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

Optimal. Leaf size=123 \[ \frac {2 b \tanh ^{-1}(\cos (x))}{a^3}-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {2 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}}-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {2 b \tanh ^{-1}(\cos (x))}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Sin[x])^2,x]

[Out]

(2*b^2*(3*a^2 - 2*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(3/2)) + (2*b*ArcTanh[Cos[x]
])/a^3 - ((a^2 - 2*b^2)*Cot[x])/(a^2*(a^2 - b^2)) - (b^2*Cot[x])/(a*(a^2 - b^2)*(a + b*Sin[x]))

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])

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 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 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 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 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 3770

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

Rubi steps

\begin {align*} \int \frac {\csc ^2(x)}{(a+b \sin (x))^2} \, dx &=-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc ^2(x) \left (a^2-2 b^2-a b \sin (x)+b^2 \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc (x) \left (-2 b \left (a^2-b^2\right )+a b^2 \sin (x)\right )}{a+b \sin (x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {(2 b) \int \csc (x) \, dx}{a^3}+\frac {\left (b^2 \left (3 a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=\frac {2 b \tanh ^{-1}(\cos (x))}{a^3}-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\left (2 b^2 \left (3 a^2-2 b^2\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^3 \left (a^2-b^2\right )}\\ &=\frac {2 b \tanh ^{-1}(\cos (x))}{a^3}-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (4 b^2 \left (3 a^2-2 b^2\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^3 \left (a^2-b^2\right )}\\ &=\frac {2 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}}+\frac {2 b \tanh ^{-1}(\cos (x))}{a^3}-\frac {\left (a^2-2 b^2\right ) \cot (x)}{a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 127, normalized size = 1.03 \[ \frac {\frac {4 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {2 a b^3 \cos (x)}{(a-b) (a+b) (a+b \sin (x))}+a \tan \left (\frac {x}{2}\right )-a \cot \left (\frac {x}{2}\right )-4 b \log \left (\sin \left (\frac {x}{2}\right )\right )+4 b \log \left (\cos \left (\frac {x}{2}\right )\right )}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.84, size = 784, normalized size = 6.37 \[ \left [-\frac {2 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \relax (x) \sin \relax (x) + {\left (3 \, a^{2} b^{3} - 2 \, b^{5} - {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \relax (x)^{2} + {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} + 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \relax (x) - 2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5} - {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \relax (x)^{2} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \relax (x)\right )}}, -\frac {{\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \relax (x) \sin \relax (x) + {\left (3 \, a^{2} b^{3} - 2 \, b^{5} - {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \relax (x)^{2} + {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \relax (x)\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \relax (x) - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6} - {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \relax (x)^{2} + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5} - {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \relax (x)^{2} + {\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(x)*sin(x) + (3*a^2*b^3 - 2*b^5 - (3*a^2*b^3 - 2*b^5)*cos(x)^2 + (3*
a^3*b^2 - 2*a*b^4)*sin(x))*sqrt(-a^2 + b^2)*log(((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*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*(a^6 - 2*a^4*b^2 + a^2*
b^4)*cos(x) - 2*(a^4*b^2 - 2*a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(x)^2 + (a^5*b - 2*a^3*b^3 + a*b^5
)*sin(x))*log(1/2*cos(x) + 1/2) + 2*(a^4*b^2 - 2*a^2*b^4 + b^6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(x)^2 + (a^5*b
- 2*a^3*b^3 + a*b^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7*b - 2*a^5*b^3 + a^3*b^5 - (a^7*b - 2*a^5*b^3 + a^3*
b^5)*cos(x)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4)*sin(x)), -((a^5*b - 3*a^3*b^3 + 2*a*b^5)*cos(x)*sin(x) + (3*a^2*b^
3 - 2*b^5 - (3*a^2*b^3 - 2*b^5)*cos(x)^2 + (3*a^3*b^2 - 2*a*b^4)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b
)/(sqrt(a^2 - b^2)*cos(x))) + (a^6 - 2*a^4*b^2 + a^2*b^4)*cos(x) - (a^4*b^2 - 2*a^2*b^4 + b^6 - (a^4*b^2 - 2*a
^2*b^4 + b^6)*cos(x)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*sin(x))*log(1/2*cos(x) + 1/2) + (a^4*b^2 - 2*a^2*b^4 + b^
6 - (a^4*b^2 - 2*a^2*b^4 + b^6)*cos(x)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7*b
- 2*a^5*b^3 + a^3*b^5 - (a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(x)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4)*sin(x))]

