Integrand size = 25, antiderivative size = 99 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \log (\tan (c+d x))}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^2(c+d x)}{d}+\frac {a \tan ^4(c+d x)}{4 d} \]
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 = 0.02 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.16 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
(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 |
\(\displaystyle \frac {a \left (\frac {1}{2} \tan ^4(c+d x)+2 \tan ^2(c+d x)+\log \left (\tan ^2(c+d x)\right )\right )}{2 d}+b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
(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)
3.15.88.3.1 Defintions of rubi rules used
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 = 0.80 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(82\) |
default | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(82\) |
parallelrisch | \(\frac {-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {3 b}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right ) a +11 b \sin \left (d x +c \right )+3 b \sin \left (3 d x +3 c \right )+7 a}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(196\) |
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 | \(\frac {\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(214\) |
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 = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.26 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {16 \, a \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \]
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. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, b \sin \left (d x + c\right )^{3} + 4 \, a \sin \left (d x + c\right )^{2} - 5 \, b \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
-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 = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {{\left (8 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (8 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (6 \, a \sin \left (d x + c\right )^{4} - 3 \, b \sin \left (d x + c\right )^{3} - 16 \, a \sin \left (d x + c\right )^{2} + 5 \, b \sin \left (d x + c\right ) + 12 \, a\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
-1/16*((8*a - 3*b)*log(abs(sin(d*x + c) + 1)) + (8*a + 3*b)*log(abs(sin(d* x + c) - 1)) - 16*a*log(abs(sin(d*x + c))) - 2*(6*a*sin(d*x + c)^4 - 3*b*s in(d*x + c)^3 - 16*a*sin(d*x + c)^2 + 5*b*sin(d*x + c) + 12*a)/(sin(d*x + c)^2 - 1)^2)/d
Time = 0.14 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {-\frac {3\,b\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {5\,b\,\sin \left (c+d\,x\right )}{8}+\frac {3\,a}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a}{2}-\frac {3\,b}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a}{2}+\frac {3\,b}{16}\right )}{d}+\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]