Integrand size = 31, antiderivative size = 43 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 B \log (1-\sin (c+d x))}{d}+\frac {a^3 (A+B)}{d (a-a \sin (c+d x))} \]
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^3 \left (\frac {B \log (1-\sin (c+d x))}{a}+\frac {A+B}{a-a \sin (c+d x)}\right )}{d} \]
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 49, 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.
\(\displaystyle \int \sec ^3(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^3 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^2 \int \left (\frac {a (A+B)}{(a-a \sin (c+d x))^2}-\frac {B}{a-a \sin (c+d x)}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\frac {a (A+B)}{a-a \sin (c+d x)}+B \log (a-a \sin (c+d x))\right )}{d}\) |
3.10.74.3.1 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 [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[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 = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(-\frac {\left (B \left (\sin \left (d x +c \right )-1\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \left (\sin \left (d x +c \right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (A +B \right ) \sin \left (d x +c \right )\right ) a^{2}}{d \left (\sin \left (d x +c \right )-1\right )}\) | \(75\) |
risch | \(-i x \,a^{2} B -\frac {2 i a^{2} B c}{d}-\frac {2 i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (A +B \right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(78\) |
derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {A \,a^{2}}{\cos \left (d x +c \right )^{2}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(189\) |
default | \(\frac {A \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+\frac {A \,a^{2}}{\cos \left (d x +c \right )^{2}}+2 B \,a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}}{d}\) | \(189\) |
norman | \(\frac {-\frac {4 A \,a^{2}+4 B \,a^{2}}{d}-\frac {\left (4 A \,a^{2}+4 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A \,a^{2}+10 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A \,a^{2}+10 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a^{2} \left (A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{2} \left (A +B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (A +B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {B \,a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(291\) |
-(B*(sin(d*x+c)-1)*ln(sec(1/2*d*x+1/2*c)^2)-2*B*(sin(d*x+c)-1)*ln(tan(1/2* d*x+1/2*c)-1)+(A+B)*sin(d*x+c))*a^2/d/(sin(d*x+c)-1)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {{\left (A + B\right )} a^{2} - {\left (B a^{2} \sin \left (d x + c\right ) - B a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{d \sin \left (d x + c\right ) - d} \]
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*sin(c + d*x)*sec(c + d *x)**3, x) + Integral(A*sin(c + d*x)**2*sec(c + d*x)**3, x) + Integral(B*s in(c + d*x)*sec(c + d*x)**3, x) + Integral(2*B*sin(c + d*x)**2*sec(c + d*x )**3, x) + Integral(B*sin(c + d*x)**3*sec(c + d*x)**3, x))
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {{\left (A + B\right )} a^{2}}{\sin \left (d x + c\right ) - 1}}{d} \]
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (45) = 90\).
Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.60 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {B a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{2}}}{d} \]
-(B*a^2*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 2*B*a^2*log(abs(tan(1/2*d*x + 1/ 2*c) - 1)) + (3*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 2*A*a^2*tan(1/2*d*x + 1/2*c ) - 8*B*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^2)/ d
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{d}-\frac {A\,a^2+B\,a^2}{d\,\left (\sin \left (c+d\,x\right )-1\right )} \]