Integrand size = 29, antiderivative size = 154 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {x}{b^2}-\frac {2 \left (a^4+a^2 b^2-2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^2 \sqrt {a^2-b^2} d}+\frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))} \]
x/b^2+2*b*arctanh(cos(d*x+c))/a^3/d+(a^2-2*b^2)*cos(d*x+c)/a^2/b/d/(a+b*si n(d*x+c))-cot(d*x+c)/a/d/(a+b*sin(d*x+c))-2*(a^4+a^2*b^2-2*b^4)*arctan((b+ a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/b^2/d/(a^2-b^2)^(1/2)
Time = 1.57 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 (c+d x)}{b^2}-\frac {4 \left (a^4+a^2 b^2-2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 b^2 \sqrt {a^2-b^2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{a^2}+\frac {4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}-\frac {4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {2 \left (a^2-b^2\right ) \cos (c+d x)}{a^2 b (a+b \sin (c+d x))}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a^2}}{2 d} \]
((2*(c + d*x))/b^2 - (4*(a^4 + a^2*b^2 - 2*b^4)*ArcTan[(b + a*Tan[(c + d*x )/2])/Sqrt[a^2 - b^2]])/(a^3*b^2*Sqrt[a^2 - b^2]) - Cot[(c + d*x)/2]/a^2 + (4*b*Log[Cos[(c + d*x)/2]])/a^3 - (4*b*Log[Sin[(c + d*x)/2]])/a^3 + (2*(a ^2 - b^2)*Cos[c + d*x])/(a^2*b*(a + b*Sin[c + d*x])) + Tan[(c + d*x)/2]/a^ 2)/(2*d)
Time = 0.74 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3369, 25, 3042, 3536, 3042, 3139, 1083, 217, 4257}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^2 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3369 |
| \(\displaystyle \frac {\int -\frac {\csc (c+d x) \left (2 b^2+a \sin (c+d x) b-a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\csc (c+d x) \left (2 b^2+a \sin (c+d x) b-a^2 \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {2 b^2+a \sin (c+d x) b-a^2 \sin (c+d x)^2}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3536 |
| \(\displaystyle -\frac {\frac {\left (a^4+a^2 b^2-2 b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {a^2 x}{b}}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\left (a^4+a^2 b^2-2 b^4\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {a^2 x}{b}}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {\frac {2 \left (a^4+a^2 b^2-2 b^4\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}+\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {a^2 x}{b}}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {-\frac {4 \left (a^4+a^2 b^2-2 b^4\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a b d}+\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {a^2 x}{b}}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {a^2 x}{b}+\frac {2 \left (a^4+a^2 b^2-2 b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d \sqrt {a^2-b^2}}}{a^2 b}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{a^2 b d (a+b \sin (c+d x))}-\frac {-\frac {a^2 x}{b}+\frac {2 \left (a^4+a^2 b^2-2 b^4\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d \sqrt {a^2-b^2}}-\frac {2 b^2 \text {arctanh}(\cos (c+d x))}{a d}}{a^2 b}-\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
-((-((a^2*x)/b) + (2*(a^4 + a^2*b^2 - 2*b^4)*ArcTan[(2*b + 2*a*Tan[(c + d* x)/2])/(2*Sqrt[a^2 - b^2])])/(a*b*Sqrt[a^2 - b^2]*d) - (2*b^2*ArcTanh[Cos[ c + d*x]])/(a*d))/(a^2*b)) + ((a^2 - 2*b^2)*Cos[c + d*x])/(a^2*b*d*(a + b* Sin[c + d*x])) - Cot[c + d*x]/(a*d*(a + b*Sin[c + d*x]))
3.12.31.3.1 Defintions of rubi rules used
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_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(d*Sin[ e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(a*d*f*(n + 1))), x] + (-Si mp[(a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*(( a + b*Sin[e + f*x])^(m + 1)/(a^2*b*d^2*f*(n + 1)*(m + 1))), x] + Simp[1/(a^ 2*b*d*(n + 1)*(m + 1)) Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^ (m + 1)*Simp[a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1 )*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) /(b*(b*c - a*d)) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)) Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a , b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.54 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {-b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{2}-b^{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{4}+a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3} b^{2}}}{d}\) | \(201\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {-b^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-b a \left (a^{2}-b^{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{4}+a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3} b^{2}}}{d}\) | \(201\) |
| risch | \(\frac {x}{b^{2}}-\frac {2 i \left (3 a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-2 i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 i b^{3}+i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-i a^{2} b -a^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2} d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{3} d}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2} a}-\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2} a}+\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{3}}\) | \(417\) |
1/d*(1/2*tan(1/2*d*x+1/2*c)/a^2-1/2/a^2/tan(1/2*d*x+1/2*c)-2/a^3*b*ln(tan( 1/2*d*x+1/2*c))+2/b^2*arctan(tan(1/2*d*x+1/2*c))-2/a^3/b^2*((-b^2*(a^2-b^2 )*tan(1/2*d*x+1/2*c)-b*a*(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)+(a^4+a^2*b^2-2*b^4)/(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))))
