Integrand size = 31, antiderivative size = 160 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {\sec ^6(c+d x) (B+A \sin (c+d x)) (a+b \sin (c+d x))^2}{6 d}+\frac {\sec ^4(c+d x) \left (2 b (4 a A-b B)+\left (5 a^2 A+3 A b^2-2 a b B\right ) \sin (c+d x)\right )}{24 d}+\frac {\left (5 a^2 A-A b^2-2 a b B\right ) \sec (c+d x) \tan (c+d x)}{16 d} \]
1/16*(5*A*a^2-A*b^2-2*B*a*b)*arctanh(sin(d*x+c))/d+1/6*sec(d*x+c)^6*(B+A*s in(d*x+c))*(a+b*sin(d*x+c))^2/d+1/24*sec(d*x+c)^4*(2*b*(4*A*a-B*b)+(5*A*a^ 2+3*A*b^2-2*B*a*b)*sin(d*x+c))/d+1/16*(5*A*a^2-A*b^2-2*B*a*b)*sec(d*x+c)*t an(d*x+c)/d
Time = 0.97 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.51 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {b \sec ^6(c+d x) (a+b \sin (c+d x))^3 (A b-a B+(-a A+b B) \sin (c+d x))+\frac {1}{4} b \sec ^4(c+d x) (a+b \sin (c+d x))^3 (3 A b+(-5 a A+2 b B) \sin (c+d x))-\frac {3 b \left (-5 a^2 A+A b^2+2 a b B\right ) \left (\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)\right )}{16 (a-b) (a+b)}}{6 b \left (-a^2+b^2\right ) d} \]
(b*Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (b*Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*(3*A*b + (-5*a*A + 2*b *B)*Sin[c + d*x]))/4 - (3*b*(-5*a^2*A + A*b^2 + 2*a*b*B)*((a^2 - b^2)^2*(L og[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + 2*a^3*b*Sec[c + d*x]^2 - 2 *(a^4 - b^4)*Sec[c + d*x]*Tan[c + d*x] + (-6*a^3*b + 4*a*b^3)*Tan[c + d*x] ^2))/(16*(a - b)*(a + b)))/(6*b*(-a^2 + b^2)*d)
Time = 0.64 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.66, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 663, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2 (A+B \sin (c+d x))}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^7 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^6 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^4}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^6 \int \left (\frac {(A-B) (a-b)^2}{16 b^3 (\sin (c+d x) b+b)^4}+\frac {(2 a A-a B-b B) (a-b)}{16 b^4 (\sin (c+d x) b+b)^3}+\frac {5 A a^2-2 b B a-A b^2}{16 b^5 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {(5 A+B) a^2+2 b (A-B) a-b^2 (A+B)}{32 b^5 (b-b \sin (c+d x))^2}-\frac {-\left ((5 A-B) a^2\right )+2 b (A+B) a+b^2 (A-B)}{32 b^5 (\sin (c+d x) b+b)^2}-\frac {(a+b) (b B-a (2 A+B))}{16 b^4 (b-b \sin (c+d x))^3}+\frac {(a+b)^2 (A+B)}{16 b^3 (b-b \sin (c+d x))^4}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^6 \left (\frac {\left (5 a^2 A-2 a b B-A b^2\right ) \text {arctanh}(\sin (c+d x))}{16 b^6}+\frac {a^2 (5 A+B)+2 a b (A-B)-b^2 (A+B)}{32 b^5 (b-b \sin (c+d x))}+\frac {-\left (a^2 (5 A-B)\right )+2 a b (A+B)+b^2 (A-B)}{32 b^5 (b \sin (c+d x)+b)}-\frac {(a-b) (2 a A-a B-b B)}{32 b^4 (b \sin (c+d x)+b)^2}-\frac {(a+b) (b B-a (2 A+B))}{32 b^4 (b-b \sin (c+d x))^2}-\frac {(a-b)^2 (A-B)}{48 b^3 (b \sin (c+d x)+b)^3}+\frac {(a+b)^2 (A+B)}{48 b^3 (b-b \sin (c+d x))^3}\right )}{d}\) |
(b^6*(((5*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/(16*b^6) + ((a + b)^2*(A + B))/(48*b^3*(b - b*Sin[c + d*x])^3) - ((a + b)*(b*B - a*(2*A + B)))/(32*b^4*(b - b*Sin[c + d*x])^2) + (2*a*b*(A - B) - b^2*(A + B) + a^2* (5*A + B))/(32*b^5*(b - b*Sin[c + d*x])) - ((a - b)^2*(A - B))/(48*b^3*(b + b*Sin[c + d*x])^3) - ((a - b)*(2*a*A - a*B - b*B))/(32*b^4*(b + b*Sin[c + d*x])^2) + (b^2*(A - B) - a^2*(5*A - B) + 2*a*b*(A + B))/(32*b^5*(b + b* Sin[c + d*x]))))/d
3.16.43.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[1/c^p Int[ExpandI ntegrand[(d + e*x)^m*(f + g*x)^n*(-q + c*x)^p*(q + c*x)^p, x], x], x] /; ! FractionalPowerFactorQ[q]] /; FreeQ[{a, c, d, e, f, g}, x] && ILtQ[p, -1] & & IntegersQ[m, n] && NiceSqrtQ[(-a)*c]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.19 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.89
