Integrand size = 25, antiderivative size = 129 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {11 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}-\frac {5 a \log (1+\sin (c+d x))}{16 d}+\frac {a^5}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac {a^5}{2 d \left (a^4-a^4 \sin (c+d x)\right )}+\frac {a^5}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-11/16*a*ln(1-sin(d*x+c))/d+a*ln(sin(d*x+c))/d-5/16*a*ln(1+sin(d*x+c))/d+1 /8*a^5/d/(a^2-a^2*sin(d*x+c))^2+1/2*a^5/d/(a^4-a^4*sin(d*x+c))+1/8*a^5/d/( a^4+a^4*sin(d*x+c))
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.89 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Input:
Integrate[Csc[c + d*x]*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
(3*a*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*a*S ec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
| \(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin (c+d x)+a}{\sin (c+d x) \cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\csc (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^6 \int \frac {\csc (c+d x)}{a (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a^6 \int \left (\frac {\csc (c+d x)}{a^6}+\frac {11}{16 a^5 (a-a \sin (c+d x))}-\frac {5}{16 a^5 (\sin (c+d x) a+a)}+\frac {1}{2 a^4 (a-a \sin (c+d x))^2}-\frac {1}{8 a^4 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^3 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^6 \left (\frac {\log (a \sin (c+d x))}{a^5}-\frac {11 \log (a-a \sin (c+d x))}{16 a^5}-\frac {5 \log (a \sin (c+d x)+a)}{16 a^5}+\frac {1}{2 a^4 (a-a \sin (c+d x))}+\frac {1}{8 a^4 (a \sin (c+d x)+a)}+\frac {1}{8 a^3 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Csc[c + d*x]*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
(a^6*(Log[a*Sin[c + d*x]]/a^5 - (11*Log[a - a*Sin[c + d*x]])/(16*a^5) - (5 *Log[a + a*Sin[c + d*x]])/(16*a^5) + 1/(8*a^3*(a - a*Sin[c + d*x])^2) + 1/ (2*a^4*(a - a*Sin[c + d*x])) + 1/(8*a^4*(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 = 0.94 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.64
| method | result | size |
| derivativedivides | \(\frac {a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
| default | \(\frac {a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
| risch | \(-\frac {i \left (2 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+18 a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 a \,{\mathrm e}^{i \left (d x +c \right )}+3 a \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {11 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(154\) |
| parallelrisch | \(-\frac {11 \left (-\frac {3}{11}+\left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {5 \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{11}+\frac {4 \left (-2+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )-2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{11}+\frac {3 \cos \left (2 d x +2 c \right )}{11}-\frac {\sin \left (d x +c \right )}{11}-\frac {3 \sin \left (3 d x +3 c \right )}{11}\right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\) | \(200\) |
| norman | \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {11 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {5 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(202\) |
Input:
int(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+t an(d*x+c)))+a*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.36 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {6 \, a \cos \left (d x + c\right )^{2} - 16 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 5 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a \sin \left (d x + c\right ) + 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/16*(6*a*cos(d*x + c)^2 - 16*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(1/2*sin(d*x + c)) + 5*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d *x + c)^2)*log(sin(d*x + c) + 1) + 11*(a*cos(d*x + c)^2*sin(d*x + c) - a*c os(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*a*sin(d*x + c) + 6*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)**5*(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {5 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/16*(5*a*log(sin(d*x + c) + 1) + 11*a*log(sin(d*x + c) - 1) - 16*a*log(s in(d*x + c)) + 2*(3*a*sin(d*x + c)^2 + a*sin(d*x + c) - 6*a)/(sin(d*x + c) ^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.70 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {1}{16} \, a {\left (\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} + \frac {11 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {16 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} + \frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 6\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/16*a*(5*log(abs(sin(d*x + c) + 1))/d + 11*log(abs(sin(d*x + c) - 1))/d - 16*log(abs(sin(d*x + c)))/d + 2*(3*sin(d*x + c)^2 + sin(d*x + c) - 6)/(d *(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2))
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.77 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {3\,a\,{\sin \left (c+d\,x\right )}^2}{8}+\frac {a\,\sin \left (c+d\,x\right )}{8}-\frac {3\,a}{4}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^3-\sin \left (c+d\,x\right )\right )}-\frac {11\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}-\frac {5\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d} \] Input:
int((a + a*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)),x)
Output:
(a*log(sin(c + d*x)))/d - ((a*sin(c + d*x))/8 - (3*a)/4 + (3*a*sin(c + d*x )^2)/8)/(d*(cos(c + d*x)^2 - sin(c + d*x) + sin(c + d*x)^3)) - (11*a*log(s in(c + d*x) - 1))/(16*d) - (5*a*log(sin(c + d*x) + 1))/(16*d)
Time = 0.17 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.20 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}+5\right )}{8 d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)*sec(d*x+c)^5*(a+a*sin(d*x+c)),x)
Output:
(a*( - 11*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 11*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2 + 11*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 11* log(tan((c + d*x)/2) - 1) - 5*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 5*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 5*log(tan((c + d*x)/2) + 1)* sin(c + d*x) - 5*log(tan((c + d*x)/2) + 1) + 8*log(tan((c + d*x)/2))*sin(c + d*x)**3 - 8*log(tan((c + d*x)/2))*sin(c + d*x)**2 - 8*log(tan((c + d*x) /2))*sin(c + d*x) + 8*log(tan((c + d*x)/2)) - sin(c + d*x)**3 - 2*sin(c + d*x)**2 + 5))/(8*d*(sin(c + d*x)**3 - sin(c + d*x)**2 - sin(c + d*x) + 1))