Integrand size = 31, antiderivative size = 178 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (5 A-3 B) \text {arctanh}(\sin (c+d x))}{32 d}+\frac {a^7 (A+B)}{16 d (a-a \sin (c+d x))^4}+\frac {a^9 A}{12 d \left (a^2-a^2 \sin (c+d x)\right )^3}+\frac {a^9 (3 A-B)}{32 d \left (a^3-a^3 \sin (c+d x)\right )^2}+\frac {a^9 (2 A-B)}{16 d \left (a^6-a^6 \sin (c+d x)\right )}-\frac {a^9 (A-B)}{32 d \left (a^6+a^6 \sin (c+d x)\right )} \] Output:
1/32*a^3*(5*A-3*B)*arctanh(sin(d*x+c))/d+1/16*a^7*(A+B)/d/(a-a*sin(d*x+c)) ^4+1/12*a^9*A/d/(a^2-a^2*sin(d*x+c))^3+1/32*a^9*(3*A-B)/d/(a^3-a^3*sin(d*x +c))^2+1/16*a^9*(2*A-B)/d/(a^6-a^6*sin(d*x+c))-1/32*a^9*(A-B)/d/(a^6+a^6*s in(d*x+c))
Time = 0.75 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^9 \left (\frac {(5 A-3 B) \text {arctanh}(\sin (c+d x))}{32 a^6}+\frac {A+B}{16 a^2 (a-a \sin (c+d x))^4}+\frac {A}{12 a^3 (a-a \sin (c+d x))^3}+\frac {3 A-B}{32 a^4 (a-a \sin (c+d x))^2}+\frac {2 A-B}{16 a^5 (a-a \sin (c+d x))}-\frac {A-B}{32 a^5 (a+a \sin (c+d x))}\right )}{d} \] Input:
Integrate[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
Output:
(a^9*(((5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*a^6) + (A + B)/(16*a^2*(a - a*Sin[c + d*x])^4) + A/(12*a^3*(a - a*Sin[c + d*x])^3) + (3*A - B)/(32*a^4 *(a - a*Sin[c + d*x])^2) + (2*A - B)/(16*a^5*(a - a*Sin[c + d*x])) - (A - B)/(32*a^5*(a + a*Sin[c + d*x]))))/d
Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.85, 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.
| \(\displaystyle \int \sec ^9(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)^9}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^9 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^8 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^8 \int \left (\frac {A}{4 a^2 (a-a \sin (c+d x))^4}+\frac {5 A-3 B}{32 a^4 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {2 A-B}{16 a^4 (a-a \sin (c+d x))^2}+\frac {A-B}{32 a^4 (\sin (c+d x) a+a)^2}+\frac {3 A-B}{16 a^3 (a-a \sin (c+d x))^3}+\frac {A+B}{4 a (a-a \sin (c+d x))^5}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^8 \left (\frac {(5 A-3 B) \text {arctanh}(\sin (c+d x))}{32 a^5}+\frac {2 A-B}{16 a^4 (a-a \sin (c+d x))}-\frac {A-B}{32 a^4 (a \sin (c+d x)+a)}+\frac {3 A-B}{32 a^3 (a-a \sin (c+d x))^2}+\frac {A}{12 a^2 (a-a \sin (c+d x))^3}+\frac {A+B}{16 a (a-a \sin (c+d x))^4}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
Output:
(a^8*(((5*A - 3*B)*ArcTanh[Sin[c + d*x]])/(32*a^5) + (A + B)/(16*a*(a - a* Sin[c + d*x])^4) + A/(12*a^2*(a - a*Sin[c + d*x])^3) + (3*A - B)/(32*a^3*( a - a*Sin[c + d*x])^2) + (2*A - B)/(16*a^4*(a - a*Sin[c + d*x])) - (A - B) /(32*a^4*(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_))*((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 = 1.53 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.64
| method | result | size |
| parallelrisch | \(-\frac {65 \left (\left (A -\frac {3 B}{5}\right ) \left (-\frac {8 \cos \left (2 d x +2 c \right )}{13}+\frac {6 \cos \left (4 d x +4 c \right )}{13}-\frac {\sin \left (5 d x +5 c \right )}{13}+\frac {14 \sin \left (d x +c \right )}{13}+\sin \left (3 d x +3 c \right )-\frac {14}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -\frac {3 B}{5}\right ) \left (-\frac {8 \cos \left (2 d x +2 c \right )}{13}+\frac {6 \cos \left (4 d x +4 c \right )}{13}-\frac {\sin \left (5 d x +5 c \right )}{13}+\frac {14 \sin \left (d x +c \right )}{13}+\sin \left (3 d x +3 c \right )-\frac {14}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {16 \left (\frac {3 A}{5}-B \right ) \cos \left (2 d x +2 c \right )}{13}+\frac {6 \left (9 A +B \right ) \cos \left (4 d x +4 c \right )}{65}+\frac {4 \left (9 B +\frac {59 A}{3}\right ) \sin \left (3 d x +3 c \right )}{65}-\frac {32 A \sin \left (5 d x +5 c \right )}{195}+\frac {4 \left (\frac {187 A}{15}-3 B \right ) \sin \left (d x +c \right )}{13}-\frac {102 A}{65}+\frac {74 B}{65}\right ) a^{3}}{32 d \left (-\sin \left (5 d x +5 c \right )+14 \sin \left (d x +c \right )+13 \sin \left (3 d x +3 c \right )+6 \cos \left (4 d x +4 c \right )-14-8 \cos \left (2 d x +2 c \right )\right )}\) | \(292\) |
