Integrand size = 25, antiderivative size = 204 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 a b \left (2 a^2+3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}-\frac {a \sec (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d} \]
3*a*b*(2*a^2+3*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2- b^2)^(7/2)/d-1/2*a*sec(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-1/2*(3*a^2+2* b^2)*sec(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))+1/2*sec(d*x+c)*(3*a*(2*a^2+ 3*b^2)-b*(11*a^2+4*b^2)*sin(d*x+c))/(a^2-b^2)^3/d
Time = 2.23 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {6 a b \left (2 a^2+3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {2}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {b^2 \cos (c+d x) \left (a \left (6 a^2+b^2\right )+b \left (5 a^2+2 b^2\right ) \sin (c+d x)\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}}{2 d} \]
((6*a*b*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ (a^2 - b^2)^(7/2) + Sin[(c + d*x)/2]*(2/((a + b)^3*(Cos[(c + d*x)/2] - Sin [(c + d*x)/2])) - 2/((a - b)^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + ( b^2*Cos[c + d*x]*(a*(6*a^2 + b^2) + b*(5*a^2 + 2*b^2)*Sin[c + d*x]))/((a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])^2))/(2*d)
Time = 0.94 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3343, 3042, 3343, 25, 3042, 3345, 27, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {\tan (c+d x) \sec (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^2 (a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3343 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) (2 b-3 a \sin (c+d x))}{(a+b \sin (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {2 b-3 a \sin (c+d x)}{\cos (c+d x)^2 (a+b \sin (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3343 |
| \(\displaystyle -\frac {\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\int -\frac {\sec ^2(c+d x) \left (5 a b-2 \left (3 a^2+2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec ^2(c+d x) \left (5 a b-2 \left (3 a^2+2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {5 a b-2 \left (3 a^2+2 b^2\right ) \sin (c+d x)}{\cos (c+d x)^2 (a+b \sin (c+d x))}dx}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle -\frac {\frac {-\frac {\int \frac {3 a b \left (2 a^2+3 b^2\right )}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {-\frac {3 a b \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {-\frac {3 a b \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {\frac {-\frac {6 a b \left (2 a^2+3 b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {\frac {\frac {12 a b \left (2 a^2+3 b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\frac {-\frac {6 a b \left (2 a^2+3 b^2\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {\sec (c+d x) \left (3 a \left (2 a^2+3 b^2\right )-b \left (11 a^2+4 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )}}{a^2-b^2}+\frac {\left (3 a^2+2 b^2\right ) \sec (c+d x)}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {a \sec (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}\) |
-1/2*(a*Sec[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (((3*a^2 + 2*b^2)*Sec[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) + ((-6*a*b*(2*a^ 2 + 3*b^2)*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (Sec[c + d*x]*(3*a*(2*a^2 + 3*b^2) - b*(11*a^2 + 4*b^2) *Sin[c + d*x]))/((a^2 - b^2)*d))/(a^2 - b^2))/(2*(a^2 - b^2))
3.15.74.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m + 1)/(f*g*(a^2 - b^2)*(m + 1))), x] + Simp[1/((a^2 - b^2)*(m + 1)) Int[(g*Cos[e + f*x])^p *(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ [a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
Time = 2.18 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b \left (\frac {\frac {7 a^{2} b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {b \left (6 a^{4}+13 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b^{2} \left (17 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a b \left (6 a^{2}+b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 a \left (2 a^{2}+3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(242\) |
| default | \(\frac {\frac {2 b \left (\frac {\frac {7 a^{2} b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {b \left (6 a^{4}+13 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {b^{2} \left (17 a^{2}+4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a b \left (6 a^{2}+b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 a \left (2 a^{2}+3 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(242\) |
| risch | \(\frac {-18 i a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}-27 i a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-9 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}-26 i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+16 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+11 i a^{2} b^{3}+4 i b^{5}+38 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+7 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \left (a^{2}-b^{2}\right )^{3}}-\frac {3 i b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {9 i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 i b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {9 i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) | \(585\) |
1/d*(2*b/(a-b)^3/(a+b)^3*((7/2*a^2*b^2*tan(1/2*d*x+1/2*c)^3+1/2*b*(6*a^4+1 3*a^2*b^2+2*b^4)/a*tan(1/2*d*x+1/2*c)^2+1/2*b^2*(17*a^2+4*b^2)*tan(1/2*d*x +1/2*c)+1/2*a*b*(6*a^2+b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c )+a)^2+3/2*a*(2*a^2+3*b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2 *c)+2*b)/(a^2-b^2)^(1/2)))-1/(a+b)^3/(tan(1/2*d*x+1/2*c)-1)+1/(a-b)^3/(tan (1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (192) = 384\).
