Integrand size = 31, antiderivative size = 140 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 (2 A-B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^7 (3 A-B)}{16 d \left (a^5-a^5 \sin (c+d x)\right )}-\frac {a^7 (A-B)}{16 d \left (a^5+a^5 \sin (c+d x)\right )} \] Output:
1/8*a^2*(2*A-B)*arctanh(sin(d*x+c))/d+1/12*a^5*(A+B)/d/(a-a*sin(d*x+c))^3+ 1/8*a^4*A/d/(a-a*sin(d*x+c))^2+1/16*a^7*(3*A-B)/d/(a^5-a^5*sin(d*x+c))-1/1 6*a^7*(A-B)/d/(a^5+a^5*sin(d*x+c))
Time = 0.74 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left (6 (2 A-B) \text {arctanh}(\sin (c+d x))-\frac {4 (A+B)}{(-1+\sin (c+d x))^3}+\frac {6 A}{(-1+\sin (c+d x))^2}+\frac {-9 A+3 B}{-1+\sin (c+d x)}-\frac {3 (A-B)}{1+\sin (c+d x)}\right )}{48 d} \] Input:
Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^2*(6*(2*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^ 3 + (6*A)/(-1 + Sin[c + d*x])^2 + (-9*A + 3*B)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])))/(48*d)
Time = 0.37 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \sin (c+d x)+a)^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^2 (A+B \sin (c+d x))}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^7 \int \frac {a A+a B \sin (c+d x)}{a (a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^6 \int \frac {a A+a B \sin (c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^6 \int \left (\frac {A}{4 a^2 (a-a \sin (c+d x))^3}+\frac {2 A-B}{8 a^3 \left (a^2-a^2 \sin ^2(c+d x)\right )}+\frac {3 A-B}{16 a^3 (a-a \sin (c+d x))^2}+\frac {A-B}{16 a^3 (\sin (c+d x) a+a)^2}+\frac {A+B}{4 a (a-a \sin (c+d x))^4}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^6 \left (\frac {(2 A-B) \text {arctanh}(\sin (c+d x))}{8 a^4}+\frac {3 A-B}{16 a^3 (a-a \sin (c+d x))}-\frac {A-B}{16 a^3 (a \sin (c+d x)+a)}+\frac {A}{8 a^2 (a-a \sin (c+d x))^2}+\frac {A+B}{12 a (a-a \sin (c+d x))^3}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(a^6*(((2*A - B)*ArcTanh[Sin[c + d*x]])/(8*a^4) + (A + B)/(12*a*(a - a*Sin [c + d*x])^3) + A/(8*a^2*(a - a*Sin[c + d*x])^2) + (3*A - B)/(16*a^3*(a - a*Sin[c + d*x])) - (A - B)/(16*a^3*(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 = 1.34 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.71
| method | result | size |
| parallelrisch | \(-\frac {\left (\left (A -\frac {B}{2}\right ) \left (-\frac {5}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}-\cos \left (2 d x +2 c \right )+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A -\frac {B}{2}\right ) \left (-\frac {5}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}-\cos \left (2 d x +2 c \right )+\sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {2 \left (A -2 B \right ) \cos \left (2 d x +2 c \right )}{3}+\frac {\left (A +\frac {B}{4}\right ) \cos \left (4 d x +4 c \right )}{3}+\frac {\left (\frac {7 B}{2}+5 A \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\left (7 A -\frac {3 B}{2}\right ) \sin \left (d x +c \right )}{2}-A +\frac {5 B}{4}\right ) a^{2}}{d \left (\cos \left (4 d x +4 c \right )-5+4 \sin \left (3 d x +3 c \right )+4 \sin \left (d x +c \right )-4 \cos \left (2 d x +2 c \right )\right )}\) | \(239\) |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{2}}{3 \cos \left (d x +c \right )^{6}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{2} B}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(304\) |
| default | \(\frac {A \,a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+a^{2} B \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )+\frac {A \,a^{2}}{3 \cos \left (d x +c \right )^{6}}+2 a^{2} B \left (\frac {\sin \left (d x +c \right )^{3}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {a^{2} B}{6 \cos \left (d x +c \right )^{6}}}{d}\) | \(304\) |
| risch | \(-\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )} \left (-24 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+6 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 i B \,{\mathrm e}^{5 i \left (d x +c \right )}-3 B \,{\mathrm e}^{6 i \left (d x +c \right )}-16 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-26 A \,{\mathrm e}^{4 i \left (d x +c \right )}-40 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+13 B \,{\mathrm e}^{4 i \left (d x +c \right )}-24 i A \,{\mathrm e}^{i \left (d x +c \right )}+26 A \,{\mathrm e}^{2 i \left (d x +c \right )}+12 i B \,{\mathrm e}^{i \left (d x +c \right )}-13 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 A +3 B \right )}{12 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}\) | \(313\) |
Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
-((A-1/2*B)*(-5/4+1/4*cos(4*d*x+4*c)-cos(2*d*x+2*c)+sin(d*x+c)+sin(3*d*x+3 *c))*ln(tan(1/2*d*x+1/2*c)-1)-(A-1/2*B)*(-5/4+1/4*cos(4*d*x+4*c)-cos(2*d*x +2*c)+sin(d*x+c)+sin(3*d*x+3*c))*ln(tan(1/2*d*x+1/2*c)+1)+2/3*(A-2*B)*cos( 2*d*x+2*c)+1/3*(A+1/4*B)*cos(4*d*x+4*c)+1/6*(7/2*B+5*A)*sin(3*d*x+3*c)+1/2 *(7*A-3/2*B)*sin(d*x+c)-A+5/4*B)*a^2/d/(cos(4*d*x+4*c)-5+4*sin(3*d*x+3*c)+ 4*sin(d*x+c)-4*cos(2*d*x+2*c))
Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (133) = 266\).
