Integrand size = 27, antiderivative size = 155 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}+\frac {a^5}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac {a^5}{d \left (a^4-a^4 \sin (c+d x)\right )}+\frac {a^5}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-39/16*a*ln(1-sin(d*x+c))/d+3*a*ln(sin (d*x+c))/d-9/16*a*ln(1+sin(d*x+c))/d+1/8*a^5/d/(a^2-a^2*sin(d*x+c))^2+a^5/ d/(a^4-a^4*sin(d*x+c))+1/8*a^5/d/(a^4+a^4*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {3 a \log (\cos (c+d x))}{d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d} \] Input:
Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
-1/2*(a*Csc[c + d*x]^2)/d - (a*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2 , Sin[c + d*x]^2])/d - (3*a*Log[Cos[c + d*x]])/d + (3*a*Log[Sin[c + d*x]]) /d + (a*Sec[c + d*x]^2)/d + (a*Sec[c + d*x]^4)/(4*d)
Time = 0.35 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 99, 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.
| \(\displaystyle \int \csc ^3(c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin (c+d x)+a}{\sin (c+d x)^3 \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\csc ^3(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^8 \int \frac {\csc ^3(c+d x)}{a^3 (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a^8 \int \left (\frac {\csc ^3(c+d x)}{a^8}+\frac {\csc ^2(c+d x)}{a^8}+\frac {3 \csc (c+d x)}{a^8}+\frac {39}{16 a^7 (a-a \sin (c+d x))}-\frac {9}{16 a^7 (\sin (c+d x) a+a)}+\frac {1}{a^6 (a-a \sin (c+d x))^2}-\frac {1}{8 a^6 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^5 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^8 \left (-\frac {\csc ^2(c+d x)}{2 a^7}-\frac {\csc (c+d x)}{a^7}+\frac {3 \log (a \sin (c+d x))}{a^7}-\frac {39 \log (a-a \sin (c+d x))}{16 a^7}-\frac {9 \log (a \sin (c+d x)+a)}{16 a^7}+\frac {1}{a^6 (a-a \sin (c+d x))}+\frac {1}{8 a^6 (a \sin (c+d x)+a)}+\frac {1}{8 a^5 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
(a^8*(-(Csc[c + d*x]/a^7) - Csc[c + d*x]^2/(2*a^7) + (3*Log[a*Sin[c + d*x] ])/a^7 - (39*Log[a - a*Sin[c + d*x]])/(16*a^7) - (9*Log[a + a*Sin[c + d*x] ])/(16*a^7) + 1/(8*a^5*(a - a*Sin[c + d*x])^2) + 1/(a^6*(a - a*Sin[c + d*x ])) + 1/(8*a^6*(a + a*Sin[c + d*x]))))/d
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(129\) |
| default | \(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(129\) |
| risch | \(-\frac {i \left (-6 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+15 a \,{\mathrm e}^{9 i \left (d x +c \right )}-14 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+28 a \,{\mathrm e}^{7 i \left (d x +c \right )}+14 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-22 a \,{\mathrm e}^{5 i \left (d x +c \right )}+6 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+28 a \,{\mathrm e}^{3 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {39 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(218\) |
| norman | \(\frac {-\frac {a}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}+\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8 d}-\frac {21 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}-\frac {21 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 d}+\frac {27 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}+\frac {27 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {39 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {9 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(305\) |
| parallelrisch | \(-\frac {3 \left (26 \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (-2+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )-2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8 \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-5\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+\frac {17 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{12}+\frac {17 \cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{12}+\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {32 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\cos \left (d x +c \right )-\frac {\cos \left (2 d x +2 c \right )}{4}+\frac {1}{2}\right )}{3}\right ) a}{16 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\) | \(305\) |
Input:
