Integrand size = 29, antiderivative size = 153 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b^2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 d}-\frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 a^4 d}+\frac {\left (a^2-3 b^2\right ) \cot (c+d x)}{3 a^3 d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d} \] Output:
-2*b^2*(a^2-b^2)^(1/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^ 4/d-1/2*b*(a^2-2*b^2)*arctanh(cos(d*x+c))/a^4/d+1/3*(a^2-3*b^2)*cot(d*x+c) /a^3/d+1/2*b*cot(d*x+c)*csc(d*x+c)/a^2/d-1/3*cot(d*x+c)*csc(d*x+c)^2/a/d
Leaf count is larger than twice the leaf count of optimal. \(351\) vs. \(2(153)=306\).
Time = 7.29 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.29 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 b^2 \sqrt {a^2-b^2} \arctan \left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (b \cos \left (\frac {1}{2} (c+d x)\right )+a \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 d}+\frac {\left (a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right ) \csc \left (\frac {1}{2} (c+d x)\right )}{6 a^3 d}+\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d}+\frac {\left (-a^2 b+2 b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\left (a^2 b-2 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (-a^2 \sin \left (\frac {1}{2} (c+d x)\right )+3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 a d} \] Input:
Integrate[(Cot[c + d*x]^2*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(-2*b^2*Sqrt[a^2 - b^2]*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*S in[(c + d*x)/2]))/Sqrt[a^2 - b^2]])/(a^4*d) + ((a^2*Cos[(c + d*x)/2] - 3*b ^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^3*d) + (b*Csc[(c + d*x)/2]^2)/ (8*a^2*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a*d) + ((-(a^2*b) + 2*b^3)*Log[Cos[(c + d*x)/2]])/(2*a^4*d) + ((a^2*b - 2*b^3)*Log[Sin[(c + d* x)/2]])/(2*a^4*d) - (b*Sec[(c + d*x)/2]^2)/(8*a^2*d) + (Sec[(c + d*x)/2]*( -(a^2*Sin[(c + d*x)/2]) + 3*b^2*Sin[(c + d*x)/2]))/(6*a^3*d) + (Sec[(c + d *x)/2]^2*Tan[(c + d*x)/2])/(24*a*d)
Time = 1.33 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.16, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3368, 3042, 3535, 25, 3042, 3534, 3042, 3534, 27, 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.
| \(\displaystyle \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2}{\sin (c+d x)^4 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3368 |
| \(\displaystyle \int \frac {\left (1-\sin ^2(c+d x)\right ) \csc ^4(c+d x)}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1-\sin (c+d x)^2}{\sin (c+d x)^4 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {\int -\frac {\csc ^3(c+d x) \left (-2 b \sin ^2(c+d x)+a \sin (c+d x)+3 b\right )}{a+b \sin (c+d x)}dx}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\csc ^3(c+d x) \left (-2 b \sin ^2(c+d x)+a \sin (c+d x)+3 b\right )}{a+b \sin (c+d x)}dx}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-2 b \sin (c+d x)^2+a \sin (c+d x)+3 b}{\sin (c+d x)^3 (a+b \sin (c+d x))}dx}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc ^2(c+d x) \left (3 b^2 \sin ^2(c+d x)-a b \sin (c+d x)+2 \left (a^2-3 b^2\right )\right )}{a+b \sin (c+d x)}dx}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {3 b^2 \sin (c+d x)^2-a b \sin (c+d x)+2 \left (a^2-3 b^2\right )}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle -\frac {\frac {\frac {\int -\frac {3 \csc (c+d x) \left (b \left (a^2-2 b^2\right )-a b^2 \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {-\frac {3 \int \frac {\csc (c+d x) \left (b \left (a^2-2 b^2\right )-a b^2 \sin (c+d x)\right )}{a+b \sin (c+d x)}dx}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {-\frac {3 \int \frac {b \left (a^2-2 b^2\right )-a b^2 \sin (c+d x)}{\sin (c+d x) (a+b \sin (c+d x))}dx}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {2 b^2 \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {2 b^2 \left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{a}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {4 b^2 \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (\frac {8 b^2 \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}+\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (\frac {b \left (a^2-2 b^2\right ) \int \csc (c+d x)dx}{a}-\frac {4 b^2 \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {-\frac {3 \left (-\frac {4 b^2 \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a d}-\frac {b \left (a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{a d}\right )}{a}-\frac {2 \left (a^2-3 b^2\right ) \cot (c+d x)}{a d}}{2 a}-\frac {3 b \cot (c+d x) \csc (c+d x)}{2 a d}}{3 a}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 a d}\) |
Input:
Int[(Cot[c + d*x]^2*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
-1/3*(Cot[c + d*x]*Csc[c + d*x]^2)/(a*d) - (((-3*((-4*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) - (b*(a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(a*d)))/a - (2*(a^2 - 3*b^2)*Cot[c + d*x]) /(a*d))/(2*a) - (3*b*Cot[c + d*x]*Csc[c + d*x])/(2*a*d))/(3*a)
