Integrand size = 31, antiderivative size = 165 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (3 a^2 A-A b^2-2 a b B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(a+b)^2 (A+B)}{16 d (1-\sin (c+d x))^2}+\frac {(a+b) (a (3 A+B)-b (A+3 B))}{16 d (1-\sin (c+d x))}-\frac {(a-b)^2 (A-B)}{16 d (1+\sin (c+d x))^2}-\frac {(a-b) (3 a A+A b-a B-3 b B)}{16 d (1+\sin (c+d x))} \] Output:
1/8*(3*A*a^2-A*b^2-2*B*a*b)*arctanh(sin(d*x+c))/d+1/16*(a+b)^2*(A+B)/d/(1- sin(d*x+c))^2+1/16*(a+b)*(a*(3*A+B)-b*(A+3*B))/d/(1-sin(d*x+c))-1/16*(a-b) ^2*(A-B)/d/(1+sin(d*x+c))^2-1/16*(a-b)*(3*A*a+A*b-B*a-3*B*b)/d/(1+sin(d*x+ c))
Time = 1.94 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.13 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {4 \left (-a^2+b^2\right ) \sec ^4(c+d x) (a+b \sin (c+d x))^3 (A b-a B+(-a A+b B) \sin (c+d x))+\left (-3 a^2 A+A b^2+2 a b B\right ) \left (\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+2 a^3 b \sec ^2(c+d x)-2 \left (a^4-b^4\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 b+4 a b^3\right ) \tan ^2(c+d x)\right )}{16 \left (a^2-b^2\right )^2 d} \] Input:
Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
(4*(-a^2 + b^2)*Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3*(A*b - a*B + (-(a*A) + b*B)*Sin[c + d*x]) + (-3*a^2*A + A*b^2 + 2*a*b*B)*((a^2 - b^2)^2*(Log[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^2 - b^2)^2*d)
Time = 0.50 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.08, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2 (A+B \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^4 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle -\frac {b^4 \int \left (-\frac {(A-B) (a-b)^2}{8 b^2 (\sin (c+d x) b+b)^3}-\frac {(3 a A+b A-a B-3 b B) (a-b)}{16 b^3 (\sin (c+d x) b+b)^2}-\frac {3 A a^2-2 b B a-A b^2}{8 b^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {(a+b) (a (3 A+B)-b (A+3 B))}{16 b^3 (b-b \sin (c+d x))^2}-\frac {(a+b)^2 (A+B)}{8 b^2 (b-b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {b^4 \left (-\frac {\left (3 a^2 A-2 a b B-A b^2\right ) \text {arctanh}(\sin (c+d x))}{8 b^4}+\frac {(a-b) (3 a A-a B+A b-3 b B)}{16 b^3 (b \sin (c+d x)+b)}-\frac {(a+b) (a (3 A+B)-b (A+3 B))}{16 b^3 (b-b \sin (c+d x))}+\frac {(a-b)^2 (A-B)}{16 b^2 (b \sin (c+d x)+b)^2}-\frac {(a+b)^2 (A+B)}{16 b^2 (b-b \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
Output:
-((b^4*(-1/8*((3*a^2*A - A*b^2 - 2*a*b*B)*ArcTanh[Sin[c + d*x]])/b^4 - ((a + b)^2*(A + B))/(16*b^2*(b - b*Sin[c + d*x])^2) - ((a + b)*(a*(3*A + B) - b*(A + 3*B)))/(16*b^3*(b - b*Sin[c + d*x])) + ((a - b)^2*(A - B))/(16*b^2 *(b + b*Sin[c + d*x])^2) + ((a - b)*(3*a*A + A*b - a*B - 3*b*B))/(16*b^3*( b + b*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[((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 = 1.52 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.43
| method | result | size |
| derivativedivides | \(\frac {A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B}{4 \cos \left (d x +c \right )^{4}}+\frac {A a b}{2 \cos \left (d x +c \right )^{4}}+2 a b B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} A \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,b^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(236\) |
| default | \(\frac {A \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2} B}{4 \cos \left (d x +c \right )^{4}}+\frac {A a b}{2 \cos \left (d x +c \right )^{4}}+2 a b B \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b^{2} A \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {B \,b^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(236\) |
| parallelrisch | \(\frac {-6 \left (A \,a^{2}-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (A \,a^{2}-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-2 A a b -a^{2} B -B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-2 A a b -a^{2} B +B \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (3 A \,a^{2}-b^{2} A -2 a b B \right ) \sin \left (3 d x +3 c \right )+\left (11 A \,a^{2}+7 b^{2} A +14 a b B \right ) \sin \left (d x +c \right )+10 A a b +5 a^{2} B +3 B \,b^{2}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(263\) |
| risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (3 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-2 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}+11 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+7 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+14 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+16 i B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-11 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-8 i B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-14 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-8 i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 A \,a^{2}+b^{2} A +2 a b B +32 i A a b \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} A}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b B}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} A}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b B}{4 d}\) | \(398\) |
