Integrand size = 31, antiderivative size = 228 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {(a A+3 A b+2 b B) \log (1-\sin (c+d x))}{4 (a+b)^3 d}+\frac {(a A-3 A b+2 b B) \log (1+\sin (c+d x))}{4 (a-b)^3 d}+\frac {b^2 \left (4 a A b-3 a^2 B-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {b \left (a^2 A+3 A b^2-4 a b B\right )}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\sec ^2(c+d x) (A b-a B-(a A-b B) \sin (c+d x))}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]
-1/4*(A*a+3*A*b+2*B*b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/4*(A*a-3*A*b+2*B*b)*ln (1+sin(d*x+c))/(a-b)^3/d+b^2*(4*A*a*b-3*B*a^2-B*b^2)*ln(a+b*sin(d*x+c))/(a ^2-b^2)^3/d-1/2*b*(A*a^2+3*A*b^2-4*B*a*b)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))-1 /2*sec(d*x+c)^2*(A*b-B*a-(A*a-B*b)*sin(d*x+c))/(a^2-b^2)/d/(a+b*sin(d*x+c) )
Time = 1.21 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {(a A-b B) ((a-b) \log (1-\sin (c+d x))-(a+b) \log (1+\sin (c+d x))+2 b \log (a+b \sin (c+d x)))}{(a-b) (a+b)}+\frac {\sec ^2(c+d x) (A b-a B+(-a A+b B) \sin (c+d x))}{a+b \sin (c+d x)}+b \left (a^2 A+3 A b^2-4 a b B\right ) \left (-\frac {\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 b}-\frac {2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}\right )}{2 \left (-a^2+b^2\right ) d} \]
(((a*A - b*B)*((a - b)*Log[1 - Sin[c + d*x]] - (a + b)*Log[1 + Sin[c + d*x ]] + 2*b*Log[a + b*Sin[c + d*x]]))/((a - b)*(a + b)) + (Sec[c + d*x]^2*(A* b - a*B + (-(a*A) + b*B)*Sin[c + d*x]))/(a + b*Sin[c + d*x]) + b*(a^2*A + 3*A*b^2 - 4*a*b*B)*(-1/2*Log[1 - Sin[c + d*x]]/(b*(a + b)^2) + Log[1 + Sin [c + d*x]]/(2*(a - b)^2*b) - (2*a*Log[a + b*Sin[c + d*x]])/((a - b)^2*(a + b)^2) + 1/((a^2 - b^2)*(a + b*Sin[c + d*x]))))/(2*(-a^2 + b^2)*d)
Time = 0.56 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^3 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {A b+B \sin (c+d x) b}{b (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {A b+B \sin (c+d x) b}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^2 \int \left (\frac {A-B}{4 (a-b)^2 b (\sin (c+d x) b+b)^2}+\frac {a A+3 b A+2 b B}{4 b^2 (a+b)^3 (b-b \sin (c+d x))}+\frac {-3 B a^2+4 A b a-b^2 B}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {a A-3 b A+2 b B}{4 (a-b)^3 b^2 (\sin (c+d x) b+b)}+\frac {A+B}{4 b (a+b)^2 (b-b \sin (c+d x))^2}+\frac {A b-a B}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (-\frac {A b-a B}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\left (-3 a^2 B+4 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3}-\frac {(a A+3 A b+2 b B) \log (b-b \sin (c+d x))}{4 b^2 (a+b)^3}+\frac {(a A-3 A b+2 b B) \log (b \sin (c+d x)+b)}{4 b^2 (a-b)^3}-\frac {A-B}{4 b (a-b)^2 (b \sin (c+d x)+b)}+\frac {A+B}{4 b (a+b)^2 (b-b \sin (c+d x))}\right )}{d}\) |
(b^2*(-1/4*((a*A + 3*A*b + 2*b*B)*Log[b - b*Sin[c + d*x]])/(b^2*(a + b)^3) + ((4*a*A*b - 3*a^2*B - b^2*B)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^3 + ( (a*A - 3*A*b + 2*b*B)*Log[b + b*Sin[c + d*x]])/(4*(a - b)^3*b^2) + (A + B) /(4*b*(a + b)^2*(b - b*Sin[c + d*x])) - (A*b - a*B)/((a^2 - b^2)^2*(a + b* Sin[c + d*x])) - (A - B)/(4*(a - b)^2*b*(b + b*Sin[c + d*x]))))/d
3.16.57.3.1 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 = 2.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {-\frac {A +B}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -3 A b -2 B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {A -B}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}-\frac {b^{2} \left (A b -B a \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(191\) |
| default | \(\frac {-\frac {A +B}{4 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -3 A b -2 B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{3}}-\frac {A -B}{4 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{3}}-\frac {b^{2} \left (A b -B a \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(191\) |
