Integrand size = 29, antiderivative size = 177 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a (5 A-B) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^7 (A+B)}{24 d \left (a^2-a^2 \sin (c+d x)\right )^3}+\frac {a^7 (3 A+B)}{32 d \left (a^3-a^3 \sin (c+d x)\right )^2}-\frac {a^7 (A-B)}{32 d \left (a^3+a^3 \sin (c+d x)\right )^2}+\frac {3 a^7 A}{16 d \left (a^6-a^6 \sin (c+d x)\right )}-\frac {a^7 (2 A-B)}{16 d \left (a^6+a^6 \sin (c+d x)\right )} \] Output:
1/16*a*(5*A-B)*arctanh(sin(d*x+c))/d+1/24*a^7*(A+B)/d/(a^2-a^2*sin(d*x+c)) ^3+1/32*a^7*(3*A+B)/d/(a^3-a^3*sin(d*x+c))^2-1/32*a^7*(A-B)/d/(a^3+a^3*sin (d*x+c))^2+3/16*a^7*A/d/(a^6-a^6*sin(d*x+c))-1/16*a^7*(2*A-B)/d/(a^6+a^6*s in(d*x+c))
Time = 5.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.59 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (6 (5 A-B) \text {arctanh}(\sin (c+d x))-\frac {4 (A+B)}{(-1+\sin (c+d x))^3}+\frac {3 (3 A+B)}{(-1+\sin (c+d x))^2}-\frac {18 A}{-1+\sin (c+d x)}-\frac {3 (A-B)}{(1+\sin (c+d x))^2}+\frac {6 (-2 A+B)}{1+\sin (c+d x)}\right )}{96 d} \] Input:
Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a*(6*(5*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^3 + (3*(3*A + B))/(-1 + Sin[c + d*x])^2 - (18*A)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])^2 + (6*(-2*A + B))/(1 + Sin[c + d*x])))/(96*d)
Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \sin (c+d x)+a) (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a) (A+B \sin (c+d x))}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^7 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^6 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^6 \int \left (\frac {3 A}{16 a^4 (a-a \sin (c+d x))^2}+\frac {5 A-B}{16 a^4 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {2 A-B}{16 a^4 (\sin (c+d x) a+a)^2}+\frac {3 A+B}{16 a^3 (a-a \sin (c+d x))^3}+\frac {A-B}{16 a^3 (\sin (c+d x) a+a)^3}+\frac {A+B}{8 a^2 (a-a \sin (c+d x))^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^6 \left (\frac {(5 A-B) \text {arctanh}(\sin (c+d x))}{16 a^5}-\frac {2 A-B}{16 a^4 (a \sin (c+d x)+a)}+\frac {3 A}{16 a^4 (a-a \sin (c+d x))}+\frac {3 A+B}{32 a^3 (a-a \sin (c+d x))^2}-\frac {A-B}{32 a^3 (a \sin (c+d x)+a)^2}+\frac {A+B}{24 a^2 (a-a \sin (c+d x))^3}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]
Output:
(a^6*(((5*A - B)*ArcTanh[Sin[c + d*x]])/(16*a^5) + (A + B)/(24*a^2*(a - a* Sin[c + d*x])^3) + (3*A + B)/(32*a^3*(a - a*Sin[c + d*x])^2) + (3*A)/(16*a ^4*(a - a*Sin[c + d*x])) - (A - B)/(32*a^3*(a + a*Sin[c + d*x])^2) - (2*A - B)/(16*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 = 1.49 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {\frac {a A}{6 \cos \left (d x +c \right )^{6}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a B}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(169\) |
| default | \(\frac {\frac {a A}{6 \cos \left (d x +c \right )^{6}}+a B \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a A \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a B}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(169\) |
| parallelrisch | \(-\frac {15 \left (\left (-\frac {8 \cos \left (2 d x +2 c \right )}{3}-\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {\sin \left (5 d x +5 c \right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}+\sin \left (3 d x +3 c \right )-2\right ) \left (A -\frac {B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (-\frac {8 \cos \left (2 d x +2 c \right )}{3}-\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {\sin \left (5 d x +5 c \right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}+\sin \left (3 d x +3 c \right )-2\right ) \left (A -\frac {B}{5}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \left (A -\frac {B}{5}\right ) \cos \left (4 d x +4 c \right )}{9}+2 \left (A -\frac {B}{5}\right ) \sin \left (3 d x +3 c \right )+\frac {2 \left (A -\frac {B}{5}\right ) \sin \left (5 d x +5 c \right )}{9}+\frac {16 \left (A -\frac {B}{5}\right ) \sin \left (d x +c \right )}{3}-\frac {14 A}{15}+\frac {18 B}{5}\right ) a}{16 d \left (-6+\sin \left (5 d x +5 c \right )-8 \cos \left (2 d x +2 c \right )+2 \sin \left (d x +c \right )+3 \sin \left (3 d x +3 c \right )-2 \cos \left (4 d x +4 c \right )\right )}\) | \(270\) |
