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

Optimal. Leaf size=187 \[ \frac{3 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2}}-\frac{\left (-11 a^2 b^2+2 a^4+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac{3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{3 b \tanh ^{-1}(\cos (x))}{a^4} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.639396, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2802, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{3 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{5/2}}-\frac{\left (-11 a^2 b^2+2 a^4+6 b^4\right ) \cot (x)}{2 a^3 \left (a^2-b^2\right )^2}-\frac{3 b^2 \left (2 a^2-b^2\right ) \cot (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{b^2 \cot (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{3 b \tanh ^{-1}(\cos (x))}{a^4} \]

Antiderivative was successfully verified.

[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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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 3770

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

Rule 2660

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 618

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc ^2(x)}{(a+b \sin (x))^3} \, dx &=-\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 \tanh ^{-1}(\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 ) \operatorname{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 \tanh ^{-1}(\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 ) \operatorname{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 ) \tan ^{-1}\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 \tanh ^{-1}(\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]  time = 1.24753, size = 174, normalized size = 0.93 \[ \frac{\frac{6 b^2 \left (-5 a^2 b^2+4 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^3 \left (7 a^2-4 b^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}+\frac{a^2 b^3 \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}+a \tan \left (\frac{x}{2}\right )-a \cot \left (\frac{x}{2}\right )-6 b \log \left (\sin \left (\frac{x}{2}\right )\right )+6 b \log \left (\cos \left (\frac{x}{2}\right )\right )}{2 a^4} \]

Antiderivative was successfully verified.

[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 [B]  time = 0.085, size = 641, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^3*tan(1/2*x)+9/a*b^4/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-6/a^3*b^6/(tan
(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3+8*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^
4-2*a^2*b^2+b^4)*tan(1/2*x)^2+11/a^2*b^5/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-
10/a^4*b^7/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2+23/a*b^4/(tan(1/2*x)^2*a+2*tan
(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-14/a^3*b^6/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^
4)*tan(1/2*x)+8*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)-5/a^2*b^5/(tan(1/2*x)^2*a+2*tan(1/
2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)+12/(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))*b^2-15/a^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))+
6/a^4*b^6/(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))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 8.01966, size = 3114, normalized size = 16.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (x \right )}}{\left (a + b \sin{\left (x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.62451, size = 378, normalized size = 2.02 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))