Time = 0.31 (sec) , antiderivative size = 895, normalized size of antiderivative = 4.39 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\left [-\frac {4 \, a^{7} - 12 \, a^{5} b^{2} + 12 \, a^{3} b^{4} - 4 \, a b^{6} + 2 \, {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{5} b + 5 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} - {\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}, -\frac {2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} + {\left (16 \, a^{5} b^{2} - 17 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left ({\left (2 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{5} b + 5 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (2 \, a^{6} b - 6 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - 2 \, b^{7} - {\left (11 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} b^{2} - 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - 4 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b - 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} - 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{10} - 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} + 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )}}\right ] \]
[-1/4*(4*a^7 - 12*a^5*b^2 + 12*a^3*b^4 - 4*a*b^6 + 2*(16*a^5*b^2 - 17*a^3* b^4 + a*b^6)*cos(d*x + c)^2 - 3*((2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2* (2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)*sin(d*x + c) - (2*a^5*b + 5*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^ 2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*co s(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a ^2 - b^2)) - 2*(2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7* a^2*b^5 - 4*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a ^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5 *b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)), -1/2*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + (16*a^5*b^2 - 17*a^3*b^4 + a*b^6)*cos(d *x + c)^2 + 3*((2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 - 2*(2*a^4*b^2 + 3*a^2 *b^4)*cos(d*x + c)*sin(d*x + c) - (2*a^5*b + 5*a^3*b^3 + 3*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d* x + c))) - (2*a^6*b - 6*a^4*b^3 + 6*a^2*b^5 - 2*b^7 - (11*a^4*b^3 - 7*a^2* b^5 - 4*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b ^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2* a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c))]
\[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.41 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.79 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {3 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {7 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 17 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{4} b^{2} + a^{2} b^{4}}{{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]
(3*(2*a^3*b + 3*a*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan(( a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^ 4 - b^6)*sqrt(a^2 - b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2* d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/ 2*d*x + 1/2*c)^2 - 1)) + (7*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*a^4*b^2*tan (1/2*d*x + 1/2*c)^2 + 13*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^6*tan(1/2*d* x + 1/2*c)^2 + 17*a^3*b^3*tan(1/2*d*x + 1/2*c) + 4*a*b^5*tan(1/2*d*x + 1/2 *c) + 6*a^4*b^2 + a^2*b^4)/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*(a*tan(1 /2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2))/d
Time = 17.57 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.06 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3\,a\,b\,\mathrm {atan}\left (\frac {\frac {3\,a\,b\,\left (2\,a^2+3\,b^2\right )\,\left (2\,a^6\,b-6\,a^4\,b^3+6\,a^2\,b^5-2\,b^7\right )}{2\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+3\,b^2\right )\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}}{6\,a^3\,b+9\,a\,b^3}\right )\,\left (2\,a^2+3\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {2\,a^5+12\,a^3\,b^2+a\,b^4}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^4\,b+7\,a^2\,b^3+6\,b^5\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-2\,a^6+24\,a^4\,b^2+21\,a^2\,b^4+2\,b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^4+39\,a^2\,b^2+4\,b^4\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^6-2\,a^4\,b^2+14\,a^2\,b^4+b^6\right )}{a\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {3\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^2+3\,b^2\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-a^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2+4\,b^2\right )+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]
(3*a*b*atan(((3*a*b*(2*a^2 + 3*b^2)*(2*a^6*b - 2*b^7 + 6*a^2*b^5 - 6*a^4*b ^3))/(2*(a + b)^(7/2)*(a - b)^(7/2)) + (3*a^2*b*tan(c/2 + (d*x)/2)*(2*a^2 + 3*b^2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/((a + b)^(7/2)*(a - b)^(7/2) ))/(9*a*b^3 + 6*a^3*b))*(2*a^2 + 3*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((a*b^4 + 2*a^5 + 12*a^3*b^2)/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (2*ta n(c/2 + (d*x)/2)^3*(2*a^4*b + 6*b^5 + 7*a^2*b^3))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^4*(2*b^6 - 2*a^6 + 21*a^2*b^4 + 24*a^4*b ^2))/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (b*tan(c/2 + (d*x)/2)*(2*a^ 4 + 4*b^4 + 39*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)^2*(2*a^6 + b^6 + 14*a^2*b^4 - 2*a^4*b^2))/(a*(a^6 - b^6 + 3*a^2 *b^4 - 3*a^4*b^2)) - (3*a^2*b*tan(c/2 + (d*x)/2)^5*(2*a^2 + 3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2 - tan(c/ 2 + (d*x)/2)^2*(a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b* tan(c/2 + (d*x)/2)^5 - 4*a*b*tan(c/2 + (d*x)/2)))