\(\int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx\) [200]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 187 \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\frac {3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2}}+\frac {3 b \text {arctanh}(\cos (x))}{a^4}-\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \]

[Out]

3*b^2*(4*a^4-5*a^2*b^2+2*b^4)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(5/2)+3*b*arctanh(cos(x))
/a^4-1/2*(2*a^4-11*a^2*b^2+6*b^4)*cot(x)/a^3/(a^2-b^2)^2-1/2*b^2*cot(x)/a/(a^2-b^2)/(a+b*sin(x))^2-3/2*b^2*(2*
a^2-b^2)*cot(x)/a^2/(a^2-b^2)^2/(a+b*sin(x))

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2881, 3134, 3080, 3855, 2739, 632, 210} \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\frac {3 b \text {arctanh}(\cos (x))}{a^4}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2}}-\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2} \]

[In]

Int[Csc[x]^2/(a + b*Sin[x])^3,x]

[Out]

(3*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(5/2)) + (3*b*Ar
cTanh[Cos[x]])/a^4 - ((2*a^4 - 11*a^2*b^2 + 6*b^4)*Cot[x])/(2*a^3*(a^2 - b^2)^2) - (b^2*Cot[x])/(2*a*(a^2 - b^
2)*(a + b*Sin[x])^2) - (3*b^2*(2*a^2 - b^2)*Cot[x])/(2*a^2*(a^2 - b^2)^2*(a + b*Sin[x]))

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*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] + Dist[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*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3080

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(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]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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] + D
ist[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])))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac {\int \frac {\csc ^2(x) \left (2 a^2-3 b^2-2 a b \sin (x)+2 b^2 \sin ^2(x)\right )}{(a+b \sin (x))^2} \, dx}{2 a \left (a^2-b^2\right )} \\ & = -\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\int \frac {\csc ^2(x) \left (2 a^4-11 a^2 b^2+6 b^4-a b \left (4 a^2-b^2\right ) \sin (x)+3 b^2 \left (2 a^2-b^2\right ) \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\int \frac {\csc (x) \left (-6 b \left (a^2-b^2\right )^2+3 a b^2 \left (2 a^2-b^2\right ) \sin (x)\right )}{a+b \sin (x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {(3 b) \int \csc (x) \, dx}{a^4}+\frac {\left (3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2} \\ & = \frac {3 b \text {arctanh}(\cos (x))}{a^4}-\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac {\left (3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^4 \left (a^2-b^2\right )^2} \\ & = \frac {3 b \text {arctanh}(\cos (x))}{a^4}-\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac {\left (6 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^4 \left (a^2-b^2\right )^2} \\ & = \frac {3 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2}}+\frac {3 b \text {arctanh}(\cos (x))}{a^4}-\frac {\left (2 a^4-11 a^2 b^2+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac {b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac {3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.93 \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\frac {\frac {6 b^2 \left (4 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-a \cot \left (\frac {x}{2}\right )+6 b \log \left (\cos \left (\frac {x}{2}\right )\right )-6 b \log \left (\sin \left (\frac {x}{2}\right )\right )+\frac {a^2 b^3 \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+\frac {a b^3 \left (7 a^2-4 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}+a \tan \left (\frac {x}{2}\right )}{2 a^4} \]

[In]

Integrate[Csc[x]^2/(a + b*Sin[x])^3,x]

[Out]

((6*b^2*(4*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - a*Cot[x/2] +
 6*b*Log[Cos[x/2]] - 6*b*Log[Sin[x/2]] + (a^2*b^3*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x])^2) + (a*b^3*(7*a^2 -
 4*b^2)*Cos[x])/((a - b)^2*(a + b)^2*(a + b*Sin[x])) + a*Tan[x/2])/(2*a^4)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.57