| risch | \(-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )} \left (30 i A \,{\mathrm e}^{i \left (d x +c \right )}+15 A \,{\mathrm e}^{8 i \left (d x +c \right )}+6 i B \,{\mathrm e}^{7 i \left (d x +c \right )}-3 B \,{\mathrm e}^{8 i \left (d x +c \right )}+22 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+40 A \,{\mathrm e}^{6 i \left (d x +c \right )}-6 i B \,{\mathrm e}^{i \left (d x +c \right )}-8 B \,{\mathrm e}^{6 i \left (d x +c \right )}-110 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-30 i A \,{\mathrm e}^{7 i \left (d x +c \right )}+150 B \,{\mathrm e}^{4 i \left (d x +c \right )}-22 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+110 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-8 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A -3 B \right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}-\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{16 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{16 d}+\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{16 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{16 d}\) | \(361\) |
| norman | \(\frac {\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}+\frac {4 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 \left (a A +a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {10 \left (4 a A +4 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}+\frac {2 \left (13 a A +13 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {2 \left (13 a A +13 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d}+\frac {a \left (11 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {a \left (11 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{8 d}+\frac {7 a \left (19 A +25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{24 d}+\frac {7 a \left (19 A +25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{24 d}+\frac {a \left (71 A +53 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 d}+\frac {a \left (71 A +53 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{24 d}+\frac {a \left (275 A +281 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{24 d}+\frac {a \left (275 A +281 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {a \left (5 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {a \left (5 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) | \(437\) |
Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d*(1/6*a*A/cos(d*x+c)^6+a*B*(1/6*sin(d*x+c)^3/cos(d*x+c)^6+1/8*sin(d*x+c )^3/cos(d*x+c)^4+1/16*sin(d*x+c)^3/cos(d*x+c)^2+1/16*sin(d*x+c)-1/16*ln(se c(d*x+c)+tan(d*x+c)))+a*A*(-(-1/6*sec(d*x+c)^5-5/24*sec(d*x+c)^3-5/16*sec( d*x+c))*tan(d*x+c)+5/16*ln(sec(d*x+c)+tan(d*x+c)))+1/6*a*B/cos(d*x+c)^6)
Time = 0.09 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {6 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} - 4 \, {\left (A - 5 \, B\right )} a - 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - B\right )} a\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{4}\right )}} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/96*(6*(5*A - B)*a*cos(d*x + c)^4 - 2*(5*A - B)*a*cos(d*x + c)^2 - 4*(A - 5*B)*a - 3*((5*A - B)*a*cos(d*x + c)^4*sin(d*x + c) - (5*A - B)*a*cos(d* x + c)^4)*log(sin(d*x + c) + 1) + 3*((5*A - B)*a*cos(d*x + c)^4*sin(d*x + c) - (5*A - B)*a*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(3*(5*A - B)*a *cos(d*x + c)^2 + 2*(5*A - B)*a)*sin(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c) - d*cos(d*x + c)^4)
