Integrand size = 31, antiderivative size = 165 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {(a A+b (2 A+B)) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {(a A-b (2 A-B)) \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {b^2 (A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {A+B}{4 (a+b) d (1-\sin (c+d x))}-\frac {A-B}{4 (a-b) d (1+\sin (c+d x))} \] Output:
-1/4*(a*A+b*(2*A+B))*ln(1-sin(d*x+c))/(a+b)^2/d+1/4*(a*A-b*(2*A-B))*ln(1+s in(d*x+c))/(a-b)^2/d+b^2*(A*b-B*a)*ln(a+b*sin(d*x+c))/(a^2-b^2)^2/d+1/4*(A +B)/(a+b)/d/(1-sin(d*x+c))-1/4*(A-B)/(a-b)/d/(1+sin(d*x+c))
Time = 0.66 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {b \left ((a-b)^2 (a A+b (2 A+B)) \log (1-\sin (c+d x))-(a+b)^2 (a A+b (-2 A+B)) \log (1+\sin (c+d x))-4 b^2 (A b-a B) \log (a+b \sin (c+d x))\right )}{2 (a-b) (a+b)}+b \sec ^2(c+d x) (A b-a B+(-a A+b B) \sin (c+d x))}{2 b \left (-a^2+b^2\right ) d} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
((b*((a - b)^2*(a*A + b*(2*A + B))*Log[1 - Sin[c + d*x]] - (a + b)^2*(a*A + b*(-2*A + B))*Log[1 + Sin[c + d*x]] - 4*b^2*(A*b - a*B)*Log[a + b*Sin[c + d*x]]))/(2*(a - b)*(a + b)) + b*Sec[c + d*x]^2*(A*b - a*B + (-(a*A) + b* B)*Sin[c + d*x]))/(2*b*(-a^2 + b^2)*d)
Time = 0.51 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (c+d x)}{\cos (c+d x)^3 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {A b+B \sin (c+d x) b}{b (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {A b+B \sin (c+d x) b}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^2 \int \left (\frac {a A-b (2 A-B)}{4 (a-b)^2 b^2 (\sin (c+d x) b+b)}+\frac {a A+b (2 A+B)}{4 b^2 (a+b)^2 (b-b \sin (c+d x))}+\frac {A b-a B}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {A+B}{4 b (a+b) (b-b \sin (c+d x))^2}+\frac {A-B}{4 (a-b) b (\sin (c+d x) b+b)^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (\frac {(A b-a B) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}-\frac {(a A+b (2 A+B)) \log (b-b \sin (c+d x))}{4 b^2 (a+b)^2}+\frac {(a A-b (2 A-B)) \log (b \sin (c+d x)+b)}{4 b^2 (a-b)^2}-\frac {A-B}{4 b (a-b) (b \sin (c+d x)+b)}+\frac {A+B}{4 b (a+b) (b-b \sin (c+d x))}\right )}{d}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
(b^2*(-1/4*((a*A + b*(2*A + B))*Log[b - b*Sin[c + d*x]])/(b^2*(a + b)^2) + ((A*b - a*B)*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + ((a*A - b*(2*A - B) )*Log[b + b*Sin[c + d*x]])/(4*(a - b)^2*b^2) + (A + B)/(4*b*(a + b)*(b - b *Sin[c + d*x])) - (A - B)/(4*(a - b)*b*(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 = 1.61 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {-\frac {A +B}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -2 A b -B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {\left (A b -B a \right ) b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {A -B}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -2 A b +B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(149\) |
| default | \(\frac {-\frac {A +B}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a A -2 A b -B b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}+\frac {\left (A b -B a \right ) b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {A -B}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a A -2 A b +B b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\) | \(149\) |
| parallelrisch | \(\frac {2 b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{2} \left (a A +b \left (2 A +B \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a +b \right ) \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (B -2 A \right ) b +a A \right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (a -b \right ) \left (\frac {\left (A b -B a \right ) \cos \left (2 d x +2 c \right )}{2}+\left (a A -B b \right ) \sin \left (d x +c \right )-\frac {A b}{2}+\frac {B a}{2}\right )\right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(219\) |
| norman | \(\frac {\frac {\left (a A -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}+\frac {\left (a A -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d \left (a^{2}-b^{2}\right )}+\frac {2 \left (a A -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d \left (a^{2}-b^{2}\right )}-\frac {\left (2 A b -2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d \left (a^{2}-b^{2}\right )}-\frac {\left (2 A b -2 B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {b^{2} \left (A b -B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (a A -2 A b +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (a A +2 A b +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(349\) |