method result size
default \(\frac {\tan \left (\frac {x}{2}\right )}{2 a^{3}}+\frac {4 b^{2} \left (\frac {\frac {3 a \,b^{2} \left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {b \left (8 a^{4}+11 a^{2} b^{2}-10 b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4 a^{4}-8 a^{2} b^{2}+4 b^{4}}+\frac {a \,b^{2} \left (23 a^{2}-14 b^{2}\right ) \tan \left (\frac {x}{2}\right )}{4 a^{4}-8 a^{2} b^{2}+4 b^{4}}+\frac {a^{2} b \left (8 a^{2}-5 b^{2}\right )}{4 a^{4}-8 a^{2} b^{2}+4 b^{4}}}{{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}^{2}}+\frac {3 \left (4 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}\right )}{a^{4}}-\frac {1}{2 a^{3} \tan \left (\frac {x}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{4}}\) \(294\)
risch \(-\frac {i \left (-24 i a \,b^{5} {\mathrm e}^{3 i x}+21 i a \,b^{5} {\mathrm e}^{i x}-38 i a^{3} {\mathrm e}^{i x} b^{3}+44 i a^{3} b^{3} {\mathrm e}^{3 i x}+8 i a^{5} b \,{\mathrm e}^{i x}+12 a^{4} b^{2} {\mathrm e}^{4 i x}+3 a^{2} b^{4} {\mathrm e}^{4 i x}-6 b^{6} {\mathrm e}^{4 i x}-8 i a^{5} b \,{\mathrm e}^{3 i x}+3 i a \,b^{5} {\mathrm e}^{5 i x}-6 i a^{3} b^{3} {\mathrm e}^{5 i x}+8 a^{6} {\mathrm e}^{2 i x}-26 a^{4} b^{2} {\mathrm e}^{2 i x}-6 a^{2} b^{4} {\mathrm e}^{2 i x}+12 b^{6} {\mathrm e}^{2 i x}-2 a^{4} b^{2}+11 a^{2} b^{4}-6 b^{6}\right )}{\left ({\mathrm e}^{2 i x}-1\right ) \left (-i b \,{\mathrm e}^{2 i x}+i b +2 a \,{\mathrm e}^{i x}\right )^{2} \left (a^{2}-b^{2}\right )^{2} a^{3}}+\frac {3 b \ln \left ({\mathrm e}^{i x}+1\right )}{a^{4}}+\frac {6 \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {15 b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}+\frac {3 b^{6} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a^{4}}-\frac {6 \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {15 b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a^{2}}-\frac {3 b^{6} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} a^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i x}-1\right )}{a^{4}}\) \(742\)

[In]

int(csc(x)^2/(a+b*sin(x))^3,x,method=_RETURNVERBOSE)

[Out]

1/2/a^3*tan(1/2*x)+4/a^4*b^2*((3/4*a*b^2*(3*a^2-2*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3+1/4*b*(8*a^4+11*a^2*b^
2-10*b^4)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2+1/4*a*b^2*(23*a^2-14*b^2)/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)+1/4*a^2*b*
(8*a^2-5*b^2)/(a^4-2*a^2*b^2+b^4))/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)^2+3/4*(4*a^4-5*a^2*b^2+2*b^4)/(a^4-2*a^2*
b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2)))-1/2/a^3/tan(1/2*x)-3/a^4*b*ln(tan(1
/2*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 686 vs. \(2 (175) = 350\).

Time = 0.96 (sec) , antiderivative size = 1436, normalized size of antiderivative = 7.68 \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(csc(x)^2/(a+b*sin(x))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(2*a^7*b^2 - 13*a^5*b^4 + 17*a^3*b^6 - 6*a*b^8)*cos(x)^3 - 2*(4*a^8*b - 20*a^6*b^3 + 25*a^4*b^5 - 9*a^
2*b^7)*cos(x)*sin(x) - 3*(8*a^5*b^3 - 10*a^3*b^5 + 4*a*b^7 - 2*(4*a^5*b^3 - 5*a^3*b^5 + 2*a*b^7)*cos(x)^2 + (4
*a^6*b^2 - a^4*b^4 - 3*a^2*b^6 + 2*b^8 - (4*a^4*b^4 - 5*a^2*b^6 + 2*b^8)*cos(x)^2)*sin(x))*sqrt(-a^2 + b^2)*lo
g(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*c
os(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*(2*a^9 - 4*a^7*b^2 - 7*a^5*b^4 + 15*a^3*b^6 - 6*a*b^8)*cos(x) + 6*(2*
a^7*b^2 - 6*a^5*b^4 + 6*a^3*b^6 - 2*a*b^8 - 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cos(x)^2 + (a^8*b - 2*
a^6*b^3 + 2*a^2*b^7 - b^9 - (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2) -
6*(2*a^7*b^2 - 6*a^5*b^4 + 6*a^3*b^6 - 2*a*b^8 - 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cos(x)^2 + (a^8*b
 - 2*a^6*b^3 + 2*a^2*b^7 - b^9 - (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(x)^2)*sin(x))*log(-1/2*cos(x) + 1
/2))/(2*a^11*b - 6*a^9*b^3 + 6*a^7*b^5 - 2*a^5*b^7 - 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*cos(x)^2 + (
a^12 - 2*a^10*b^2 + 2*a^6*b^6 - a^4*b^8 - (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*cos(x)^2)*sin(x)), 1/2*
((2*a^7*b^2 - 13*a^5*b^4 + 17*a^3*b^6 - 6*a*b^8)*cos(x)^3 - (4*a^8*b - 20*a^6*b^3 + 25*a^4*b^5 - 9*a^2*b^7)*co
s(x)*sin(x) - 3*(8*a^5*b^3 - 10*a^3*b^5 + 4*a*b^7 - 2*(4*a^5*b^3 - 5*a^3*b^5 + 2*a*b^7)*cos(x)^2 + (4*a^6*b^2
- a^4*b^4 - 3*a^2*b^6 + 2*b^8 - (4*a^4*b^4 - 5*a^2*b^6 + 2*b^8)*cos(x)^2)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*s
in(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - (2*a^9 - 4*a^7*b^2 - 7*a^5*b^4 + 15*a^3*b^6 - 6*a*b^8)*cos(x) + 3*(2*a^
7*b^2 - 6*a^5*b^4 + 6*a^3*b^6 - 2*a*b^8 - 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cos(x)^2 + (a^8*b - 2*a^
6*b^3 + 2*a^2*b^7 - b^9 - (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2) - 3*
(2*a^7*b^2 - 6*a^5*b^4 + 6*a^3*b^6 - 2*a*b^8 - 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*cos(x)^2 + (a^8*b -
 2*a^6*b^3 + 2*a^2*b^7 - b^9 - (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(x)^2)*sin(x))*log(-1/2*cos(x) + 1/2
))/(2*a^11*b - 6*a^9*b^3 + 6*a^7*b^5 - 2*a^5*b^7 - 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*cos(x)^2 + (a^
12 - 2*a^10*b^2 + 2*a^6*b^6 - a^4*b^8 - (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*cos(x)^2)*sin(x))]