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giac [A]  time = 0.15, size = 234, normalized size = 1.90 \[ \frac {2 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {4 \, a^{3} b \tan \left (\frac {1}{2} \, x\right )^{3} - 4 \, a b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 11 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 4 \, b^{4} \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, a^{3} b \tan \left (\frac {1}{2} \, x\right ) + 14 \, a b^{3} \tan \left (\frac {1}{2} \, x\right ) - 3 \, a^{4} + 3 \, a^{2} b^{2}}{6 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} + a \tan \left (\frac {1}{2} \, x\right )\right )}} - \frac {2 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{3}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 201, normalized size = 1.63 \[ \frac {\tan \left (\frac {x}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {x}{2}\right )}-\frac {2 b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}+\frac {2 b^{4} \tan \left (\frac {x}{2}\right )}{a^{3} \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}+\frac {2 b^{3}}{a^{2} \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}+\frac {6 b^{2} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {4 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{3} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^2*tan(1/2*x)-1/2/a^2/tan(1/2*x)-2/a^3*b*ln(tan(1/2*x))+2/a^3*b^4/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-
b^2)*tan(1/2*x)+2/a^2*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)+6/a*b^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*
tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))-4/a^3*b^4/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2/(a+b*sin(x))^2,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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.63, size = 1471, normalized size = 11.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a + b*sin(x))^2),x)

[Out]

tan(x/2)/(2*a^2) - (a - (tan(x/2)^2*(4*b^4 - a^4 + a^2*b^2))/(a*(a^2 - b^2)) + (2*b*tan(x/2)*(a^2 - 3*b^2))/(a
^2 - b^2))/(2*a^3*tan(x/2) + 2*a^3*tan(x/2)^3 + 4*a^2*b*tan(x/2)^2) - (2*b*log(tan(x/2)))/a^3 - (b^2*atan(((b^
2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(8*a*b^7 - 2*a^7*b - 20*a^3*b^5 + 14*a^5*b^3))/(a^
7 + a^3*b^4 - 2*a^5*b^2) - (2*(4*a^3*b^4 - 5*a^5*b^2))/(a^6 - a^4*b^2) + (b^2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a -
b)^3)^(1/2)*((2*(a^8*b - a^6*b^3))/(a^6 - a^4*b^2) - (2*tan(x/2)*(3*a^10 - 4*a^4*b^6 + 11*a^6*b^4 - 10*a^8*b^
2))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*
a^7*b^2) - (b^2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(4*a^3*b^4 - 5*a^5*b^2))/(a^6 - a^4*b^2) - (2
*tan(x/2)*(8*a*b^7 - 2*a^7*b - 20*a^3*b^5 + 14*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b^2*(3*a^2 - 2*b^2)*(-
(a + b)^3*(a - b)^3)^(1/2)*((2*(a^8*b - a^6*b^3))/(a^6 - a^4*b^2) - (2*tan(x/2)*(3*a^10 - 4*a^4*b^6 + 11*a^6*b
^4 - 10*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*1i)/(a^9 - a^3*b^6 +
3*a^5*b^4 - 3*a^7*b^2))/((4*tan(x/2)*(4*b^6 - 6*a^2*b^4))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (4*(4*b^5 - 6*a^2*b^3)
)/(a^6 - a^4*b^2) + (b^2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*tan(x/2)*(8*a*b^7 - 2*a^7*b - 20*a^3
*b^5 + 14*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) - (2*(4*a^3*b^4 - 5*a^5*b^2))/(a^6 - a^4*b^2) + (b^2*(3*a^2 -
2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(a^8*b - a^6*b^3))/(a^6 - a^4*b^2) - (2*tan(x/2)*(3*a^10 - 4*a^4*b^6 +
11*a^6*b^4 - 10*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))/(a^9 - a^3*
b^6 + 3*a^5*b^4 - 3*a^7*b^2) + (b^2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(4*a^3*b^4 - 5*a^5*b^2))/
(a^6 - a^4*b^2) - (2*tan(x/2)*(8*a*b^7 - 2*a^7*b - 20*a^3*b^5 + 14*a^5*b^3))/(a^7 + a^3*b^4 - 2*a^5*b^2) + (b^
2*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*(a^8*b - a^6*b^3))/(a^6 - a^4*b^2) - (2*tan(x/2)*(3*a^10 -
4*a^4*b^6 + 11*a^6*b^4 - 10*a^8*b^2))/(a^7 + a^3*b^4 - 2*a^5*b^2)))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))/
(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*(3*a^2 - 2*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(a^9 - a^3*b^6 + 3*
a^5*b^4 - 3*a^7*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\relax (x )}}{\left (a + b \sin {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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