| risch | \(-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )} \left (150 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+15 A \,{\mathrm e}^{8 i \left (d x +c \right )}+90 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-9 B \,{\mathrm e}^{8 i \left (d x +c \right )}+90 i A \,{\mathrm e}^{i \left (d x +c \right )}-200 A \,{\mathrm e}^{6 i \left (d x +c \right )}-150 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+120 B \,{\mathrm e}^{6 i \left (d x +c \right )}-90 i A \,{\mathrm e}^{7 i \left (d x +c \right )}-142 A \,{\mathrm e}^{4 i \left (d x +c \right )}-54 i B \,{\mathrm e}^{i \left (d x +c \right )}-222 B \,{\mathrm e}^{4 i \left (d x +c \right )}-90 i B \,{\mathrm e}^{5 i \left (d x +c \right )}-200 A \,{\mathrm e}^{2 i \left (d x +c \right )}+54 i B \,{\mathrm e}^{7 i \left (d x +c \right )}+120 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A -9 B \right )}{48 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{8} d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{32 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{32 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{32 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{32 d}\) | \(371\) |
| derivativedivides | \(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{5}}{16 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{5}}{64 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{128 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{128}-\frac {3 \sin \left (d x +c \right )}{128}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{8}}+\frac {5 \sin \left (d x +c \right )^{3}}{48 \cos \left (d x +c \right )^{6}}+\frac {5 \sin \left (d x +c \right )^{3}}{64 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{3}}{128 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )}{128}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+\frac {3 a^{3} A}{8 \cos \left (d x +c \right )^{8}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{8}}+\frac {5 \sin \left (d x +c \right )^{3}}{48 \cos \left (d x +c \right )^{6}}+\frac {5 \sin \left (d x +c \right )^{3}}{64 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{3}}{128 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )}{128}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{7}}{8}-\frac {7 \sec \left (d x +c \right )^{5}}{48}-\frac {35 \sec \left (d x +c \right )^{3}}{192}-\frac {35 \sec \left (d x +c \right )}{128}\right ) \tan \left (d x +c \right )+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+\frac {a^{3} B}{8 \cos \left (d x +c \right )^{8}}}{d}\) | \(542\) |
| default | \(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{5}}{16 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{5}}{64 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{128 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{128}-\frac {3 \sin \left (d x +c \right )}{128}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{8}}+\frac {5 \sin \left (d x +c \right )^{3}}{48 \cos \left (d x +c \right )^{6}}+\frac {5 \sin \left (d x +c \right )^{3}}{64 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{3}}{128 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )}{128}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+\frac {3 a^{3} A}{8 \cos \left (d x +c \right )^{8}}+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{8}}+\frac {5 \sin \left (d x +c \right )^{3}}{48 \cos \left (d x +c \right )^{6}}+\frac {5 \sin \left (d x +c \right )^{3}}{64 \cos \left (d x +c \right )^{4}}+\frac {5 \sin \left (d x +c \right )^{3}}{128 \cos \left (d x +c \right )^{2}}+\frac {5 \sin \left (d x +c \right )}{128}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{7}}{8}-\frac {7 \sec \left (d x +c \right )^{5}}{48}-\frac {35 \sec \left (d x +c \right )^{3}}{192}-\frac {35 \sec \left (d x +c \right )}{128}\right ) \tan \left (d x +c \right )+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{128}\right )+\frac {a^{3} B}{8 \cos \left (d x +c \right )^{8}}}{d}\) | \(542\) |
Input:
int(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-65/32*((A-3/5*B)*(-8/13*cos(2*d*x+2*c)+6/13*cos(4*d*x+4*c)-1/13*sin(5*d*x +5*c)+14/13*sin(d*x+c)+sin(3*d*x+3*c)-14/13)*ln(tan(1/2*d*x+1/2*c)-1)-(A-3 /5*B)*(-8/13*cos(2*d*x+2*c)+6/13*cos(4*d*x+4*c)-1/13*sin(5*d*x+5*c)+14/13* sin(d*x+c)+sin(3*d*x+3*c)-14/13)*ln(tan(1/2*d*x+1/2*c)+1)+16/13*(3/5*A-B)* cos(2*d*x+2*c)+6/65*(9*A+B)*cos(4*d*x+4*c)+4/65*(9*B+59/3*A)*sin(3*d*x+3*c )-32/195*A*sin(5*d*x+5*c)+4/13*(187/15*A-3*B)*sin(d*x+c)-102/65*A+74/65*B) *a^3/d/(-sin(5*d*x+5*c)+14*sin(d*x+c)+13*sin(3*d*x+3*c)+6*cos(4*d*x+4*c)-1 4-8*cos(2*d*x+2*c))
Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (170) = 340\).
Time = 0.11 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.98 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {6 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 26 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 12 \, {\left (3 \, A - 5 \, B\right )} a^{3} + 3 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - {\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - {\left ({\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A - 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{192 \, {\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/192*(6*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 26*(5*A - 3*B)*a^3*cos(d*x + c)^ 2 + 12*(3*A - 5*B)*a^3 + 3*(3*(5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3* B)*a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^ 3*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3*(3*(5*A - 3*B)*a ^3*cos(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2 - ((5*A - 3*B)*a^3*co s(d*x + c)^4 - 4*(5*A - 3*B)*a^3*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d* x + c) + 1) + 6*(3*(5*A - 3*B)*a^3*cos(d*x + c)^2 - 2*(5*A - 3*B)*a^3)*sin (d*x + c))/(3*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2)*sin(d*x + c))
Timed out. \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**9*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.04 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} - 9 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 7 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{2} + 3 \, {\left (5 \, A - 3 \, B\right )} a^{3} \sin \left (d x + c\right ) - 32 \, A a^{3}\right )}}{\sin \left (d x + c\right )^{5} - 3 \, \sin \left (d x + c\right )^{4} + 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) + 1}}{192 \, d} \] Input:
integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/192*(3*(5*A - 3*B)*a^3*log(sin(d*x + c) + 1) - 3*(5*A - 3*B)*a^3*log(sin (d*x + c) - 1) - 2*(3*(5*A - 3*B)*a^3*sin(d*x + c)^4 - 9*(5*A - 3*B)*a^3*s in(d*x + c)^3 + 7*(5*A - 3*B)*a^3*sin(d*x + c)^2 + 3*(5*A - 3*B)*a^3*sin(d *x + c) - 32*A*a^3)/(sin(d*x + c)^5 - 3*sin(d*x + c)^4 + 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 - 3*sin(d*x + c) + 1))/d
