Integrand size = 27, antiderivative size = 114 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 b \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 d}+\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^3 d}+\frac {b \cot (c+d x)}{a^2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]
1/2*(a^2-2*b^2)*arctanh(cos(d*x+c))/a^3/d+b*cot(d*x+c)/a^2/d-1/2*cot(d*x+c )*csc(d*x+c)/a/d+2*b*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))*(a^2 -b^2)^(1/2)/a^3/d
Time = 1.19 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.59 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 b \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+4 a b \cot \left (\frac {1}{2} (c+d x)\right )-a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )+4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )-4 a b \tan \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \]
(16*b*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + 4 *a*b*Cot[(c + d*x)/2] - a^2*Csc[(c + d*x)/2]^2 + 4*a^2*Log[Cos[(c + d*x)/2 ]] - 8*b^2*Log[Cos[(c + d*x)/2]] - 4*a^2*Log[Sin[(c + d*x)/2]] + 8*b^2*Log [Sin[(c + d*x)/2]] + a^2*Sec[(c + d*x)/2]^2 - 4*a*b*Tan[(c + d*x)/2])/(8*a ^3*d)
Time = 0.91 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.16, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3368, 3042, 3535, 25, 3042, 3534, 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.
\(\displaystyle \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2}{\sin (c+d x)^3 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3368 |
\(\displaystyle \int \frac {\left (1-\sin ^2(c+d x)\right ) \csc ^3(c+d x)}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1-\sin (c+d x)^2}{\sin (c+d x)^3 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3535 |
\(\displaystyle \frac {\int -\frac {\csc ^2(c+d x) \left (-b \sin ^2(c+d x)+a \sin (c+d x)+2 b\right )}{a+b \sin (c+d x)}dx}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\csc ^2(c+d x) \left (-b \sin ^2(c+d x)+a \sin (c+d x)+2 b\right )}{a+b \sin (c+d x)}dx}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {-b \sin (c+d x)^2+a \sin (c+d x)+2 b}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle -\frac {\frac {\int \frac {\csc (c+d x) \left (a^2-b \sin (c+d x) a-2 b^2\right )}{a+b \sin (c+d x)}dx}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {a^2-b \sin (c+d x) a-2 b^2}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle -\frac {\frac {\frac {\left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {2 b \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {2 b \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {\frac {\frac {\left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {4 b \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {\frac {\frac {8 b \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {\left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\frac {\frac {\left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {4 b \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\frac {-\frac {4 b \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}-\frac {\left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{a d}}{a}-\frac {2 b \cot (c+d x)}{a d}}{2 a}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\) |
-1/2*(((-4*b*Sqrt[a^2 - b^2]*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a ^2 - b^2])])/(a*d) - ((a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a*d))/a - (2*b *Cot[c + d*x])/(a*d))/a - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)
3.13.91.3.1 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_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n }, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[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[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, 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[((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 = 0.36 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) | \(144\) |
default | \(\frac {\frac {\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{2}-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2}}-\frac {1}{8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (-2 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{3}}+\frac {b}{2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 b \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3}}}{d}\) | \(144\) |
risch | \(\frac {i \left (-i a \,{\mathrm e}^{3 i \left (d x +c \right )}-i a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{a^{3} d}-\frac {i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,a^{3}}+\frac {i \sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}\) | \(255\) |
1/d*(1/4/a^2*(1/2*tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2*c))-1/8/a/tan (1/2*d*x+1/2*c)^2+1/4/a^3*(-2*a^2+4*b^2)*ln(tan(1/2*d*x+1/2*c))+1/2*b/a^2/ tan(1/2*d*x+1/2*c)+2*b/a^3*(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 = 0.42 (sec) , antiderivative size = 472, normalized size of antiderivative = 4.14 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) - 2 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}}, -\frac {4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) + 4 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left ({\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}}\right ] \]
[-1/4*(4*a*b*cos(d*x + c)*sin(d*x + c) - 2*a^2*cos(d*x + c) - 2*(b*cos(d*x + c)^2 - b)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*s in(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)) - ((a^2 - 2*b^2)*cos(d*x + c)^2 - a^2 + 2*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((a^2 - 2*b^2)*cos(d*x + c)^2 - a^2 + 2*b^2)*log(-1/2*cos(d*x + c) + 1/ 2))/(a^3*d*cos(d*x + c)^2 - a^3*d), -1/4*(4*a*b*cos(d*x + c)*sin(d*x + c) - 2*a^2*cos(d*x + c) + 4*(b*cos(d*x + c)^2 - b)*sqrt(a^2 - b^2)*arctan(-(a *sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - ((a^2 - 2*b^2)*cos(d* x + c)^2 - a^2 + 2*b^2)*log(1/2*cos(d*x + c) + 1/2) + ((a^2 - 2*b^2)*cos(d *x + c)^2 - a^2 + 2*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(a^3*d*cos(d*x + c) ^2 - a^3*d)]
\[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 198, normalized size of antiderivative = 1.74 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {4 \, {\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac {16 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]
1/8*((a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 16*(a^2*b - 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)))/(sqrt(a^2 - b^2)*a^3) + (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*t an(1/2*d*x + 1/2*c)^2 + 4*a*b*tan(1/2*d*x + 1/2*c) - a^2)/(a^3*tan(1/2*d*x + 1/2*c)^2))/d
Time = 12.60 (sec) , antiderivative size = 790, normalized size of antiderivative = 6.93 \[ \int \frac {\cot ^2(c+d x) \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,\left (\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}-\frac {a^2\,\left (\frac {\cos \left (c+d\,x\right )}{2}+\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4}-\frac {\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{4}\right )}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}}-\frac {b^2\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2\,\left (\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}+\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,\left (\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}+\frac {b\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}-a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}-a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}}-\frac {b\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}-a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}-a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-5\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-12\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\frac {a^3\,d}{2}-\frac {a^3\,d\,\cos \left (2\,c+2\,d\,x\right )}{2}} \]
(b^2*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(2*((a^3*d)/2 - (a^3*d*co s(2*c + 2*d*x))/2)) - (a^2*(cos(c + d*x)/2 + log(sin(c/2 + (d*x)/2)/cos(c/ 2 + (d*x)/2))/4 - (log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2* d*x))/4))/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) - (b^2*log(sin(c/2 + (d *x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/(2*((a^3*d)/2 - (a^3*d*cos(2* c + 2*d*x))/2)) + (a*b*sin(2*c + 2*d*x))/(2*((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2)) + (b*atan((a^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + b^4*s in(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i - a^2*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^3*b*cos (c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*3i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin( c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) - 5*a^3*b^2*cos(c/2 + (d*x)/2) - 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(b^2 - a^2 )^(1/2)*1i)/((a^3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2) - (b*cos(2*c + 2*d*x) *atan((a^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + b^4*sin(c/2 + (d*x)/2 )*(b^2 - a^2)^(1/2)*8i - a^2*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*8i + a*b^3*cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*4i - a^3*b*cos(c/2 + (d*x)/2)* (b^2 - a^2)^(1/2)*3i)/(a^5*cos(c/2 + (d*x)/2) + 8*b^5*sin(c/2 + (d*x)/2) + 4*a*b^4*cos(c/2 + (d*x)/2) + 4*a^4*b*sin(c/2 + (d*x)/2) - 5*a^3*b^2*cos(c /2 + (d*x)/2) - 12*a^2*b^3*sin(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2)*1i)/((a^ 3*d)/2 - (a^3*d*cos(2*c + 2*d*x))/2)