| risch | \(\frac {2 i b^{2} B a c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i a A x}{2 a^{2}+4 a b +2 b^{2}}+\frac {i B b x}{2 a^{2}+4 a b +2 b^{2}}+\frac {i B b c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {i a A c}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {i B b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i b^{3} A c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 i b^{2} B a x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B a}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i A b c}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {i A b c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a A c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {i \left (a A \,{\mathrm e}^{3 i \left (d x +c \right )}-B b \,{\mathrm e}^{3 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}-2 i A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B b \,{\mathrm e}^{i \left (d x +c \right )}+2 i B a \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i A b x}{a^{2}-2 a b +b^{2}}-\frac {i a A x}{2 \left (a^{2}-2 a b +b^{2}\right )}-\frac {i B b x}{2 \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i b^{3} A x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {i A b x}{a^{2}+2 a b +b^{2}}\) | \(771\) |
Input:
int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d*(-(A+B)/(4*a+4*b)/(sin(d*x+c)-1)+1/4/(a+b)^2*(-A*a-2*A*b-B*b)*ln(sin(d *x+c)-1)+(A*b-B*a)*b^2/(a+b)^2/(a-b)^2*ln(a+b*sin(d*x+c))-(A-B)/(4*a-4*b)/ (1+sin(d*x+c))+1/4*(A*a-2*A*b+B*b)/(a-b)^2*ln(1+sin(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.42 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {2 \, B a^{3} - 2 \, A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3} - 4 \, {\left (B a b^{2} - A b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (A a^{3} + B a^{2} b - {\left (3 \, A - 2 \, B\right )} a b^{2} - {\left (2 \, A - B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{3} + B a^{2} b - {\left (3 \, A + 2 \, B\right )} a b^{2} + {\left (2 \, A + B\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="fri cas")
Output:
1/4*(2*B*a^3 - 2*A*a^2*b - 2*B*a*b^2 + 2*A*b^3 - 4*(B*a*b^2 - A*b^3)*cos(d *x + c)^2*log(b*sin(d*x + c) + a) + (A*a^3 + B*a^2*b - (3*A - 2*B)*a*b^2 - (2*A - B)*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (A*a^3 + B*a^2*b - (3*A + 2*B)*a*b^2 + (2*A + B)*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^3 - B*a^2*b - A*a*b^2 + B*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b ^4)*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\int \frac {\left (A + B \sin {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
Output:
Integral((A + B*sin(c + d*x))*sec(c + d*x)**3/(a + b*sin(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, {\left (B a b^{2} - A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (A a - {\left (2 \, A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (A a + {\left (2 \, A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (B a - A b + {\left (A a - B b\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="max ima")
Output:
-1/4*(4*(B*a*b^2 - A*b^3)*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) - (A*a - (2*A - B)*b)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b + b^2) + (A*a + ( 2*A + B)*b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) + 2*(B*a - A*b + (A* a - B*b)*sin(d*x + c))/((a^2 - b^2)*sin(d*x + c)^2 - a^2 + b^2))/d
Time = 0.42 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {{\left (B a b^{3} - A b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} - \frac {{\left (A a + 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} + \frac {{\left (A a - 2 \, A b + B b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} - \frac {B a^{3} - A a^{2} b - B a b^{2} + A b^{3} + {\left (A a^{3} - B a^{2} b - A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="gia c")
Output:
-(B*a*b^3 - A*b^4)*log(abs(b*sin(d*x + c) + a))/(a^4*b*d - 2*a^2*b^3*d + b ^5*d) - 1/4*(A*a + 2*A*b + B*b)*log(abs(-sin(d*x + c) + 1))/(a^2*d + 2*a*b *d + b^2*d) + 1/4*(A*a - 2*A*b + B*b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) - 1/2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3 + (A*a^3 - B*a^ 2*b - A*a*b^2 + B*b^3)*sin(d*x + c))/((a + b)^2*(a - b)^2*d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1))
Time = 0.58 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.19 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b^3-B\,a\,b^2\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {\frac {A\,b-B\,a}{2\,\left (a^2-b^2\right )}-\frac {\sin \left (c+d\,x\right )\,\left (A\,a-B\,b\right )}{2\,\left (a^2-b^2\right )}}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A\,a-b\,\left (2\,A-B\right )\right )}{d\,\left (4\,a^2-8\,a\,b+4\,b^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+b\,\left (2\,A+B\right )\right )}{d\,\left (4\,a^2+8\,a\,b+4\,b^2\right )} \] Input:
int((A + B*sin(c + d*x))/(cos(c + d*x)^3*(a + b*sin(c + d*x))),x)
Output:
(log(a + b*sin(c + d*x))*(A*b^3 - B*a*b^2))/(d*(a^4 + b^4 - 2*a^2*b^2)) - ((A*b - B*a)/(2*(a^2 - b^2)) - (sin(c + d*x)*(A*a - B*b))/(2*(a^2 - b^2))) /(d*cos(c + d*x)^2) + (log(sin(c + d*x) + 1)*(A*a - b*(2*A - B)))/(d*(4*a^ 2 - 8*a*b + 4*b^2)) - (log(sin(c + d*x) - 1)*(A*a + b*(2*A + B)))/(d*(8*a* b + 4*a^2 + 4*b^2))
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^3*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
Output:
( - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + log(tan((c + d*x)/2) - 1) + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - log(tan((c + d*x)/2) + 1) - sin(c + d*x))/(2*d*(sin(c + d*x)**2 - 1))