Integrand size = 21, antiderivative size = 115 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \sqrt {a^2-b^2} d}+\frac {2 b \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))} \] Output:
-2*(a^2-2*b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b ^2)^(1/2)/d+2*b*arctanh(cos(d*x+c))/a^3/d-2*cot(d*x+c)/a^2/d+cot(d*x+c)/a/ d/(a+b*sin(d*x+c))
Time = 0.82 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {4 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a \cot \left (\frac {1}{2} (c+d x)\right )-4 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 a b \cos (c+d x)}{a+b \sin (c+d x)}-a \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^3 d} \] Input:
Integrate[Cot[c + d*x]^2/(a + b*Sin[c + d*x])^2,x]
Output:
-1/2*((4*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/S qrt[a^2 - b^2] + a*Cot[(c + d*x)/2] - 4*b*Log[Cos[(c + d*x)/2]] + 4*b*Log[ Sin[(c + d*x)/2]] + (2*a*b*Cos[c + d*x])/(a + b*Sin[c + d*x]) - a*Tan[(c + d*x)/2])/(a^3*d)
Time = 1.02 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.40, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3202, 3042, 3535, 3042, 3535, 25, 3042, 3480, 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.
| \(\displaystyle \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3202 |
| \(\displaystyle \int \frac {\left (1-\sin ^2(c+d x)\right ) \csc ^2(c+d x)}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1-\sin (c+d x)^2}{\sin (c+d x)^2 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) \left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \sin (c+d x)^2}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\frac {\int -\frac {\csc (c+d x) \left (2 b \left (a^2-b^2\right )+a \sin (c+d x) \left (a^2-b^2\right )\right )}{a+b \sin (c+d x)}dx}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc (c+d x) \left (2 b \left (a^2-b^2\right )+a \sin (c+d x) \left (a^2-b^2\right )\right )}{a+b \sin (c+d x)}dx}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {2 b \left (a^2-b^2\right )+a \sin (c+d x) \left (a^2-b^2\right )}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {-\frac {\frac {\left (a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}+\frac {2 b \left (a^2-b^2\right ) \int \csc (c+d x)dx}{a}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\left (a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}+\frac {2 b \left (a^2-b^2\right ) \int \csc (c+d x)dx}{a}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {-\frac {\frac {2 \left (a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}+\frac {2 b \left (a^2-b^2\right ) \int \csc (c+d x)dx}{a}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {-\frac {\frac {2 b \left (a^2-b^2\right ) \int \csc (c+d x)dx}{a}-\frac {4 \left (a^2-2 b^2\right ) \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {-\frac {\frac {2 b \left (a^2-b^2\right ) \int \csc (c+d x)dx}{a}+\frac {2 \left (a^2-2 b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {2 \left (a^2-2 b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}-\frac {2 b \left (a^2-b^2\right ) \text {arctanh}(\cos (c+d x))}{a d}}{a}-\frac {2 \left (a^2-b^2\right ) \cot (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {\cot (c+d x)}{a d (a+b \sin (c+d x))}\) |
Input:
Int[Cot[c + d*x]^2/(a + b*Sin[c + d*x])^2,x]
Output:
(-(((2*(a^2 - 2*b^2)*Sqrt[a^2 - b^2]*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/( 2*Sqrt[a^2 - b^2])])/(a*d) - (2*b*(a^2 - b^2)*ArcTanh[Cos[c + d*x]])/(a*d) )/a) - (2*(a^2 - b^2)*Cot[c + d*x])/(a*d))/(a*(a^2 - b^2)) + Cot[c + d*x]/ (a*d*(a + b*Sin[c + d*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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((1 - Sin[e + f*x]^2)/Sin[e + f*x]^ 2), x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.44 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.36
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) | \(156\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}-\frac {2 \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) | \(156\) |
| risch | \(-\frac {2 \left (-3 a \,{\mathrm e}^{i \left (d x +c \right )}+2 i {\mathrm e}^{2 i \left (d x +c \right )} b -2 i b +a \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right ) a^{2} d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{3} d}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}\) | \(417\) |
Input:
int(cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/a^2*tan(1/2*d*x+1/2*c)-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/a^3*((b^2*tan(1/2*d*x+1/2*c)+a*b)/(tan(1/2*d*x+1/2*c)^2* a+2*b*tan(1/2*d*x+1/2*c)+a)+(a^2-2*b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*ta n(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (110) = 220\).
