Integrand size = 31, antiderivative size = 105 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))} \]
1/8*a^3*(A-B)*arctanh(sin(d*x+c))/d+1/6*a^6*(A+B)/d/(a-a*sin(d*x+c))^3+1/8 *a^5*(A-B)/d/(a-a*sin(d*x+c))^2+1/8*a^4*(A-B)/d/(a-a*sin(d*x+c))
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 \left (2 (-5 A+B)-3 (A-B) \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6+9 (A-B) \sin (c+d x)-3 (A-B) \sin ^2(c+d x)\right )}{24 d (-1+\sin (c+d x))^3} \]
(a^3*(2*(-5*A + B) - 3*(A - B)*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - S in[(c + d*x)/2])^6 + 9*(A - B)*Sin[c + d*x] - 3*(A - B)*Sin[c + d*x]^2))/( 24*d*(-1 + Sin[c + d*x])^3)
Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, 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, 86, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^7 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^4 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^6 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^6 \int \left (\frac {A-B}{8 a^2 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {A-B}{8 a^2 (a-a \sin (c+d x))^2}+\frac {A-B}{4 a (a-a \sin (c+d x))^3}+\frac {A+B}{2 (a-a \sin (c+d x))^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^6 \left (\frac {(A-B) \text {arctanh}(\sin (c+d x))}{8 a^3}+\frac {A-B}{8 a^2 (a-a \sin (c+d x))}+\frac {A-B}{8 a (a-a \sin (c+d x))^2}+\frac {A+B}{6 (a-a \sin (c+d x))^3}\right )}{d}\) |
(a^6*(((A - B)*ArcTanh[Sin[c + d*x]])/(8*a^3) + (A + B)/(6*(a - a*Sin[c + d*x])^3) + (A - B)/(8*a*(a - a*Sin[c + d*x])^2) + (A - B)/(8*a^2*(a - a*Si n[c + d*x]))))/d
3.10.93.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.56 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.83
| method | result | size |
| parallelrisch | \(-\frac {3 \left (\left (A -B \right ) \left (\cos \left (2 d x +2 c \right )+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -B \right ) \left (\cos \left (2 d x +2 c \right )+\frac {5 \sin \left (d x +c \right )}{2}-\frac {\sin \left (3 d x +3 c \right )}{6}-\frac {5}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \cos \left (2 d x +2 c \right ) \left (A -B \right )}{3}+\frac {\left (-A +B \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (d x +c \right ) \left (A -B \right )}{2}+\frac {8 A}{9}+\frac {8 B}{9}\right ) a^{3}}{4 d \left (-10+15 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )+6 \cos \left (2 d x +2 c \right )\right )}\) | \(192\) |
| risch | \(-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )} \left (-18 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+18 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-3 B \,{\mathrm e}^{4 i \left (d x +c \right )}+18 i A \,{\mathrm e}^{i \left (d x +c \right )}-46 A \,{\mathrm e}^{2 i \left (d x +c \right )}-18 i B \,{\mathrm e}^{i \left (d x +c \right )}+14 B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A -3 B \right )}{12 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}\) | \(241\) |
| derivativedivides | \(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{5}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{48 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{3}}{2 \cos \left (d x +c \right )^{6}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{3}}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(442\) |
| default | \(\frac {A \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{5}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{48 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{48}-\frac {\sin \left (d x +c \right )}{16}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 A \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{3}}{2 \cos \left (d x +c \right )^{6}}+3 B \,a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{3}}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(442\) |
-3/4*((A-B)*(cos(2*d*x+2*c)+5/2*sin(d*x+c)-1/6*sin(3*d*x+3*c)-5/3)*ln(tan( 1/2*d*x+1/2*c)-1)-(A-B)*(cos(2*d*x+2*c)+5/2*sin(d*x+c)-1/6*sin(3*d*x+3*c)- 5/3)*ln(tan(1/2*d*x+1/2*c)+1)+2/3*cos(2*d*x+2*c)*(A-B)+1/6*(-A+B)*sin(3*d* x+3*c)+1/2*sin(d*x+c)*(A-B)+8/9*A+8/9*B)*a^3/d/(-10+15*sin(d*x+c)-sin(3*d* x+3*c)+6*cos(2*d*x+2*c))
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (100) = 200\).
Time = 0.29 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.30 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} + 18 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) - 2 \, {\left (13 \, A - 5 \, B\right )} a^{3} + 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]
1/48*(6*(A - B)*a^3*cos(d*x + c)^2 + 18*(A - B)*a^3*sin(d*x + c) - 2*(13*A - 5*B)*a^3 + 3*(3*(A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a ^3*cos(d*x + c)^2 - 4*(A - B)*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3 *(3*(A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a^3*cos(d*x + c) ^2 - 4*(A - B)*a^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/(3*d*cos(d*x + c )^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.17 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{2} - 9 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) + 2 \, {\left (5 \, A - B\right )} a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]
1/48*(3*(A - B)*a^3*log(sin(d*x + c) + 1) - 3*(A - B)*a^3*log(sin(d*x + c) - 1) - 2*(3*(A - B)*a^3*sin(d*x + c)^2 - 9*(A - B)*a^3*sin(d*x + c) + 2*( 5*A - B)*a^3)/(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c) - 1))/d
Time = 0.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {11 \, A a^{3} \sin \left (d x + c\right )^{3} - 11 \, B a^{3} \sin \left (d x + c\right )^{3} - 45 \, A a^{3} \sin \left (d x + c\right )^{2} + 45 \, B a^{3} \sin \left (d x + c\right )^{2} + 69 \, A a^{3} \sin \left (d x + c\right ) - 69 \, B a^{3} \sin \left (d x + c\right ) - 51 \, A a^{3} + 19 \, B a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]
1/96*(6*(A*a^3 - B*a^3)*log(abs(sin(d*x + c) + 1)) - 6*(A*a^3 - B*a^3)*log (abs(sin(d*x + c) - 1)) + (11*A*a^3*sin(d*x + c)^3 - 11*B*a^3*sin(d*x + c) ^3 - 45*A*a^3*sin(d*x + c)^2 + 45*B*a^3*sin(d*x + c)^2 + 69*A*a^3*sin(d*x + c) - 69*B*a^3*sin(d*x + c) - 51*A*a^3 + 19*B*a^3)/(sin(d*x + c) - 1)^3)/ d
Time = 9.61 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^3}{8}-\frac {B\,a^3}{8}\right )+\frac {5\,A\,a^3}{12}-\frac {B\,a^3}{12}-\sin \left (c+d\,x\right )\,\left (\frac {3\,A\,a^3}{8}-\frac {3\,B\,a^3}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^3-3\,{\sin \left (c+d\,x\right )}^2+3\,\sin \left (c+d\,x\right )-1\right )} \]