Integrand size = 31, antiderivative size = 112 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {(a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))}{4 d}+\frac {(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x)) (A b+a B+(a A+b B) \sin (c+d x))}{2 d} \]
-1/4*(a+b)*(a*A-b*(A+2*B))*ln(1-sin(d*x+c))/d+1/4*(a-b)*(a*A+b*(A-2*B))*ln (1+sin(d*x+c))/d+1/2*sec(d*x+c)^2*(a+b*sin(d*x+c))*(A*b+B*a+(A*a+B*b)*sin( d*x+c))/d
Time = 1.02 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.55 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\left (a^2-b^2\right ) ((a+b) (a A-b (A+2 B)) \log (1-\sin (c+d x))-(a-b) (a A+b (A-2 B)) \log (1+\sin (c+d x)))-2 a^3 (-A b+a B) \sec ^2(c+d x)-2 \left (a^2-b^2\right ) \left (a^2 A+A b^2+2 a b B\right ) \sec (c+d x) \tan (c+d x)+\left (-6 a^3 A b+4 a A b^3+2 b^4 B\right ) \tan ^2(c+d x)}{4 \left (-a^2+b^2\right ) d} \]
((a^2 - b^2)*((a + b)*(a*A - b*(A + 2*B))*Log[1 - Sin[c + d*x]] - (a - b)* (a*A + b*(A - 2*B))*Log[1 + Sin[c + d*x]]) - 2*a^3*(-(A*b) + a*B)*Sec[c + d*x]^2 - 2*(a^2 - b^2)*(a^2*A + A*b^2 + 2*a*b*B)*Sec[c + d*x]*Tan[c + d*x] + (-6*a^3*A*b + 4*a*A*b^3 + 2*b^4*B)*Tan[c + d*x]^2)/(4*(-a^2 + b^2)*d)
Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.21, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^2 (A+B \sin (c+d x))}{\cos (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^3 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{b \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^2 \int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 663 |
| \(\displaystyle \frac {b^2 \int \left (\frac {(A-B) (a-b)^2}{4 b (\sin (c+d x) b+b)^2}+\frac {(a A+b (A-2 B)) (a-b)}{4 b^2 (\sin (c+d x) b+b)}+\frac {(a+b) (a A-b (A+2 B))}{4 b^2 (b-b \sin (c+d x))}+\frac {(a+b)^2 (A+B)}{4 b (b-b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \left (\frac {(a-b) (a A+b (A-2 B)) \log (b \sin (c+d x)+b)}{4 b^2}-\frac {(a+b) (a A-b (A+2 B)) \log (b-b \sin (c+d x))}{4 b^2}-\frac {(a-b)^2 (A-B)}{4 b (b \sin (c+d x)+b)}+\frac {(a+b)^2 (A+B)}{4 b (b-b \sin (c+d x))}\right )}{d}\) |
(b^2*(-1/4*((a + b)*(a*A - b*(A + 2*B))*Log[b - b*Sin[c + d*x]])/b^2 + ((a - b)*(a*A + b*(A - 2*B))*Log[b + b*Sin[c + d*x]])/(4*b^2) + ((a + b)^2*(A + B))/(4*b*(b - b*Sin[c + d*x])) - ((a - b)^2*(A - B))/(4*b*(b + b*Sin[c + d*x]))))/d
3.16.41.3.1 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 = 0.87 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.67
| method | result | size |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(187\) |
| default | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {B \,a^{2}}{2 \cos \left (d x +c \right )^{2}}+\frac {A a b}{\cos \left (d x +c \right )^{2}}+2 B a b \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,b^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(187\) |
| parallelrisch | \(\frac {-2 B \,b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (-A -2 B \right ) b +a A \right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (a A +b \left (A -2 B \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-2 A a b -B \,a^{2}-B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (2 A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \sin \left (d x +c \right )+2 A a b +B \,a^{2}+B \,b^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(198\) |
| risch | \(-i B \,b^{2} x -\frac {2 i B \,b^{2} c}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,a^{2}-A \,b^{2}+4 i A a b \,{\mathrm e}^{i \left (d x +c \right )}-2 B a b +2 i B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}\) | \(335\) |
| norman | \(\frac {\frac {\left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 A a b +2 B \,a^{2}+2 B \,b^{2}}{d}+\frac {4 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (A \,a^{2}+A \,b^{2}+2 B a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 A a b +2 B \,a^{2}+2 B \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A a b +5 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (10 A a b +5 B \,a^{2}+5 B \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b -2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b +2 B \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}-\frac {B \,b^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(418\) |
1/d*(A*a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+1/2*B *a^2/cos(d*x+c)^2+A*a*b/cos(d*x+c)^2+2*B*a*b*(1/2*sin(d*x+c)^3/cos(d*x+c)^ 2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+A*b^2*(1/2*sin(d*x+c)^3/co s(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+tan(d*x+c)))+B*b^2*(1/2*tan(d* x+c)^2+ln(cos(d*x+c))))
Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a^{2} + 4 \, A a b + 2 \, B b^{2} + 2 \, {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*((A*a^2 - 2*B*a*b - (A - 2*B)*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1 ) - (A*a^2 - 2*B*a*b - (A + 2*B)*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1 ) + 2*B*a^2 + 4*A*a*b + 2*B*b^2 + 2*(A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c) )/(d*cos(d*x + c)^2)
\[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\int \left (A + B \sin {\left (c + d x \right )}\right ) \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - {\left (A - 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - 2 \, B a b - {\left (A + 2 \, B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (B a^{2} + 2 \, A a b + B b^{2} + {\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((A*a^2 - 2*B*a*b - (A - 2*B)*b^2)*log(sin(d*x + c) + 1) - (A*a^2 - 2* B*a*b - (A + 2*B)*b^2)*log(sin(d*x + c) - 1) - 2*(B*a^2 + 2*A*a*b + B*b^2 + (A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c))/(sin(d*x + c)^2 - 1))/d
Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.30 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\left (A a^{2} - 2 \, B a b - A b^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a^{2} - 2 \, B a b - A b^{2} - 2 \, B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (B b^{2} \sin \left (d x + c\right )^{2} + A a^{2} \sin \left (d x + c\right ) + 2 \, B a b \sin \left (d x + c\right ) + A b^{2} \sin \left (d x + c\right ) + B a^{2} + 2 \, A a b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]
1/4*((A*a^2 - 2*B*a*b - A*b^2 + 2*B*b^2)*log(abs(sin(d*x + c) + 1)) - (A*a ^2 - 2*B*a*b - A*b^2 - 2*B*b^2)*log(abs(sin(d*x + c) - 1)) - 2*(B*b^2*sin( d*x + c)^2 + A*a^2*sin(d*x + c) + 2*B*a*b*sin(d*x + c) + A*b^2*sin(d*x + c ) + B*a^2 + 2*A*a*b)/(sin(d*x + c)^2 - 1))/d
Time = 12.65 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (A\,a+A\,b-2\,B\,b\right )}{4\,d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{2}+B\,a\,b+\frac {A\,b^2}{2}\right )+\frac {B\,a^2}{2}+\frac {B\,b^2}{2}+A\,a\,b}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A\,b-A\,a+2\,B\,b\right )}{4\,d} \]