Integrand size = 27, antiderivative size = 101 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))} \]
-7/8*a^2*ln(1-sin(d*x+c))/d+a^2*ln(sin(d*x+c))/d-1/8*a^2*ln(1+sin(d*x+c))/ d+1/4*a^4/d/(a-a*sin(d*x+c))^2+3/4*a^3/d/(a-a*sin(d*x+c))
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \left (7 \log (1-\sin (c+d x))-8 \log (\sin (c+d x))+\log (1+\sin (c+d x))-\frac {2}{(-1+\sin (c+d x))^2}+\frac {6}{-1+\sin (c+d x)}\right )}{8 d} \]
-1/8*(a^2*(7*Log[1 - Sin[c + d*x]] - 8*Log[Sin[c + d*x]] + Log[1 + Sin[c + d*x]] - 2/(-1 + Sin[c + d*x])^2 + 6/(-1 + Sin[c + d*x])))/d
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97, 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, 93, 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.
\(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^2}{\sin (c+d x) \cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\csc (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^6 \int \frac {\csc (c+d x)}{a (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \frac {a^6 \int \left (\frac {\csc (c+d x)}{a^5}+\frac {7}{8 a^4 (a-a \sin (c+d x))}-\frac {1}{8 a^4 (\sin (c+d x) a+a)}+\frac {3}{4 a^3 (a-a \sin (c+d x))^2}+\frac {1}{2 a^2 (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^6 \left (\frac {\log (a \sin (c+d x))}{a^4}-\frac {7 \log (a-a \sin (c+d x))}{8 a^4}-\frac {\log (a \sin (c+d x)+a)}{8 a^4}+\frac {3}{4 a^3 (a-a \sin (c+d x))}+\frac {1}{4 a^2 (a-a \sin (c+d x))^2}\right )}{d}\) |
(a^6*(Log[a*Sin[c + d*x]]/a^4 - (7*Log[a - a*Sin[c + d*x]])/(8*a^4) - Log[ a + a*Sin[c + d*x]]/(8*a^4) + 1/(4*a^2*(a - a*Sin[c + d*x])^2) + 3/(4*a^3* (a - a*Sin[c + d*x]))))/d
3.9.66.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
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 = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(100\) |
default | \(\frac {\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(100\) |
risch | \(-\frac {i \left (-3 a^{2} {\mathrm e}^{i \left (d x +c \right )}-8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(127\) |
parallelrisch | \(-\frac {\left (\left (7 \cos \left (2 d x +2 c \right )+28 \sin \left (d x +c \right )-21\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-4 \cos \left (2 d x +2 c \right )-16 \sin \left (d x +c \right )+12\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \cos \left (2 d x +2 c \right )+10 \sin \left (d x +c \right )-4\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(149\) |
norman | \(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {4 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(298\) |
1/d*(1/4*a^2/cos(d*x+c)^4+2*a^2*(-(-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)))+a^2*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^ 2+ln(tan(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.64 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + 8 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
1/8*(6*a^2*sin(d*x + c) - 8*a^2 + 8*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(1/2*sin(d*x + c)) - (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(sin(d*x + c) + 1) - 7*(a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]
-1/8*(a^2*log(sin(d*x + c) + 1) + 7*a^2*log(sin(d*x + c) - 1) - 8*a^2*log( sin(d*x + c)) + 2*(3*a^2*sin(d*x + c) - 4*a^2)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 14 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {21 \, a^{2} \sin \left (d x + c\right )^{2} - 54 \, a^{2} \sin \left (d x + c\right ) + 37 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]
-1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) + 14*a^2*log(abs(sin(d*x + c) - 1) ) - 16*a^2*log(abs(sin(d*x + c))) - (21*a^2*sin(d*x + c)^2 - 54*a^2*sin(d* x + c) + 37*a^2)/(sin(d*x + c) - 1)^2)/d
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {\frac {3\,a^2\,\sin \left (c+d\,x\right )}{4}-a^2}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}-\frac {7\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d} \]