| norman | \(\frac {\frac {\left (7 A \,a^{2}+11 b^{2} A +22 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (4 A a b +2 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {\left (4 A a b +2 a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {\left (12 A a b +6 a^{2} B +4 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (12 A a b +6 a^{2} B +4 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {\left (16 A a b +8 a^{2} B +12 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {\left (16 A a b +8 a^{2} B +12 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {\left (5 A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (5 A \,a^{2}+b^{2} A +2 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (9 A \,a^{2}+5 b^{2} A +10 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 d}+\frac {\left (9 A \,a^{2}+5 b^{2} A +10 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 d}+\frac {\left (27 A \,a^{2}+31 b^{2} A +62 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {\left (27 A \,a^{2}+31 b^{2} A +62 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {\left (3 A \,a^{2}-b^{2} A -2 a b B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A \,a^{2}-b^{2} A -2 a b B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(523\) |
Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO SE)
Output:
1/d*(A*a^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+ c)+tan(d*x+c)))+1/4*a^2*B/cos(d*x+c)^4+1/2*A*a*b/cos(d*x+c)^4+2*a*b*B*(1/4 *sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/ 8*ln(sec(d*x+c)+tan(d*x+c)))+b^2*A*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin( d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c)))+1/4*B* b^2*sin(d*x+c)^4/cos(d*x+c)^4)
Time = 0.09 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 4 \, B a^{2} + 8 \, A a b + 4 \, B b^{2} + 2 \, {\left (2 \, A a^{2} + 4 \, B a b + 2 \, A b^{2} + {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f ricas")
Output:
1/16*((3*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - ( 3*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 8*B*b^2 *cos(d*x + c)^2 + 4*B*a^2 + 8*A*a*b + 4*B*b^2 + 2*(2*A*a^2 + 4*B*a*b + 2*A *b^2 + (3*A*a^2 - 2*B*a*b - A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d* x + c)^4)
Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.04 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (4 \, B b^{2} \sin \left (d x + c\right )^{2} - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 2 \, B a^{2} + 4 \, A a b - 2 \, B b^{2} + {\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m axima")
Output:
1/16*((3*A*a^2 - 2*B*a*b - A*b^2)*log(sin(d*x + c) + 1) - (3*A*a^2 - 2*B*a *b - A*b^2)*log(sin(d*x + c) - 1) + 2*(4*B*b^2*sin(d*x + c)^2 - (3*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 + 2*B*a^2 + 4*A*a*b - 2*B*b^2 + (5*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/ d
Time = 0.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.16 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {3 \, A a^{2} \sin \left (d x + c\right )^{3} - 2 \, B a b \sin \left (d x + c\right )^{3} - A b^{2} \sin \left (d x + c\right )^{3} - 4 \, B b^{2} \sin \left (d x + c\right )^{2} - 5 \, A a^{2} \sin \left (d x + c\right ) - 2 \, B a b \sin \left (d x + c\right ) - A b^{2} \sin \left (d x + c\right ) - 2 \, B a^{2} - 4 \, A a b + 2 \, B b^{2}}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:
integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g iac")
Output:
1/16*(3*A*a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) + 1))/d - 1/16*(3*A* a^2 - 2*B*a*b - A*b^2)*log(abs(sin(d*x + c) - 1))/d - 1/8*(3*A*a^2*sin(d*x + c)^3 - 2*B*a*b*sin(d*x + c)^3 - A*b^2*sin(d*x + c)^3 - 4*B*b^2*sin(d*x + c)^2 - 5*A*a^2*sin(d*x + c) - 2*B*a*b*sin(d*x + c) - A*b^2*sin(d*x + c) - 2*B*a^2 - 4*A*a*b + 2*B*b^2)/((sin(d*x + c)^2 - 1)^2*d)
Time = 33.78 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.10 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,a^2}{4}-\frac {B\,b^2}{4}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {A\,a\,b}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {3\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{-\frac {3\,A\,a^2}{4}+\frac {B\,a\,b}{2}+\frac {A\,b^2}{4}}\right )\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )}{d} \] Input:
int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x)^5,x)
Output:
(sin(c + d*x)*((5*A*a^2)/8 + (A*b^2)/8 + (B*a*b)/4) + (B*a^2)/4 - (B*b^2)/ 4 + sin(c + d*x)^3*((A*b^2)/8 - (3*A*a^2)/8 + (B*a*b)/4) + (B*b^2*sin(c + d*x)^2)/2 + (A*a*b)/2)/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) - (atan h((4*sin(c + d*x)*((A*b^2)/16 - (3*A*a^2)/16 + (B*a*b)/8))/((A*b^2)/4 - (3 *A*a^2)/4 + (B*a*b)/2))*((A*b^2)/8 - (3*A*a^2)/8 + (B*a*b)/4))/d
Time = 0.18 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.38 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a \,b^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a \,b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-6 \sin \left (d x +c \right )^{4} a^{2} b +2 \sin \left (d x +c \right )^{4} b^{3}-3 \sin \left (d x +c \right )^{3} a^{3}+3 \sin \left (d x +c \right )^{3} a \,b^{2}+12 \sin \left (d x +c \right )^{2} a^{2} b +5 \sin \left (d x +c \right ) a^{3}+3 \sin \left (d x +c \right ) a \,b^{2}}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
Output:
( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 + 3*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**4*a*b**2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*a**3 - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - 3*log(tan( (c + d*x)/2) - 1)*a**3 + 3*log(tan((c + d*x)/2) - 1)*a*b**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 + 6*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 + 3*log(tan((c + d*x)/2) + 1 )*a**3 - 3*log(tan((c + d*x)/2) + 1)*a*b**2 - 6*sin(c + d*x)**4*a**2*b + 2 *sin(c + d*x)**4*b**3 - 3*sin(c + d*x)**3*a**3 + 3*sin(c + d*x)**3*a*b**2 + 12*sin(c + d*x)**2*a**2*b + 5*sin(c + d*x)*a**3 + 3*sin(c + d*x)*a*b**2) /(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))