| parallelrisch | \(\frac {8 \left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) b^{2} \left (A a b -\frac {3}{4} B \,a^{2}-\frac {1}{4} B \,b^{2}\right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) a \left (a A +3 b \left (A +\frac {2 B}{3}\right )\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right ) \left (a A -3 b \left (A -\frac {2 B}{3}\right )\right ) a \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (3 a \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right ) \cos \left (2 d x +2 c \right )+b \left (A \,a^{2} b -3 A \,b^{3}-\frac {3}{2} B \,a^{3}+\frac {7}{2} B a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (-2 A \,a^{4}+3 A \,a^{2} b^{2}-3 A \,b^{4}+\frac {1}{2} B \,a^{3} b +\frac {3}{2} B a \,b^{3}\right ) \sin \left (d x +c \right )+5 a \left (a -b \right ) \left (a +b \right ) \left (A b -B a \right )\right ) \left (a -b \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} a d \left (b \sin \left (3 d x +3 c \right )+2 a \cos \left (2 d x +2 c \right )+b \sin \left (d x +c \right )+2 a \right )}\) | \(419\) |
| norman | \(\frac {\frac {\left (3 A \,a^{4}-5 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +6 B a \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (3 A \,a^{4}-5 A \,a^{2} b^{2}-2 A \,b^{4}-2 B \,a^{3} b +6 B a \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (A \,a^{4}+A \,a^{2} b^{2}+2 A \,b^{4}-2 B \,a^{3} b -2 B a \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (A \,a^{4}+A \,a^{2} b^{2}+2 A \,b^{4}-2 B \,a^{3} b -2 B a \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (A b -B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}-\frac {2 \left (A b -B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}-\frac {2 \left (2 A b -2 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {b^{2} \left (4 A a b -3 B \,a^{2}-B \,b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {\left (a A -3 A b +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (a A +3 A b +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(629\) |
| risch | \(\text {Expression too large to display}\) | \(1389\) |
1/d*(-1/4*(A+B)/(a+b)^2/(sin(d*x+c)-1)+1/4/(a+b)^3*(-A*a-3*A*b-2*B*b)*ln(s in(d*x+c)-1)-1/4*(A-B)/(a-b)^2/(1+sin(d*x+c))+1/4*(A*a-3*A*b+2*B*b)/(a-b)^ 3*ln(1+sin(d*x+c))-b^2*(A*b-B*a)/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))+b^2*(4*A *a*b-3*B*a^2-B*b^2)/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (219) = 438\).
Time = 0.76 (sec) , antiderivative size = 598, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 4 \, A a^{2} b^{3} + 2 \, B a b^{4} - 2 \, A b^{5} - 2 \, {\left (A a^{4} b - 4 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + 4 \, B a b^{4} - 3 \, A b^{5}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (3 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3} + B a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (A a^{4} b + 2 \, B a^{3} b^{2} - 6 \, {\left (A - B\right )} a^{2} b^{3} - 2 \, {\left (4 \, A - 3 \, B\right )} a b^{4} - {\left (3 \, A - 2 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (A a^{5} + 2 \, B a^{4} b - 6 \, {\left (A - B\right )} a^{3} b^{2} - 2 \, {\left (4 \, A - 3 \, B\right )} a^{2} b^{3} - {\left (3 \, A - 2 \, B\right )} a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A a^{4} b + 2 \, B a^{3} b^{2} - 6 \, {\left (A + B\right )} a^{2} b^{3} + 2 \, {\left (4 \, A + 3 \, B\right )} a b^{4} - {\left (3 \, A + 2 \, B\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (A a^{5} + 2 \, B a^{4} b - 6 \, {\left (A + B\right )} a^{3} b^{2} + 2 \, {\left (4 \, A + 3 \, B\right )} a^{2} b^{3} - {\left (3 \, A + 2 \, B\right )} a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{5} - B a^{4} b - 2 \, A a^{3} b^{2} + 2 \, B a^{2} b^{3} + A a b^{4} - B b^{5}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2}\right )}} \]
1/4*(2*B*a^5 - 2*A*a^4*b - 4*B*a^3*b^2 + 4*A*a^2*b^3 + 2*B*a*b^4 - 2*A*b^5 - 2*(A*a^4*b - 4*B*a^3*b^2 + 2*A*a^2*b^3 + 4*B*a*b^4 - 3*A*b^5)*cos(d*x + c)^2 - 4*((3*B*a^2*b^3 - 4*A*a*b^4 + B*b^5)*cos(d*x + c)^2*sin(d*x + c) + (3*B*a^3*b^2 - 4*A*a^2*b^3 + B*a*b^4)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) + ((A*a^4*b + 2*B*a^3*b^2 - 6*(A - B)*a^2*b^3 - 2*(4*A - 3*B)*a*b^4 - (3*A - 2*B)*b^5)*cos(d*x + c)^2*sin(d*x + c) + (A*a^5 + 2*B*a^4*b - 6*(A - B)*a^3*b^2 - 2*(4*A - 3*B)*a^2*b^3 - (3*A - 2*B)*a*b^4)*cos(d*x + c)^2)* log(sin(d*x + c) + 1) - ((A*a^4*b + 2*B*a^3*b^2 - 6*(A + B)*a^2*b^3 + 2*(4 *A + 3*B)*a*b^4 - (3*A + 2*B)*b^5)*cos(d*x + c)^2*sin(d*x + c) + (A*a^5 + 2*B*a^4*b - 6*(A + B)*a^3*b^2 + 2*(4*A + 3*B)*a^2*b^3 - (3*A + 2*B)*a*b^4) *cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(A*a^5 - B*a^4*b - 2*A*a^3*b^2 + 2*B*a^2*b^3 + A*a*b^4 - B*b^5)*sin(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^ 2*b^5 - b^7)*d*cos(d*x + c)^2*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (A a - {\left (3 \, A - 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + {\left (3 \, A + 2 \, B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left (B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3} + {\left (A a^{2} b - 4 \, B a b^{2} + 3 \, A b^{3}\right )} \sin \left (d x + c\right )^{2} + {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}}{4 \, d} \]
-1/4*(4*(3*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(b*sin(d*x + c) + a)/(a^6 - 3 *a^4*b^2 + 3*a^2*b^4 - b^6) - (A*a - (3*A - 2*B)*b)*log(sin(d*x + c) + 1)/ (a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (A*a + (3*A + 2*B)*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(B*a^3 - 2*A*a^2*b + 3*B*a*b^2 - 2 *A*b^3 + (A*a^2*b - 4*B*a*b^2 + 3*A*b^3)*sin(d*x + c)^2 + (A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*sin(d*x + c))/(a^5 - 2*a^3*b^2 + a*b^4 - (a^4*b - 2*a^ 2*b^3 + b^5)*sin(d*x + c)^3 - (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^2 + ( a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)))/d
Time = 0.39 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{2} b^{3} - 4 \, A a b^{4} + B b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} - \frac {{\left (A a - 3 \, A b + 2 \, B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (A a + 3 \, A b + 2 \, B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (A a^{2} b \sin \left (d x + c\right )^{2} - 4 \, B a b^{2} \sin \left (d x + c\right )^{2} + 3 \, A b^{3} \sin \left (d x + c\right )^{2} + A a^{3} \sin \left (d x + c\right ) - B a^{2} b \sin \left (d x + c\right ) - A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + B a^{3} - 2 \, A a^{2} b + 3 \, B a b^{2} - 2 \, A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (b \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right ) - a\right )}}}{4 \, d} \]
-1/4*(4*(3*B*a^2*b^3 - 4*A*a*b^4 + B*b^5)*log(abs(b*sin(d*x + c) + a))/(a^ 6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7) - (A*a - 3*A*b + 2*B*b)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (A*a + 3*A*b + 2*B*b)*log(ab s(-sin(d*x + c) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(A*a^2*b*sin(d*x + c)^2 - 4*B*a*b^2*sin(d*x + c)^2 + 3*A*b^3*sin(d*x + c)^2 + A*a^3*sin(d* x + c) - B*a^2*b*sin(d*x + c) - A*a*b^2*sin(d*x + c) + B*b^3*sin(d*x + c) + B*a^3 - 2*A*a^2*b + 3*B*a*b^2 - 2*A*b^3)/((a^4 - 2*a^2*b^2 + b^4)*(b*sin (d*x + c)^3 + a*sin(d*x + c)^2 - b*sin(d*x + c) - a)))/d
Time = 12.16 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2\,\left (A\,a^2\,b-4\,B\,a\,b^2+3\,A\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {-B\,a^3+2\,A\,a^2\,b-3\,B\,a\,b^2+2\,A\,b^3}{2\,{\left (a^2-b^2\right )}^2}+\frac {\sin \left (c+d\,x\right )\,\left (A\,a-B\,b\right )}{2\,\left (a^2-b^2\right )}}{d\,\left (-b\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+b\,\sin \left (c+d\,x\right )+a\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+b\,\left (3\,A+2\,B\right )\right )}{d\,\left (4\,a^3+12\,a^2\,b+12\,a\,b^2+4\,b^3\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A\,a-b\,\left (3\,A-2\,B\right )\right )}{d\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (3\,B\,a^2\,b^2-4\,A\,a\,b^3+B\,b^4\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \]
((sin(c + d*x)^2*(3*A*b^3 + A*a^2*b - 4*B*a*b^2))/(2*(a^4 + b^4 - 2*a^2*b^ 2)) - (2*A*b^3 - B*a^3 + 2*A*a^2*b - 3*B*a*b^2)/(2*(a^2 - b^2)^2) + (sin(c + d*x)*(A*a - B*b))/(2*(a^2 - b^2)))/(d*(a + b*sin(c + d*x) - a*sin(c + d *x)^2 - b*sin(c + d*x)^3)) - (log(sin(c + d*x) - 1)*(A*a + b*(3*A + 2*B))) /(d*(12*a*b^2 + 12*a^2*b + 4*a^3 + 4*b^3)) + (log(sin(c + d*x) + 1)*(A*a - b*(3*A - 2*B)))/(d*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3)) - (log(a + b*si n(c + d*x))*(B*b^4 + 3*B*a^2*b^2 - 4*A*a*b^3))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))