| method | result | size |
| derivativedivides | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}+\frac {A a b}{3 \cos \left (d x +c \right )^{6}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\) | \(302\) |
| default | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (d x +c \right )\right )}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+\frac {B \,a^{2}}{6 \cos \left (d x +c \right )^{6}}+\frac {A a b}{3 \cos \left (d x +c \right )^{6}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{16 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{16}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )+B \,b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )}{d}\) | \(302\) |
| parallelrisch | \(\frac {-15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (A \,a^{2}-\frac {1}{5} A \,b^{2}-\frac {2}{5} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right ) \left (A \,a^{2}-\frac {1}{5} A \,b^{2}-\frac {2}{5} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-240 A a b -120 B \,a^{2}-132 B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-96 A a b -48 B \,a^{2}+24 B \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (-16 A a b -8 B \,a^{2}+4 B \,b^{2}\right ) \cos \left (6 d x +6 c \right )+\left (170 A \,a^{2}-34 A \,b^{2}-68 B a b \right ) \sin \left (3 d x +3 c \right )+\left (30 A \,a^{2}-6 A \,b^{2}-12 B a b \right ) \sin \left (5 d x +5 c \right )+\left (396 A \,a^{2}+228 A \,b^{2}+456 B a b \right ) \sin \left (d x +c \right )+352 A a b +176 B \,a^{2}+104 B \,b^{2}}{48 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) | \(352\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (15 A \,a^{2} {\mathrm e}^{10 i \left (d x +c \right )}-3 A \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-6 B a b \,{\mathrm e}^{10 i \left (d x +c \right )}+85 A \,a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-17 A \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-34 B a b \,{\mathrm e}^{8 i \left (d x +c \right )}+198 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+114 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+228 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}+256 i B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-198 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-114 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-96 i B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-228 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+512 i A a b \,{\mathrm e}^{5 i \left (d x +c \right )}-96 i B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-85 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+17 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+34 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+64 i B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-15 A \,a^{2}+3 A \,b^{2}+6 B a b \right )}{24 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{8 d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{8 d}\) | \(519\) |
1/d*(A*a^2*(-(-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*B*a^2/cos(d*x+c)^6+1/3*A*a*b/cos(d *x+c)^6+2*B*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(sec(d*x+c)+tan(d *x+c)))+A*b^2*(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(sec(d*x+c)+tan(d*x +c)))+B*b^2*(1/6*sin(d*x+c)^4/cos(d*x+c)^6+1/12*sin(d*x+c)^4/cos(d*x+c)^4) )
Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.27 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 24 \, B b^{2} \cos \left (d x + c\right )^{2} + 16 \, B a^{2} + 32 \, A a b + 16 \, B b^{2} + 2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, A a^{2} + 16 \, B a b + 8 \, A b^{2} + 2 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, d \cos \left (d x + c\right )^{6}} \]
1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) - 24* B*b^2*cos(d*x + c)^2 + 16*B*a^2 + 32*A*a*b + 16*B*b^2 + 2*(3*(5*A*a^2 - 2* B*a*b - A*b^2)*cos(d*x + c)^4 + 8*A*a^2 + 16*B*a*b + 8*A*b^2 + 2*(5*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)
Timed out. \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.32 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} - 8 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2} + 3 \, {\left (11 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1}}{96 \, d} \]
1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) + 1) - 3*(5*A*a^2 - 2 *B*a*b - A*b^2)*log(sin(d*x + c) - 1) - 2*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*s in(d*x + c)^5 + 12*B*b^2*sin(d*x + c)^2 - 8*(5*A*a^2 - 2*B*a*b - A*b^2)*si n(d*x + c)^3 + 8*B*a^2 + 16*A*a*b - 4*B*b^2 + 3*(11*A*a^2 + 2*B*a*b + A*b^ 2)*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1 ))/d
Time = 0.64 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.43 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{2} \sin \left (d x + c\right )^{5} - 6 \, B a b \sin \left (d x + c\right )^{5} - 3 \, A b^{2} \sin \left (d x + c\right )^{5} - 40 \, A a^{2} \sin \left (d x + c\right )^{3} + 16 \, B a b \sin \left (d x + c\right )^{3} + 8 \, A b^{2} \sin \left (d x + c\right )^{3} + 12 \, B b^{2} \sin \left (d x + c\right )^{2} + 33 \, A a^{2} \sin \left (d x + c\right ) + 6 \, B a b \sin \left (d x + c\right ) + 3 \, A b^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} + 16 \, A a b - 4 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{96 \, d} \]
1/96*(3*(5*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*A*a^ 2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) - 1)) - 2*(15*A*a^2*sin(d*x + c) ^5 - 6*B*a*b*sin(d*x + c)^5 - 3*A*b^2*sin(d*x + c)^5 - 40*A*a^2*sin(d*x + c)^3 + 16*B*a*b*sin(d*x + c)^3 + 8*A*b^2*sin(d*x + c)^3 + 12*B*b^2*sin(d*x + c)^2 + 33*A*a^2*sin(d*x + c) + 6*B*a*b*sin(d*x + c) + 3*A*b^2*sin(d*x + c) + 8*B*a^2 + 16*A*a*b - 4*B*b^2)/(sin(d*x + c)^2 - 1)^3)/d
Time = 12.36 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.38 \[ \int \sec ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {5\,A\,a^2}{32}+\frac {B\,a\,b}{16}+\frac {A\,b^2}{32}\right )}{-\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}}\right )\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {11\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,a^2}{6}-\frac {B\,b^2}{12}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {5\,A\,a^2}{6}+\frac {B\,a\,b}{3}+\frac {A\,b^2}{6}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {5\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{4}+\frac {A\,a\,b}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^6-3\,{\sin \left (c+d\,x\right )}^4+3\,{\sin \left (c+d\,x\right )}^2-1\right )} \]
- (atanh((4*sin(c + d*x)*((A*b^2)/32 - (5*A*a^2)/32 + (B*a*b)/16))/((A*b^2 )/8 - (5*A*a^2)/8 + (B*a*b)/4))*((A*b^2)/16 - (5*A*a^2)/16 + (B*a*b)/8))/d - (sin(c + d*x)*((11*A*a^2)/16 + (A*b^2)/16 + (B*a*b)/8) + (B*a^2)/6 - (B *b^2)/12 + sin(c + d*x)^3*((A*b^2)/6 - (5*A*a^2)/6 + (B*a*b)/3) - sin(c + d*x)^5*((A*b^2)/16 - (5*A*a^2)/16 + (B*a*b)/8) + (B*b^2*sin(c + d*x)^2)/4 + (A*a*b)/3)/(d*(3*sin(c + d*x)^2 - 3*sin(c + d*x)^4 + sin(c + d*x)^6 - 1) )