Sympy [F]

\[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\int \frac {\csc ^{2}{\left (x \right )}}{\left (a + b \sin {\left (x \right )}\right )^{3}}\, dx \]

[In]

integrate(csc(x)**2/(a+b*sin(x))**3,x)

[Out]

Integral(csc(x)**2/(a + b*sin(x))**3, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(x)^2/(a+b*sin(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
 for more de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.50 \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\frac {3 \, {\left (4 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + 2 \, b^{6}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {9 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a b^{6} \tan \left (\frac {1}{2} \, x\right )^{3} + 8 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 11 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x\right )^{2} - 10 \, b^{7} \tan \left (\frac {1}{2} \, x\right )^{2} + 23 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x\right ) - 14 \, a b^{6} \tan \left (\frac {1}{2} \, x\right ) + 8 \, a^{4} b^{3} - 5 \, a^{2} b^{5}}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}^{2}} - \frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{4}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a^{3}} + \frac {6 \, b \tan \left (\frac {1}{2} \, x\right ) - a}{2 \, a^{4} \tan \left (\frac {1}{2} \, x\right )} \]

[In]

integrate(csc(x)^2/(a+b*sin(x))^3,x, algorithm="giac")

[Out]

3*(4*a^4*b^2 - 5*a^2*b^4 + 2*b^6)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)
))/((a^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(a^2 - b^2)) + (9*a^3*b^4*tan(1/2*x)^3 - 6*a*b^6*tan(1/2*x)^3 + 8*a^4*b^3*
tan(1/2*x)^2 + 11*a^2*b^5*tan(1/2*x)^2 - 10*b^7*tan(1/2*x)^2 + 23*a^3*b^4*tan(1/2*x) - 14*a*b^6*tan(1/2*x) + 8
*a^4*b^3 - 5*a^2*b^5)/((a^8 - 2*a^6*b^2 + a^4*b^4)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) - 3*b*log(abs(tan(
1/2*x)))/a^4 + 1/2*tan(1/2*x)/a^3 + 1/2*(6*b*tan(1/2*x) - a)/(a^4*tan(1/2*x))

Mupad [B] (verification not implemented)

Time = 8.70 (sec) , antiderivative size = 2295, normalized size of antiderivative = 12.27 \[ \int \frac {\csc ^2(x)}{(a+b \sin (x))^3} \, dx=\text {Too large to display} \]

[In]

int(1/(sin(x)^2*(a + b*sin(x))^3),x)

[Out]