int(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4/sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin(d *x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)))+a*(1/4/sin(d*x+c)^2/cos(d*x+c)^4+3/4 /sin(d*x+c)^2/cos(d*x+c)^2-3/2/sin(d*x+c)^2+3*ln(tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.90 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {30 \, a \cos \left (d x + c\right )^{4} - 16 \, a \cos \left (d x + c\right )^{2} + 48 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 9 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 39 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/16*(30*a*cos(d*x + c)^4 - 16*a*cos(d*x + c)^2 + 48*(a*cos(d*x + c)^4 - a *cos(d*x + c)^2 - (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log( 1/2*sin(d*x + c)) - 9*(a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 39*(a*cos(d *x + c)^4 - a*cos(d*x + c)^2 - (a*cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d *x + c))*log(-sin(d*x + c) + 1) + 2*(3*a*cos(d*x + c)^2 + a)*sin(d*x + c) - 6*a)/(d*cos(d*x + c)^4 - d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - d*cos(d* x + c)^2)*sin(d*x + c))
Timed out. \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**3*sec(d*x+c)**5*(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.82 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {9 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 39 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{4} - 3 \, a \sin \left (d x + c\right )^{3} - 22 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 4 \, a\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
-1/16*(9*a*log(sin(d*x + c) + 1) + 39*a*log(sin(d*x + c) - 1) - 48*a*log(s in(d*x + c)) + 2*(15*a*sin(d*x + c)^4 - 3*a*sin(d*x + c)^3 - 22*a*sin(d*x + c)^2 + 4*a*sin(d*x + c) + 4*a)/(sin(d*x + c)^5 - sin(d*x + c)^4 - sin(d* x + c)^3 + sin(d*x + c)^2))/d
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.77 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {1}{16} \, a {\left (\frac {9 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} + \frac {39 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {48 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} + \frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} - 3 \, \sin \left (d x + c\right )^{3} - 22 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) + 4\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2} \sin \left (d x + c\right )^{2}}\right )} \] Input:
integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/16*a*(9*log(abs(sin(d*x + c) + 1))/d + 39*log(abs(sin(d*x + c) - 1))/d - 48*log(abs(sin(d*x + c)))/d + 2*(15*sin(d*x + c)^4 - 3*sin(d*x + c)^3 - 22*sin(d*x + c)^2 + 4*sin(d*x + c) + 4)/(d*(sin(d*x + c) + 1)*(sin(d*x + c ) - 1)^2*sin(d*x + c)^2))
Time = 32.37 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.86 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {39\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}-\frac {9\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^4}{8}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {11\,a\,{\sin \left (c+d\,x\right )}^2}{4}+\frac {a\,\sin \left (c+d\,x\right )}{2}+\frac {a}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2\right )} \] Input:
int((a + a*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^3),x)
Output:
(3*a*log(sin(c + d*x)))/d - (39*a*log(sin(c + d*x) - 1))/(16*d) - (9*a*log (sin(c + d*x) + 1))/(16*d) - (a/2 + (a*sin(c + d*x))/2 - (11*a*sin(c + d*x )^2)/4 - (3*a*sin(c + d*x)^3)/8 + (15*a*sin(c + d*x)^4)/8)/(d*(sin(c + d*x )^2 - sin(c + d*x)^3 - sin(c + d*x)^4 + sin(c + d*x)^5))
Time = 0.19 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.26 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{5}+39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}-39 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{5}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{5}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )^{5}-14 \sin \left (d x +c \right )^{4}+4 \sin \left (d x +c \right )^{3}+21 \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )-4\right )}{8 \sin \left (d x +c \right )^{2} d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c)),x)
Output:
(a*( - 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 + 39*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**4 + 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 - 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 9*log(tan((c + d*x)/2) + 1) *sin(c + d*x)**5 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 9*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**3 - 9*log(tan((c + d*x)/2) + 1)*sin(c + d *x)**2 + 24*log(tan((c + d*x)/2))*sin(c + d*x)**5 - 24*log(tan((c + d*x)/2 ))*sin(c + d*x)**4 - 24*log(tan((c + d*x)/2))*sin(c + d*x)**3 + 24*log(tan ((c + d*x)/2))*sin(c + d*x)**2 - sin(c + d*x)**5 - 14*sin(c + d*x)**4 + 4* sin(c + d*x)**3 + 21*sin(c + d*x)**2 - 4*sin(c + d*x) - 4))/(8*sin(c + d*x )**2*d*(sin(c + d*x)**3 - sin(c + d*x)**2 - sin(c + d*x) + 1))