50*a^4*tan(c/2 + (d*x)/2)^11 + 50*a^4*tan(c/2 + (d*x)/2))/(d*(10*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 20*tan(c/ 2 + (d*x)/2)^3 + 31*tan(c/2 + (d*x)/2)^4 - 40*tan(c/2 + (d*x)/2)^5 + 44*ta n(c/2 + (d*x)/2)^6 - 40*tan(c/2 + (d*x)/2)^7 + 31*tan(c/2 + (d*x)/2)^8 - 2 0*tan(c/2 + (d*x)/2)^9 + 10*tan(c/2 + (d*x)/2)^10 - 4*tan(c/2 + (d*x)/2)^1 1 + tan(c/2 + (d*x)/2)^12 + 1))
Time = 0.17 (sec) , antiderivative size = 436, normalized size of antiderivative = 3.38 \[ \int (a+a \sin (c+d x))^4 \tan ^5(c+d x) \, dx=\frac {a^{4} \left (-16 \sin \left (d x +c \right )^{7}-48 \sin \left (d x +c \right )^{6}-160 \sin \left (d x +c \right )^{5}-6 \tan \left (d x +c \right )^{2}+6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+3 \tan \left (d x +c \right )^{4}-144 \sin \left (d x +c \right )^{2}+288 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{4}+6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sin \left (d x +c \right )^{4}-12 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-588 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+1176 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-6 \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}-6 \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{4}+12 \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}+288 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-588 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+216 \sin \left (d x +c \right )^{4}+500 \sin \left (d x +c \right )^{3}-576 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-300 \sin \left (d x +c \right )+3 \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \sin \left (d x +c \right )^{8}\right )}{12 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a+a*sin(d*x+c))^4*tan(d*x+c)^5,x)
Output:
(a**4*(6*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**4 - 12*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2 + 6*log(tan(c + d*x)**2 + 1) + 288*log(tan((c + d*x) /2)**2 + 1)*sin(c + d*x)**4 - 576*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x )**2 + 288*log(tan((c + d*x)/2)**2 + 1) - 588*log(tan((c + d*x)/2) - 1)*si n(c + d*x)**4 + 1176*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 588*log(t an((c + d*x)/2) - 1) + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 24*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 12*log(tan((c + d*x)/2) + 1) - 3*sin(c + d*x)**8 - 16*sin(c + d*x)**7 - 48*sin(c + d*x)**6 - 160*sin(c + d*x)**5 + 3*sin(c + d*x)**4*tan(c + d*x)**4 - 6*sin(c + d*x)**4*tan(c + d* x)**2 + 216*sin(c + d*x)**4 + 500*sin(c + d*x)**3 - 6*sin(c + d*x)**2*tan( c + d*x)**4 + 12*sin(c + d*x)**2*tan(c + d*x)**2 - 144*sin(c + d*x)**2 - 3 00*sin(c + d*x) + 3*tan(c + d*x)**4 - 6*tan(c + d*x)**2))/(12*d*(sin(c + d *x)**4 - 2*sin(c + d*x)**2 + 1))