Time = 0.25 (sec) , antiderivative size = 768, normalized size of antiderivative = 6.68 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
[1/2*(4*(a^3*b - a*b^3)*cos(d*x + c)*sin(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*(a^4 - a^2*b^2)*cos(d*x + c) - 2*(a^2*b^2 - b^4 - (a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*s in(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 2*(a^2*b^2 - b^4 - (a^2*b^2 - b ^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^5*b - a^3*b^3)*d*cos(d*x + c)^2 - (a^6 - a^4*b^2)*d*sin(d*x + c ) - (a^5*b - a^3*b^3)*d), (2*(a^3*b - a*b^3)*cos(d*x + c)*sin(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))) + (a^4 - a^2*b^2)*cos(d*x + c) - (a^2*b^2 - b^4 - (a^2*b^2 - b^4)* cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^2*b^2 - b^4 - (a^2*b^2 - b^4)*cos(d*x + c)^2 + (a^3*b - a*b^3)*sin(d *x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^5*b - a^3*b^3)*d*cos(d*x + c)^2 - (a^6 - a^4*b^2)*d*sin(d*x + c) - (a^5*b - a^3*b^3)*d)]
\[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cot ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(cot(d*x+c)**2/(a+b*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**2/(a + b*sin(c + d*x))**2, x)
Exception generated. \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
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 = 0.16 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.90 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 14 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{3}}}{6 \, d} \] Input:
integrate(cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/6*(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*(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^2 - 2*b^2)/(sqrt(a^2 - b^2)*a^3) - (4*a*b* tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 - 4*b^2*tan(1/2*d*x + 1/2*c)^2 - 14*a*b*tan(1/2*d*x + 1/2*c) - 3*a^2)/((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))/d
Time = 19.03 (sec) , antiderivative size = 1616, normalized size of antiderivative = 14.05 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(cot(c + d*x)^2/(a + b*sin(c + d*x))^2,x)
Output:
-(a^4*cos(c + d*x) - b^4/2 - b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2) ) + (a^2*b^2)/2 + (b^4*cos(2*c + 2*d*x))/2 - a^2*b^2*cos(c + d*x) - a*b^3* sin(2*c + 2*d*x) + a^3*b*sin(2*c + 2*d*x) + a^2*b^2*log(sin(c/2 + (d*x)/2) /cos(c/2 + (d*x)/2)) + b^4*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos( 2*c + 2*d*x) + b^3*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3 *sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c /2 + (d*x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3* a^3*b*cos(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2 )*2i - (a^2*b^2*cos(2*c + 2*d*x))/2 - a*b^3*sin(c + d*x) + a^3*b*sin(c + d *x) - a^2*b*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3*sin(c/ 2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - a^2)^( 1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c/2 + (d *x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3*a^3*b*c os(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*1i - a^3*sin(c + d*x)*atan((a^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - b^3*s in(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a*b^2*cos(c/2 + (d*x)/2)*(b^2 - a ^2)^(1/2)*4i + a^2*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i)/(a^4*sin(c/2 + (d*x)/2) + 8*b^4*sin(c/2 + (d*x)/2) + 4*a*b^3*cos(c/2 + (d*x)/2) - 3*a^ 3*b*cos(c/2 + (d*x)/2) - 8*a^2*b^2*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/...
Time = 0.17 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.49 \[ \int \frac {\cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} a^{2} b +4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{2} b^{3}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) a \,b^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-\cos \left (d x +c \right ) a^{4}+\cos \left (d x +c \right ) a^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a^{3} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a \,b^{3}}{\sin \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right ) a^{2} b -\sin \left (d x +c \right ) b^{3}+a^{3}-a \,b^{2}\right )} \] Input:
int(cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*s in(c + d*x)**2*a**2*b + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/ sqrt(a**2 - b**2))*sin(c + d*x)**2*b**3 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3 + 4*sqrt(a**2 - b** 2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b**2 - 2*cos(c + d*x)*sin(c + d*x)*a**3*b + 2*cos(c + d*x)*sin(c + d*x)*a*b**3 - cos(c + d*x)*a**4 + cos(c + d*x)*a**2*b**2 - 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b**2 + 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**4 - 2*lo g(tan((c + d*x)/2))*sin(c + d*x)*a**3*b + 2*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**3)/(sin(c + d*x)*a**3*d*(sin(c + d*x)*a**2*b - sin(c + d*x)*b**3 + a**3 - a*b**2))