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

Optimal. Leaf size=45 \[ -\frac {10 \cot (x)}{3 a^2}+\frac {2 \tanh ^{-1}(\cos (x))}{a^2}+\frac {2 \cot (x)}{a^2 (\sin (x)+1)}+\frac {\cot (x)}{3 (a \sin (x)+a)^2} \]

[Out]

2*arctanh(cos(x))/a^2-10/3*cot(x)/a^2+2*cot(x)/a^2/(1+sin(x))+1/3*cot(x)/(a+a*sin(x))^2

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2766, 2978, 2748, 3767, 8, 3770} \[ -\frac {10 \cot (x)}{3 a^2}+\frac {2 \tanh ^{-1}(\cos (x))}{a^2}+\frac {2 \cot (x)}{a^2 (\sin (x)+1)}+\frac {\cot (x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + a*Sin[x])^2,x]

[Out]

(2*ArcTanh[Cos[x]])/a^2 - (10*Cot[x])/(3*a^2) + (2*Cot[x])/(a^2*(1 + Sin[x])) + Cot[x]/(3*(a + a*Sin[x])^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2766

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

Rule 2978

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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+a \sin (x))^2} \, dx &=\frac {\cot (x)}{3 (a+a \sin (x))^2}+\frac {\int \frac {\csc ^2(x) (4 a-2 a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac {2 \cot (x)}{a^2 (1+\sin (x))}+\frac {\cot (x)}{3 (a+a \sin (x))^2}+\frac {\int \csc ^2(x) \left (10 a^2-6 a^2 \sin (x)\right ) \, dx}{3 a^4}\\ &=\frac {2 \cot (x)}{a^2 (1+\sin (x))}+\frac {\cot (x)}{3 (a+a \sin (x))^2}-\frac {2 \int \csc (x) \, dx}{a^2}+\frac {10 \int \csc ^2(x) \, dx}{3 a^2}\\ &=\frac {2 \tanh ^{-1}(\cos (x))}{a^2}+\frac {2 \cot (x)}{a^2 (1+\sin (x))}+\frac {\cot (x)}{3 (a+a \sin (x))^2}-\frac {10 \operatorname {Subst}(\int 1 \, dx,x,\cot (x))}{3 a^2}\\ &=\frac {2 \tanh ^{-1}(\cos (x))}{a^2}-\frac {10 \cot (x)}{3 a^2}+\frac {2 \cot (x)}{a^2 (1+\sin (x))}+\frac {\cot (x)}{3 (a+a \sin (x))^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.37, size = 166, normalized size = 3.69 \[ \frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (4 \sin \left (\frac {x}{2}\right )+28 \sin \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2-2 \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )+12 \log \left (\cos \left (\frac {x}{2}\right )\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3-12 \log \left (\sin \left (\frac {x}{2}\right )\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3+3 \tan \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3-3 \cot \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3\right )}{6 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[x/2] + Sin[x/2])*(4*Sin[x/2] - 2*(Cos[x/2] + Sin[x/2]) + 28*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 - 3*Cot[x/2
]*(Cos[x/2] + Sin[x/2])^3 + 12*Log[Cos[x/2]]*(Cos[x/2] + Sin[x/2])^3 - 12*Log[Sin[x/2]]*(Cos[x/2] + Sin[x/2])^
3 + 3*(Cos[x/2] + Sin[x/2])^3*Tan[x/2]))/(6*(a + a*Sin[x])^2)

