Integrand size = 27, antiderivative size = 77 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))} \]
-a^3*ln(1-sin(d*x+c))/d+a^3*ln(sin(d*x+c))/d+1/2*a^5/d/(a-a*sin(d*x+c))^2+ a^4/d/(a-a*sin(d*x+c))
Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-2 \log (1-\sin (c+d x))+2 \log (\sin (c+d x))+\frac {3-2 \sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{2 d} \]
(a^3*(-2*Log[1 - Sin[c + d*x]] + 2*Log[Sin[c + d*x]] + (3 - 2*Sin[c + d*x] )/(-1 + Sin[c + d*x])^2))/(2*d)
Time = 0.29 (sec) , antiderivative size = 75, 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, 54, 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.
\(\displaystyle \int \csc (c+d x) \sec ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\sin (c+d x) \cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\csc (c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^6 \int \frac {\csc (c+d x)}{a (a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^6 \int \left (\frac {\csc (c+d x)}{a^4}+\frac {1}{a^3 (a-a \sin (c+d x))}+\frac {1}{a^2 (a-a \sin (c+d x))^2}+\frac {1}{a (a-a \sin (c+d x))^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^6 \left (\frac {\log (a \sin (c+d x))}{a^3}-\frac {\log (a-a \sin (c+d x))}{a^3}+\frac {1}{a^2 (a-a \sin (c+d x))}+\frac {1}{2 a (a-a \sin (c+d x))^2}\right )}{d}\) |
(a^6*(Log[a*Sin[c + d*x]]/a^3 - Log[a - a*Sin[c + d*x]]/a^3 + 1/(2*a*(a - a*Sin[c + d*x])^2) + 1/(a^2*(a - a*Sin[c + d*x]))))/d
3.9.75.3.1 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[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )+\frac {1}{\sin \left (d x +c \right )-1}\right )}{d}\) | \(50\) |
default | \(-\frac {a^{3} \left (-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )+\frac {1}{\sin \left (d x +c \right )-1}\right )}{d}\) | \(50\) |
risch | \(-\frac {2 i \left (-a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(105\) |
parallelrisch | \(\frac {\left (\left (-2 \cos \left (2 d x +2 c \right )+6-8 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \cos \left (2 d x +2 c \right )}{2}-4 \sin \left (d x +c \right )+\frac {3}{2}\right ) a^{3}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(114\) |
norman | \(\frac {\frac {10 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {26 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {26 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {16 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(316\) |
Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.64 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
1/2*(2*a^3*sin(d*x + c) - 3*a^3 + 2*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*log(1/2*sin(d*x + c)) - 2*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {2 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \]
-1/2*(2*a^3*log(sin(d*x + c) - 1) - 2*a^3*log(sin(d*x + c)) + (2*a^3*sin(d *x + c) - 3*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.60 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 114 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{6 \, d} \]
-1/6*(12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*a^3*log(abs(tan(1/2*d* x + 1/2*c))) - (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 76*a^3*tan(1/2*d*x + 1/2*c )^3 + 114*a^3*tan(1/2*d*x + 1/2*c)^2 - 76*a^3*tan(1/2*d*x + 1/2*c) + 25*a^ 3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d
Time = 9.91 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {a^3\,\sin \left (c+d\,x\right )-\frac {3\,a^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )} \]