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_)]^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 = 0.63 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.34
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3}}-\frac {2 b^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{4}}}{d}\) | \(205\) |
| default | \(\frac {\frac {\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}+4 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3}}-\frac {2 b^{2} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}-\frac {1}{24 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {-a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {b \left (a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{4}}}{d}\) | \(205\) |
| risch | \(-\frac {-6 i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 i b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-12 i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 i a^{2}+6 i b^{2}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 a^{2} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{4}}\) | \(301\) |
Input:
int(cot(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/8/a^3*(1/3*a^2*tan(1/2*d*x+1/2*c)^3-a*b*tan(1/2*d*x+1/2*c)^2-tan(1/ 2*d*x+1/2*c)*a^2+4*b^2*tan(1/2*d*x+1/2*c))-2*b^2*(a^2-b^2)^(1/2)/a^4*arcta n(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/24/a/tan(1/2*d*x+1/2 *c)^3-1/8*(-a^2+4*b^2)/a^3/tan(1/2*d*x+1/2*c)+1/8/a^2*b/tan(1/2*d*x+1/2*c) ^2+1/2/a^4*b*(a^2-2*b^2)*ln(tan(1/2*d*x+1/2*c)))
Time = 0.19 (sec) , antiderivative size = 591, normalized size of antiderivative = 3.86 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[-1/12*(6*a^2*b*cos(d*x + c)*sin(d*x + c) - 12*a*b^2*cos(d*x + c) - 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 - 6*(b^2*cos(d*x + c)^2 - b^2)*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))*sin(d*x + c) - 3*(a^2*b - 2*b^ 3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c))/((a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x + c)), -1 /12*(6*a^2*b*cos(d*x + c)*sin(d*x + c) - 12*a*b^2*cos(d*x + c) - 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 - 12*(b^2*cos(d*x + c)^2 - b^2)*sqrt(a^2 - b^2)*ar ctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) - 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(a^2*b - 2*b^3 - (a^2*b - 2*b^3)*cos(d*x + c)^2)*log (-1/2*cos(d*x + c) + 1/2)*sin(d*x + c))/((a^4*d*cos(d*x + c)^2 - a^4*d)*si n(d*x + c))]
\[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cot ^{2}{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**2*csc(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**2*csc(c + d*x)**2/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c)),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 = 0.21 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.76 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {12 \, {\left (a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{2} b^{2} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {22 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/24*((a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*t an(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/a^3 + 12*(a^2*b - 2*b^3 )*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 48*(a^2*b^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)))/(sqrt(a^2 - b^2)*a^4) - (22*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 44*b^3*t an(1/2*d*x + 1/2*c)^3 - 3*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/2*d* x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(a^4*tan(1/2*d*x + 1/2* c)^3))/d
Time = 21.34 (sec) , antiderivative size = 749, normalized size of antiderivative = 4.90 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
int(cot(c + d*x)^2/(sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
Output:
(a^3*(cos(c + d*x)/8 + cos(3*c + 3*d*x)/24) - a^2*((b*sin(2*c + 2*d*x))/8 - (b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/16 + (3* b*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/16) + a*((b^2*c os(c + d*x))/8 - (b^2*cos(3*c + 3*d*x))/8) + (3*b^3*sin(c + d*x)*log(sin(c /2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 - (b^3*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/8 + (b^2*sin(c + 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)*3i)/4 - (b^2*sin(3*c + 3*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)/4)/(( a^4*d*sin(3*c + 3*d*x))/8 - (3*a^4*d*sin(c + d*x))/8)
Time = 0.18 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.19 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-12 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -2 \cos \left (d x +c \right ) a^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{2} b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} b^{3}}{6 \sin \left (d x +c \right )^{3} a^{4} d} \] Input:
int(cot(d*x+c)^2*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* sin(c + d*x)**3*b**2 + 2*cos(c + d*x)*sin(c + d*x)**2*a**3 - 6*cos(c + d*x )*sin(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*sin(c + d*x)*a**2*b - 2*cos(c + d*x)*a**3 + 3*log(tan((c + d*x)/2))*sin(c + d*x)**3*a**2*b - 6*log(tan((c + d*x)/2))*sin(c + d*x)**3*b**3)/(6*sin(c + d*x)**3*a**4*d)