Time = 0.40 (sec) , antiderivative size = 675, normalized size of antiderivative = 4.38 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {2 \, a^{3} b d x \cos \left (d x + c\right )^{2} - 2 \, a^{3} b d x + 2 \, a^{2} b^{2} \cos \left (d x + c\right ) - {\left (a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} + 2 \, a b^{2}\right )} \sin \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 (b^{4} \cos \left (d x + c\right )^{2} - a b^{3} \sin \left (d x + c\right ) - b^{4}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - a b^{3} \sin \left (d x + c\right ) - b^{4}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (a^{4} d x + {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{3} b^{3} d \cos \left (d x + c\right )^{2} - a^{4} b^{2} d \sin \left (d x + c\right ) - a^{3} b^{3} d\right )}}, \frac {a^{3} b d x \cos \left (d x + c\right )^{2} - a^{3} b d x + a^{2} b^{2} \cos \left (d x + c\right ) - {\left (a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} + 2 \, a b^{2}\right )} \sin \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 (b^{4} \cos \left (d x + c\right )^{2} - a b^{3} \sin \left (d x + c\right ) - b^{4}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (b^{4} \cos \left (d x + c\right )^{2} - a b^{3} \sin \left (d x + c\right ) - b^{4}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a^{4} d x + {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{a^{3} b^{3} d \cos \left (d x + c\right )^{2} - a^{4} b^{2} d \sin \left (d x + c\right ) - a^{3} b^{3} d}\right ] \]
[1/2*(2*a^3*b*d*x*cos(d*x + c)^2 - 2*a^3*b*d*x + 2*a^2*b^2*cos(d*x + c) - (a^2*b + 2*b^3 - (a^2*b + 2*b^3)*cos(d*x + c)^2 + (a^3 + 2*a*b^2)*sin(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*cos(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*(b^4 *cos(d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(1/2*cos(d*x + c) + 1/2) - 2*(b^4*cos(d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^4*d*x + (a^3*b - 2*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3 *d*cos(d*x + c)^2 - a^4*b^2*d*sin(d*x + c) - a^3*b^3*d), (a^3*b*d*x*cos(d* x + c)^2 - a^3*b*d*x + a^2*b^2*cos(d*x + c) - (a^2*b + 2*b^3 - (a^2*b + 2* b^3)*cos(d*x + c)^2 + (a^3 + 2*a*b^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan (-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (b^4*cos(d*x + c) ^2 - a*b^3*sin(d*x + c) - b^4)*log(1/2*cos(d*x + c) + 1/2) - (b^4*cos(d*x + c)^2 - a*b^3*sin(d*x + c) - b^4)*log(-1/2*cos(d*x + c) + 1/2) - (a^4*d*x + (a^3*b - 2*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^4*b^2*d*sin(d*x + c) - a^3*b^3*d)]
\[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.42 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {6 \, {\left (d x + c\right )}}{b^{2}} - \frac {12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {12 \, {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3} b^{2}} + \frac {4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3} b}}{6 \, d} \]
1/6*(6*(d*x + c)/b^2 - 12*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 3*tan(1/2 *d*x + 1/2*c)/a^2 - 12*(a^4 + a^2*b^2 - 2*b^4)*(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)))/(sqr t(a^2 - b^2)*a^3*b^2) + (4*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*b*tan(1/2* d*x + 1/2*c)^2 - 4*b^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c ) - 14*a*b^2*tan(1/2*d*x + 1/2*c) - 3*a^2*b)/((a*tan(1/2*d*x + 1/2*c)^3 + 2*b*tan(1/2*d*x + 1/2*c)^2 + a*tan(1/2*d*x + 1/2*c))*a^3*b))/d
Time = 13.83 (sec) , antiderivative size = 6377, normalized size of antiderivative = 41.41 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]
(2*cos(c/2 + (d*x)/2)^2)/(b*d*(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/ 2)^2 + 2*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))) + sin(c/2 + (d*x)/2)^3/ (2*a*d*cos(c/2 + (d*x)/2)*(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))) - cos(c/2 + (d*x)/2)^3/(2*a *d*sin(c/2 + (d*x)/2)*(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2 *b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))) - (3*b*cos(c/2 + (d*x)/2)^2)/(a ^2*d*(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2*b*cos(c/2 + (d*x )/2)*sin(c/2 + (d*x)/2))) + (2*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a*d *(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2*b*cos(c/2 + (d*x)/2) *sin(c/2 + (d*x)/2))) + (b*sin(c/2 + (d*x)/2)^2)/(a^2*d*(a*cos(c/2 + (d*x) /2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2*b*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2) )) - (2*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^2) /(a^2*d*(a*cos(c/2 + (d*x)/2)^2 + a*sin(c/2 + (d*x)/2)^2 + 2*b*cos(c/2 + ( d*x)/2)*sin(c/2 + (d*x)/2))) + (atan((40*a^5*b^5*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 4*a^12*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 128*b^10*sin(c /2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 16*a^3*b^7*cos(c/2 + (d*x)/2)*(b^2 - a^2 )^(3/2) - 4*a^10*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) + 4*a^5*b^7*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 20*a^7*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^( 3/2) - 20*a^7*b^5*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 9*a^9*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*a^2*b^8*sin(c/2 + (d*x)/2)*(b^2 - a^2...