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.97 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{4} - 3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{2} + 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right ) + 8 \, {\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 1}}{96 \, d} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="max ima")
Output:
1/96*(3*(5*A - B)*a*log(sin(d*x + c) + 1) - 3*(5*A - B)*a*log(sin(d*x + c) - 1) - 2*(3*(5*A - B)*a*sin(d*x + c)^4 - 3*(5*A - B)*a*sin(d*x + c)^3 - 5 *(5*A - B)*a*sin(d*x + c)^2 + 5*(5*A - B)*a*sin(d*x + c) + 8*(A + B)*a)/(s in(d*x + c)^5 - sin(d*x + c)^4 - 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 + sin (d*x + c) - 1))/d
Time = 0.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.89 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {{\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, d} - \frac {{\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, d} - \frac {3 \, {\left (5 \, A a - B a\right )} \sin \left (d x + c\right )^{4} - 3 \, {\left (5 \, A a - B a\right )} \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A a - B a\right )} \sin \left (d x + c\right )^{2} + 8 \, A a + 8 \, B a + 5 \, {\left (5 \, A a - B a\right )} \sin \left (d x + c\right )}{48 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="gia c")
Output:
1/32*(5*A*a - B*a)*log(abs(sin(d*x + c) + 1))/d - 1/32*(5*A*a - B*a)*log(a bs(sin(d*x + c) - 1))/d - 1/48*(3*(5*A*a - B*a)*sin(d*x + c)^4 - 3*(5*A*a - B*a)*sin(d*x + c)^3 - 5*(5*A*a - B*a)*sin(d*x + c)^2 + 8*A*a + 8*B*a + 5 *(5*A*a - B*a)*sin(d*x + c))/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^3)
Time = 33.81 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.88 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A-B\right )}{16\,d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {B\,a}{16}-\frac {5\,A\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,B\,a}{48}-\frac {25\,A\,a}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {25\,A\,a}{48}-\frac {5\,B\,a}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A\,a}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x)))/cos(c + d*x)^7,x)
Output:
(a*atanh(sin(c + d*x))*(5*A - B))/(16*d) - ((A*a)/6 + (B*a)/6 + sin(c + d* x)*((25*A*a)/48 - (5*B*a)/48) - sin(c + d*x)^3*((5*A*a)/16 - (B*a)/16) + s in(c + d*x)^4*((5*A*a)/16 - (B*a)/16) - sin(c + d*x)^2*((25*A*a)/48 - (5*B *a)/48))/(d*(sin(c + d*x) + 2*sin(c + d*x)^2 - 2*sin(c + d*x)^3 - sin(c + d*x)^4 + sin(c + d*x)^5 - 1))
Time = 0.18 (sec) , antiderivative size = 661, normalized size of antiderivative = 3.73 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)
Output:
(a*( - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a + 3*log(tan((c + d*x )/2) - 1)*sin(c + d*x)**5*b + 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 *a - 3*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 - 6*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 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a + 3*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)*b + 15*log(tan((c + d*x)/2) - 1)*a - 3*log(tan((c + d*x)/2) - 1)*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **5*a - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*b - 15*log(tan((c + d* x)/2) + 1)*sin(c + d*x)**4*a + 3*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 + 6*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 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 15*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b - 15*log (tan((c + d*x)/2) + 1)*a + 3*log(tan((c + d*x)/2) + 1)*b + 25*sin(c + d*x) **5*a - 5*sin(c + d*x)**5*b - 40*sin(c + d*x)**4*a + 8*sin(c + d*x)**4*b - 35*sin(c + d*x)**3*a + 7*sin(c + d*x)**3*b + 75*sin(c + d*x)**2*a - 15*si n(c + d*x)**2*b - 33*a - 3*b))/(48*d*(sin(c + d*x)**5 - sin(c + d*x)**4 - 2*sin(c + d*x)**3 + 2*sin(c + d*x)**2 + sin(c + d*x) - 1))