Time = 0.09 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.94 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {12 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 8 \, {\left (A - 2 \, B\right )} a^{2} - 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, A - B\right )} a^{2}\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
-1/48*(12*(2*A - B)*a^2*cos(d*x + c)^2 - 8*(A - 2*B)*a^2 - 3*((2*A - B)*a^ 2*cos(d*x + c)^4 + 2*(2*A - B)*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*((2*A - B)*a^2*cos(d*x + c)^4 + 2*(2*A - B)*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d *x + c)^2)*log(-sin(d*x + c) + 1) - 2*(3*(2*A - B)*a^2*cos(d*x + c)^2 - 4* (2*A - B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2*sin(d* x + c) - 2*d*cos(d*x + c)^2)
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**7*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right ) + 2 \, {\left (4 \, A + B\right )} a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/48*(3*(2*A - B)*a^2*log(sin(d*x + c) + 1) - 3*(2*A - B)*a^2*log(sin(d*x + c) - 1) - 2*(3*(2*A - B)*a^2*sin(d*x + c)^3 - 6*(2*A - B)*a^2*sin(d*x + c)^2 + (2*A - B)*a^2*sin(d*x + c) + 2*(4*A + B)*a^2)/(sin(d*x + c)^4 - 2*s in(d*x + c)^3 + 2*sin(d*x + c) - 1))/d
Time = 0.17 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.15 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {3 \, {\left (2 \, A a^{2} - B a^{2}\right )} \sin \left (d x + c\right )^{3} + 8 \, A a^{2} + 2 \, B a^{2} - 6 \, {\left (2 \, A a^{2} - B a^{2}\right )} \sin \left (d x + c\right )^{2} + {\left (2 \, A a^{2} - B a^{2}\right )} \sin \left (d x + c\right )}{24 \, d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/16*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1))/d - 1/16*(2*A*a^2 - B*a^ 2)*log(abs(sin(d*x + c) - 1))/d - 1/24*(3*(2*A*a^2 - B*a^2)*sin(d*x + c)^3 + 8*A*a^2 + 2*B*a^2 - 6*(2*A*a^2 - B*a^2)*sin(d*x + c)^2 + (2*A*a^2 - B*a ^2)*sin(d*x + c))/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^3)
Time = 34.00 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.97 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (2\,A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^2}{2}-\frac {B\,a^2}{4}\right )+\frac {A\,a^2}{3}+\frac {B\,a^2}{12}+\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{12}-\frac {B\,a^2}{24}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,\sin \left (c+d\,x\right )-1\right )} \] Input:
int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^7,x)
Output:
(a^2*atanh(sin(c + d*x))*(2*A - B))/(8*d) - (sin(c + d*x)^3*((A*a^2)/4 - ( B*a^2)/8) - sin(c + d*x)^2*((A*a^2)/2 - (B*a^2)/4) + (A*a^2)/3 + (B*a^2)/1 2 + sin(c + d*x)*((A*a^2)/12 - (B*a^2)/24))/(d*(2*sin(c + d*x) - 2*sin(c + d*x)^3 + sin(c + d*x)^4 - 1))
Time = 0.17 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.14 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} b -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} b +24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +2 \sin \left (d x +c \right )^{4} a -\sin \left (d x +c \right )^{4} b -16 \sin \left (d x +c \right )^{3} a +8 \sin \left (d x +c \right )^{3} b +24 \sin \left (d x +c \right )^{2} a -12 \sin \left (d x +c \right )^{2} b -18 a -3 b \right )}{48 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
(a**2*( - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **3*a - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b - 24*log(tan((c + d *x)/2) - 1)*sin(c + d*x)*a + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b + 12*log(tan((c + d*x)/2) - 1)*a - 6*log(tan((c + d*x)/2) - 1)*b + 12*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**4*a - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a + 12*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**3*b + 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*b - 12*log(tan((c + d *x)/2) + 1)*a + 6*log(tan((c + d*x)/2) + 1)*b + 2*sin(c + d*x)**4*a - sin( c + d*x)**4*b - 16*sin(c + d*x)**3*a + 8*sin(c + d*x)**3*b + 24*sin(c + d* x)**2*a - 12*sin(c + d*x)**2*b - 18*a - 3*b))/(48*d*(sin(c + d*x)**4 - 2*s in(c + d*x)**3 + 2*sin(c + d*x) - 1))