Integrand size = 25, antiderivative size = 99 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \log (\tan (c+d x))}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^2(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \] Output:
3/8*b*arctanh(sin(d*x+c))/d+a*ln(tan(d*x+c))/d+3/8*b*sec(d*x+c)*tan(d*x+c) /d+1/4*b*sec(d*x+c)^3*tan(d*x+c)/d+a*tan(d*x+c)^2/d+1/4*a*tan(d*x+c)^4/d
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.16 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Input:
Integrate[Csc[c + d*x]*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]
Output:
(3*b*ArcTanh[Sin[c + d*x]])/(8*d) - (a*Log[Cos[c + d*x]])/d + (a*Log[Sin[c + d*x]])/d + (a*Sec[c + d*x]^2)/(2*d) + (a*Sec[c + d*x]^4)/(4*d) + (3*b*S ec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.56 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3313, 3042, 3100, 243, 49, 2009, 4255, 3042, 4255, 3042, 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.
| \(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \sin (c+d x)}{\sin (c+d x) \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \csc (c+d x) \sec ^5(c+d x)dx+b \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc (c+d x) \sec (c+d x)^5dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 3100 |
| \(\displaystyle \frac {a \int \cot (c+d x) \left (\tan ^2(c+d x)+1\right )^2d\tan (c+d x)}{d}+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {a \int \cot (c+d x) \left (\tan ^2(c+d x)+1\right )^2d\tan ^2(c+d x)}{2 d}+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {a \int \left (\tan ^2(c+d x)+\cot (c+d x)+2\right )d\tan ^2(c+d x)}{2 d}+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+\frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}+b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
Input:
Int[Csc[c + d*x]*Sec[c + d*x]^5*(a + b*Sin[c + d*x]),x]
Output:
(a*(Log[Tan[c + d*x]^2] + 2*Tan[c + d*x]^2 + Tan[c + d*x]^4/2))/(2*d) + b* ((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + ( Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(82\) |
| default | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(82\) |
| parallelrisch | \(\frac {-16 \left (a +\frac {3 b}{8}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (a -\frac {3 b}{8}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \cos \left (2 d x +2 c \right )-3 a \cos \left (4 d x +4 c \right )+11 b \sin \left (d x +c \right )+3 b \sin \left (3 d x +3 c \right )+7 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(196\) |
| risch | \(-\frac {i \left (8 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{7 i \left (d x +c \right )}+32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+11 b \,{\mathrm e}^{5 i \left (d x +c \right )}+8 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-11 b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(202\) |
| norman | \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(214\) |
Input:
int(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+b*(-(-1/4*sec(d* x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.26 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \, a \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/16*(16*a*cos(d*x + c)^4*log(1/2*sin(d*x + c)) - (8*a - 3*b)*cos(d*x + c) ^4*log(sin(d*x + c) + 1) - (8*a + 3*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*a*cos(d*x + c)^2 + 2*(3*b*cos(d*x + c)^2 + 2*b)*sin(d*x + c) + 4*a) /(d*cos(d*x + c)^4)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)**5*(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 5 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/16*((8*a - 3*b)*log(sin(d*x + c) + 1) + (8*a + 3*b)*log(sin(d*x + c) - 1) - 16*a*log(sin(d*x + c)) + 2*(3*b*sin(d*x + c)^3 + 4*a*sin(d*x + c)^2 - 5*b*sin(d*x + c) - 6*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.19 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (8 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} - \frac {3 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 5 \, b \sin \left (d x + c\right ) - 6 \, a}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-1/16*(8*a - 3*b)*log(abs(sin(d*x + c) + 1))/d - 1/16*(8*a + 3*b)*log(abs( sin(d*x + c) - 1))/d + a*log(abs(sin(d*x + c)))/d - 1/8*(3*b*sin(d*x + c)^ 3 + 4*a*sin(d*x + c)^2 - 5*b*sin(d*x + c) - 6*a)/(d*(sin(d*x + c) + 1)^2*( sin(d*x + c) - 1)^2)
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-\frac {3\,b\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {5\,b\,\sin \left (c+d\,x\right )}{8}+\frac {3\,a}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a}{2}-\frac {3\,b}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a}{2}+\frac {3\,b}{16}\right )}{d}+\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \] Input:
int((a + b*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)),x)
Output:
((3*a)/4 + (5*b*sin(c + d*x))/8 - (a*sin(c + d*x)^2)/2 - (3*b*sin(c + d*x) ^3)/8)/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) - (log(sin(c + d*x) + 1 )*(a/2 - (3*b)/16))/d - (log(sin(c + d*x) - 1)*(a/2 + (3*b)/16))/d + (a*lo g(sin(c + d*x)))/d
Time = 0.17 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.73 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b +16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a -16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -6 \sin \left (d x +c \right )^{4} a -3 \sin \left (d x +c \right )^{3} b +8 \sin \left (d x +c \right )^{2} a +5 \sin \left (d x +c \right ) b}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(csc(d*x+c)*sec(d*x+c)^5*(a+b*sin(d*x+c)),x)
Output:
( - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 16*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 8*log(tan((c + d*x)/2) - 1)*a - 3*log(tan((c + d*x)/2) - 1)*b - 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 6*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**2*b - 8*log(tan((c + d*x)/2) + 1)*a + 3*log(tan((c + d*x)/2) + 1)*b + 8*log(tan((c + d*x)/2))*sin(c + d*x)**4*a - 16*log(tan((c + d*x)/2))*sin(c + d*x)**2*a + 8*log(tan((c + d*x)/2))*a - 6*sin(c + d*x)**4*a - 3*sin(c + d*x)**3*b + 8*sin(c + d*x)**2*a + 5*sin(c + d*x)*b)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))