Integrand size = 27, antiderivative size = 141 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {a \log (\sin (c+d x))}{d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}+\frac {a^5}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac {3 a^5}{4 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {a^5}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-a*csc(d*x+c)/d-23/16*a*ln(1-sin(d*x+c))/d+a*ln(sin(d*x+c))/d+7/16*a*ln(1+ sin(d*x+c))/d+1/8*a^5/d/(a^2-a^2*sin(d*x+c))^2+3/4*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 = 86, normalized size of antiderivative = 0.61 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {a \log (\cos (c+d x))}{d}+\frac {a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{2 d}+\frac {a \sec ^4(c+d x)}{4 d} \] Input:
Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
-((a*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/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)
Time = 0.34 (sec) , antiderivative size = 128, 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.
\(\displaystyle \int \csc ^2(c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \sin (c+d x)+a}{\sin (c+d x)^2 \cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\csc ^2(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^7 \int \frac {\csc ^2(c+d x)}{a^2 (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^7 \int \left (\frac {\csc ^2(c+d x)}{a^7}+\frac {\csc (c+d x)}{a^7}+\frac {23}{16 a^6 (a-a \sin (c+d x))}+\frac {7}{16 a^6 (\sin (c+d x) a+a)}+\frac {3}{4 a^5 (a-a \sin (c+d x))^2}+\frac {1}{8 a^5 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^4 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^7 \left (-\frac {\csc (c+d x)}{a^6}+\frac {\log (a \sin (c+d x))}{a^6}-\frac {23 \log (a-a \sin (c+d x))}{16 a^6}+\frac {7 \log (a \sin (c+d x)+a)}{16 a^6}+\frac {3}{4 a^5 (a-a \sin (c+d x))}-\frac {1}{8 a^5 (a \sin (c+d x)+a)}+\frac {1}{8 a^4 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]
Output:
(a^7*(-(Csc[c + d*x]/a^6) + Log[a*Sin[c + d*x]]/a^6 - (23*Log[a - a*Sin[c + d*x]])/(16*a^6) + (7*Log[a + a*Sin[c + d*x]])/(16*a^6) + 1/(8*a^4*(a - a *Sin[c + d*x])^2) + 3/(4*a^5*(a - a*Sin[c + d*x])) - 1/(8*a^5*(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 = 0.86 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(101\) |
default | \(\frac {a \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(101\) |
risch | \(-\frac {i a \left (-22 i {\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{7 i \left (d x +c \right )}-20 i {\mathrm e}^{4 i \left (d x +c \right )}+11 \,{\mathrm e}^{5 i \left (d x +c \right )}-22 i {\mathrm e}^{2 i \left (d x +c \right )}-11 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {23 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {7 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(186\) |
norman | \(\frac {\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {4 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {a}{2 d}+\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d}+\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 d}+\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{4 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {23 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {7 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(238\) |
parallelrisch | \(-\frac {43 \left (\frac {1}{43}+\frac {23 \left (-\frac {\sin \left (3 d x +3 c \right )}{2}-\frac {\sin \left (d x +c \right )}{2}+\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{43}+\frac {7 \left (-1+\frac {\sin \left (3 d x +3 c \right )}{2}+\frac {\sin \left (d x +c \right )}{2}-\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{43}+\frac {4 \left (-2+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )-2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{43}+\left (\cos \left (d x +c \right )-\frac {14 \cos \left (2 d x +2 c \right )}{43}+\frac {7 \cos \left (3 d x +3 c \right )}{43}-\frac {36}{43}\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{43}-\frac {\cos \left (2 d x +2 c \right )}{43}\right ) a}{4 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\) | \(244\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*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)))+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))))
Time = 0.09 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.62 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {22 \, a \cos \left (d x + c\right )^{2} - 16 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 7 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 23 \, {\left (a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/16*(22*a*cos(d*x + c)^2 - 16*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d *x + c) - a*cos(d*x + c)^2)*log(1/2*sin(d*x + c)) - 7*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 23*(a*cos(d*x + c)^4 + a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*l og(-sin(d*x + c) + 1) - 2*(15*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*sin(d*x + c) - d*cos(d*x + c)^2)
Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+a*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.81 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{3} - 11 \, a \sin \left (d x + c\right )^{2} - 14 \, a \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{16 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/16*(7*a*log(sin(d*x + c) + 1) - 23*a*log(sin(d*x + c) - 1) + 16*a*log(si n(d*x + c)) - 2*(15*a*sin(d*x + c)^3 - 11*a*sin(d*x + c)^2 - 14*a*sin(d*x + c) + 8*a)/(sin(d*x + c)^4 - sin(d*x + c)^3 - sin(d*x + c)^2 + sin(d*x + c)))/d
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.78 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{16} \, a {\left (\frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {23 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {16 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{d} - \frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{3} - 11 \, \sin \left (d x + c\right )^{2} - 14 \, \sin \left (d x + c\right ) + 8\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2} \sin \left (d x + c\right )}\right )} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/16*a*(7*log(abs(sin(d*x + c) + 1))/d - 23*log(abs(sin(d*x + c) - 1))/d + 16*log(abs(sin(d*x + c)))/d - 2*(15*sin(d*x + c)^3 - 11*sin(d*x + c)^2 - 14*sin(d*x + c) + 8)/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2*sin(d*x + c)))
Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {11\,a\,{\sin \left (c+d\,x\right )}^2}{8}-\frac {7\,a\,\sin \left (c+d\,x\right )}{4}+a}{d\,\left ({\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^3-{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )}-\frac {23\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}+\frac {7\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d} \] Input:
int((a + a*sin(c + d*x))/(cos(c + d*x)^5*sin(c + d*x)^2),x)
Output:
(a*log(sin(c + d*x)))/d - (a - (7*a*sin(c + d*x))/4 - (11*a*sin(c + d*x)^2 )/8 + (15*a*sin(c + d*x)^3)/8)/(d*(sin(c + d*x) - sin(c + d*x)^2 - sin(c + d*x)^3 + sin(c + d*x)^4)) - (23*a*log(sin(c + d*x) - 1))/(16*d) + (7*a*lo g(sin(c + d*x) + 1))/(16*d)
Time = 0.22 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.37 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-23 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+23 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+23 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-23 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )+7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+3 \sin \left (d x +c \right )^{4}-18 \sin \left (d x +c \right )^{3}+8 \sin \left (d x +c \right )^{2}+17 \sin \left (d x +c \right )-8\right )}{8 \sin \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c)),x)
Output:
(a*( - 23*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 23*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**3 + 23*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 23*log(tan((c + d*x)/2) - 1)*sin(c + d*x) + 7*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**4 - 7*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 7*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 7*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 8*log(tan((c + d*x)/2))*sin(c + d*x)**4 - 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) + 3*sin(c + d*x)**4 - 18*sin(c + d*x)**3 + 8*sin(c + d*x )**2 + 17*sin(c + d*x) - 8))/(8*sin(c + d*x)*d*(sin(c + d*x)**3 - sin(c + d*x)**2 - sin(c + d*x) + 1))