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 168, normalized size = 3.73 \[ -\frac {10 \, \cos \relax (x)^{3} - 4 \, \cos \relax (x)^{2} - 3 \, {\left (\cos \relax (x)^{3} + 2 \, \cos \relax (x)^{2} + {\left (\cos \relax (x)^{2} - \cos \relax (x) - 2\right )} \sin \relax (x) - \cos \relax (x) - 2\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 3 \, {\left (\cos \relax (x)^{3} + 2 \, \cos \relax (x)^{2} + {\left (\cos \relax (x)^{2} - \cos \relax (x) - 2\right )} \sin \relax (x) - \cos \relax (x) - 2\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (10 \, \cos \relax (x)^{2} + 14 \, \cos \relax (x) + 1\right )} \sin \relax (x) - 13 \, \cos \relax (x) + 1}{3 \, {\left (a^{2} \cos \relax (x)^{3} + 2 \, a^{2} \cos \relax (x)^{2} - a^{2} \cos \relax (x) - 2 \, a^{2} + {\left (a^{2} \cos \relax (x)^{2} - a^{2} \cos \relax (x) - 2 \, a^{2}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(10*cos(x)^3 - 4*cos(x)^2 - 3*(cos(x)^3 + 2*cos(x)^2 + (cos(x)^2 - cos(x) - 2)*sin(x) - cos(x) - 2)*log(1
/2*cos(x) + 1/2) + 3*(cos(x)^3 + 2*cos(x)^2 + (cos(x)^2 - cos(x) - 2)*sin(x) - cos(x) - 2)*log(-1/2*cos(x) + 1
/2) - (10*cos(x)^2 + 14*cos(x) + 1)*sin(x) - 13*cos(x) + 1)/(a^2*cos(x)^3 + 2*a^2*cos(x)^2 - a^2*cos(x) - 2*a^
2 + (a^2*cos(x)^2 - a^2*cos(x) - 2*a^2)*sin(x))

________________________________________________________________________________________

giac [A]  time = 0.30, size = 69, normalized size = 1.53 \[ -\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a^{2}} + \frac {4 \, \tan \left (\frac {1}{2} \, x\right ) - 1}{2 \, a^{2} \tan \left (\frac {1}{2} \, x\right )} - \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, x\right ) + 8\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(abs(tan(1/2*x)))/a^2 + 1/2*tan(1/2*x)/a^2 + 1/2*(4*tan(1/2*x) - 1)/(a^2*tan(1/2*x)) - 2/3*(9*tan(1/2*x)
^2 + 15*tan(1/2*x) + 8)/(a^2*(tan(1/2*x) + 1)^3)

________________________________________________________________________________________

maple [A]  time = 0.13, size = 71, normalized size = 1.58 \[ \frac {\tan \left (\frac {x}{2}\right )}{2 a^{2}}-\frac {1}{2 a^{2} \tan \left (\frac {x}{2}\right )}-\frac {2 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {4}{3 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {6}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.92, size = 126, normalized size = 2.80 \[ -\frac {\frac {41 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {69 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {39 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + 3}{6 \, {\left (\frac {a^{2} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, a^{2} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {a^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}\right )}} - \frac {2 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} + \frac {\sin \relax (x)}{2 \, a^{2} {\left (\cos \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(41*sin(x)/(cos(x) + 1) + 69*sin(x)^2/(cos(x) + 1)^2 + 39*sin(x)^3/(cos(x) + 1)^3 + 3)/(a^2*sin(x)/(cos(x
) + 1) + 3*a^2*sin(x)^2/(cos(x) + 1)^2 + 3*a^2*sin(x)^3/(cos(x) + 1)^3 + a^2*sin(x)^4/(cos(x) + 1)^4) - 2*log(
sin(x)/(cos(x) + 1))/a^2 + 1/2*sin(x)/(a^2*(cos(x) + 1))

________________________________________________________________________________________

mupad [B]  time = 6.54, size = 91, normalized size = 2.02 \[ \frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a^2}-\frac {13\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+23\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {41\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+1}{2\,a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+6\,a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+6\,a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {2\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a + a*sin(x))^2),x)

[Out]

tan(x/2)/(2*a^2) - ((41*tan(x/2))/3 + 23*tan(x/2)^2 + 13*tan(x/2)^3 + 1)/(2*a^2*tan(x/2) + 6*a^2*tan(x/2)^2 +
6*a^2*tan(x/2)^3 + 2*a^2*tan(x/2)^4) - (2*log(tan(x/2)))/a^2

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{2}{\relax (x )}}{\sin ^{2}{\relax (x )} + 2 \sin {\relax (x )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**2/(sin(x)**2 + 2*sin(x) + 1), x)/a**2

________________________________________________________________________________________