Time = 0.17 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.01 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {{\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{64 \, d} - \frac {{\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{64 \, d} - \frac {3 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \sin \left (d x + c\right )^{4} - 32 \, A a^{3} - 9 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \sin \left (d x + c\right )^{3} + 7 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \sin \left (d x + c\right )^{2} + 3 \, {\left (5 \, A a^{3} - 3 \, B a^{3}\right )} \sin \left (d x + c\right )}{96 \, d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} \] Input:
integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/64*(5*A*a^3 - 3*B*a^3)*log(abs(sin(d*x + c) + 1))/d - 1/64*(5*A*a^3 - 3* B*a^3)*log(abs(sin(d*x + c) - 1))/d - 1/96*(3*(5*A*a^3 - 3*B*a^3)*sin(d*x + c)^4 - 32*A*a^3 - 9*(5*A*a^3 - 3*B*a^3)*sin(d*x + c)^3 + 7*(5*A*a^3 - 3* B*a^3)*sin(d*x + c)^2 + 3*(5*A*a^3 - 3*B*a^3)*sin(d*x + c))/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^4)
Time = 34.08 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.97 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A-3\,B\right )}{32\,d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {5\,A\,a^3}{32}-\frac {3\,B\,a^3}{32}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {15\,A\,a^3}{32}-\frac {9\,B\,a^3}{32}\right )+{\sin \left (c+d\,x\right )}^2\,\left (\frac {35\,A\,a^3}{96}-\frac {7\,B\,a^3}{32}\right )-\frac {A\,a^3}{3}+\sin \left (c+d\,x\right )\,\left (\frac {5\,A\,a^3}{32}-\frac {3\,B\,a^3}{32}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^5-3\,{\sin \left (c+d\,x\right )}^4+2\,{\sin \left (c+d\,x\right )}^3+2\,{\sin \left (c+d\,x\right )}^2-3\,\sin \left (c+d\,x\right )+1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^9,x)
Output:
(a^3*atanh(sin(c + d*x))*(5*A - 3*B))/(32*d) - (sin(c + d*x)^4*((5*A*a^3)/ 32 - (3*B*a^3)/32) - sin(c + d*x)^3*((15*A*a^3)/32 - (9*B*a^3)/32) + sin(c + d*x)^2*((35*A*a^3)/96 - (7*B*a^3)/32) - (A*a^3)/3 + sin(c + d*x)*((5*A* a^3)/32 - (3*B*a^3)/32))/(d*(2*sin(c + d*x)^2 - 3*sin(c + d*x) + 2*sin(c + d*x)^3 - 3*sin(c + d*x)^4 + sin(c + d*x)^5 + 1))
Time = 0.17 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.61 \[ \int \sec ^9(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^9*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
Output:
(a**3*( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a + 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*b + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **4*a - 27*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 30*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**3*a + 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *3*b - 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 18*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**2*b + 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a - 27*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b - 15*log(tan((c + d*x)/2) - 1)*a + 9*log(tan((c + d*x)/2) - 1)*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*a - 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*b - 45*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**4*a + 27*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a - 18*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**3*b + 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 45*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 27*log(tan((c + d*x)/2) + 1)*sin(c + d*x) *b + 15*log(tan((c + d*x)/2) + 1)*a - 9*log(tan((c + d*x)/2) + 1)*b - 5*si n(c + d*x)**5*a + 3*sin(c + d*x)**5*b + 35*sin(c + d*x)**3*a - 21*sin(c + d*x)**3*b - 45*sin(c + d*x)**2*a + 27*sin(c + d*x)**2*b + 27*a + 3*b))/(96 *d*(sin(c + d*x)**5 - 3*sin(c + d*x)**4 + 2*sin(c + d*x)**3 + 2*sin(c + d* x)**2 - 3*sin(c + d*x) + 1))