tan(x/2)/(2*a^3) - (a^2 + (2*tan(x/2)*(7*a*b^5 + 2*a^5*b - 12*a^3*b^3))/(a^4 + b^4 - 2*a^2*b^2) + (tan(x/2)^4*
(a^6 + 12*b^6 - 17*a^2*b^4 - 2*a^4*b^2))/(a^4 + b^4 - 2*a^2*b^2) + (2*tan(x/2)^2*(a^6 + 16*b^6 - 26*a^2*b^4))/
(a^4 + b^4 - 2*a^2*b^2) + (2*tan(x/2)^3*(2*a^6*b + 10*b^7 - 9*a^2*b^5 - 12*a^4*b^3))/(a*(a^4 + b^4 - 2*a^2*b^2
)))/(tan(x/2)^3*(4*a^5 + 8*a^3*b^2) + 2*a^5*tan(x/2) + 2*a^5*tan(x/2)^5 + 8*a^4*b*tan(x/2)^2 + 8*a^4*b*tan(x/2
)^4) - (3*b*log(tan(x/2)))/a^4 - (b^2*atan(((b^2*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2)*((12
*a^4*b^6 - 27*a^6*b^4 + 18*a^8*b^2)/(a^10 + a^6*b^4 - 2*a^8*b^2) - (tan(x/2)*(6*a^12*b - 24*a^2*b^11 + 108*a^4
*b^9 - 192*a^6*b^7 + 162*a^8*b^5 - 60*a^10*b^3))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2) + (3*b^
2*((2*a^12*b + 2*a^8*b^5 - 4*a^10*b^3)/(a^10 + a^6*b^4 - 2*a^8*b^2) - (tan(x/2)*(6*a^16 - 8*a^6*b^10 + 38*a^8*
b^8 - 72*a^10*b^6 + 68*a^12*b^4 - 32*a^14*b^2))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2))*(-(a +
b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2))/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4
- 5*a^12*b^2)))*3i)/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2)) - (b^2*(-(a + b)
^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2)*((tan(x/2)*(6*a^12*b - 24*a^2*b^11 + 108*a^4*b^9 - 192*a^6*b^7
 + 162*a^8*b^5 - 60*a^10*b^3))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2) - (12*a^4*b^6 - 27*a^6*b^
4 + 18*a^8*b^2)/(a^10 + a^6*b^4 - 2*a^8*b^2) + (3*b^2*((2*a^12*b + 2*a^8*b^5 - 4*a^10*b^3)/(a^10 + a^6*b^4 - 2
*a^8*b^2) - (tan(x/2)*(6*a^16 - 8*a^6*b^10 + 38*a^8*b^8 - 72*a^10*b^6 + 68*a^12*b^4 - 32*a^14*b^2))/(a^13 + a^
5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2))*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2))/(2*(a^1
4 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2)))*3i)/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a
^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2)))/((2*(18*b^7 - 45*a^2*b^5 + 36*a^4*b^3))/(a^10 + a^6*b^4 - 2*a^8*b^2) + (2
*tan(x/2)*(18*b^10 - 81*a^2*b^8 + 126*a^4*b^6 - 72*a^6*b^4))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*
b^2) + (3*b^2*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2)*((12*a^4*b^6 - 27*a^6*b^4 + 18*a^8*b^2)
/(a^10 + a^6*b^4 - 2*a^8*b^2) - (tan(x/2)*(6*a^12*b - 24*a^2*b^11 + 108*a^4*b^9 - 192*a^6*b^7 + 162*a^8*b^5 -
60*a^10*b^3))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2) + (3*b^2*((2*a^12*b + 2*a^8*b^5 - 4*a^10*b
^3)/(a^10 + a^6*b^4 - 2*a^8*b^2) - (tan(x/2)*(6*a^16 - 8*a^6*b^10 + 38*a^8*b^8 - 72*a^10*b^6 + 68*a^12*b^4 - 3
2*a^14*b^2))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2))*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^
4 - 5*a^2*b^2))/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2))))/(2*(a^14 - a^4*b^1
0 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2)) + (3*b^2*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 -
 5*a^2*b^2)*((tan(x/2)*(6*a^12*b - 24*a^2*b^11 + 108*a^4*b^9 - 192*a^6*b^7 + 162*a^8*b^5 - 60*a^10*b^3))/(a^13
 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^11*b^2) - (12*a^4*b^6 - 27*a^6*b^4 + 18*a^8*b^2)/(a^10 + a^6*b^4 - 2*
a^8*b^2) + (3*b^2*((2*a^12*b + 2*a^8*b^5 - 4*a^10*b^3)/(a^10 + a^6*b^4 - 2*a^8*b^2) - (tan(x/2)*(6*a^16 - 8*a^
6*b^10 + 38*a^8*b^8 - 72*a^10*b^6 + 68*a^12*b^4 - 32*a^14*b^2))/(a^13 + a^5*b^8 - 4*a^7*b^6 + 6*a^9*b^4 - 4*a^
11*b^2))*(-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2))/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^
6 + 10*a^10*b^4 - 5*a^12*b^2))))/(2*(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^10*b^4 - 5*a^12*b^2))))*(
-(a + b)^5*(a - b)^5)^(1/2)*(4*a^4 + 2*b^4 - 5*a^2*b^2)*3i)/(a^14 - a^4*b^10 + 5*a^6*b^8 - 10*a^8*b^6 + 10*a^1
0*b^4 - 5*a^12*b^2)