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

Optimal. Leaf size=38 \[ \frac {4 \cos (x)}{3 a^2 (\sin (x)+1)}-\frac {\tanh ^{-1}(\cos (x))}{a^2}+\frac {\cos (x)}{3 (a \sin (x)+a)^2} \]

[Out]

-arctanh(cos(x))/a^2+4/3*cos(x)/a^2/(1+sin(x))+1/3*cos(x)/(a+a*sin(x))^2

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Rubi [A]  time = 0.09, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2766, 2978, 12, 3770} \[ \frac {4 \cos (x)}{3 a^2 (\sin (x)+1)}-\frac {\tanh ^{-1}(\cos (x))}{a^2}+\frac {\cos (x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + a*Sin[x])^2,x]

[Out]

-(ArcTanh[Cos[x]]/a^2) + (4*Cos[x])/(3*a^2*(1 + Sin[x])) + Cos[x]/(3*(a + a*Sin[x])^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 (x)}{(a+a \sin (x))^2} \, dx &=\frac {\cos (x)}{3 (a+a \sin (x))^2}+\frac {\int \frac {\csc (x) (3 a-a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac {4 \cos (x)}{3 a^2 (1+\sin (x))}+\frac {\cos (x)}{3 (a+a \sin (x))^2}+\frac {\int 3 a^2 \csc (x) \, dx}{3 a^4}\\ &=\frac {4 \cos (x)}{3 a^2 (1+\sin (x))}+\frac {\cos (x)}{3 (a+a \sin (x))^2}+\frac {\int \csc (x) \, dx}{a^2}\\ &=-\frac {\tanh ^{-1}(\cos (x))}{a^2}+\frac {4 \cos (x)}{3 a^2 (1+\sin (x))}+\frac {\cos (x)}{3 (a+a \sin (x))^2}\\ \end {align*}

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Mathematica [B]  time = 0.14, size = 129, normalized size = 3.39 \[ \frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {3 x}{2}\right ) \left (-3 \log \left (\sin \left (\frac {x}{2}\right )\right )+3 \log \left (\cos \left (\frac {x}{2}\right )\right )+8\right )+\cos \left (\frac {x}{2}\right ) \left (9 \log \left (\sin \left (\frac {x}{2}\right )\right )-9 \log \left (\cos \left (\frac {x}{2}\right )\right )-6\right )-6 \sin \left (\frac {x}{2}\right ) \left (-2 \log \left (\sin \left (\frac {x}{2}\right )\right )+2 \log \left (\cos \left (\frac {x}{2}\right )\right )+\cos (x) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )+3\right )\right )}{6 a^2 (\sin (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + a*Sin[x])^2,x]

[Out]

((Cos[x/2] + Sin[x/2])*(Cos[(3*x)/2]*(8 + 3*Log[Cos[x/2]] - 3*Log[Sin[x/2]]) + Cos[x/2]*(-6 - 9*Log[Cos[x/2]]
+ 9*Log[Sin[x/2]]) - 6*(3 + 2*Log[Cos[x/2]] + Cos[x]*(Log[Cos[x/2]] - Log[Sin[x/2]]) - 2*Log[Sin[x/2]])*Sin[x/
2]))/(6*a^2*(1 + Sin[x])^2)

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fricas [B]  time = 0.48, size = 117, normalized size = 3.08 \[ -\frac {8 \, \cos \relax (x)^{2} + 3 \, {\left (\cos \relax (x)^{2} - {\left (\cos \relax (x) + 2\right )} \sin \relax (x) - \cos \relax (x) - 2\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 3 \, {\left (\cos \relax (x)^{2} - {\left (\cos \relax (x) + 2\right )} \sin \relax (x) - \cos \relax (x) - 2\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 2 \, {\left (4 \, \cos \relax (x) - 1\right )} \sin \relax (x) + 10 \, \cos \relax (x) + 2}{6 \, {\left (a^{2} \cos \relax (x)^{2} - a^{2} \cos \relax (x) - 2 \, a^{2} - {\left (a^{2} \cos \relax (x) + 2 \, a^{2}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(8*cos(x)^2 + 3*(cos(x)^2 - (cos(x) + 2)*sin(x) - cos(x) - 2)*log(1/2*cos(x) + 1/2) - 3*(cos(x)^2 - (cos(
x) + 2)*sin(x) - cos(x) - 2)*log(-1/2*cos(x) + 1/2) + 2*(4*cos(x) - 1)*sin(x) + 10*cos(x) + 2)/(a^2*cos(x)^2 -
a^2*cos(x) - 2*a^2 - (a^2*cos(x) + 2*a^2)*sin(x))

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giac [A]  time = 1.70, size = 40, normalized size = 1.05 \[ \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (6 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, x\right ) + 5\right )}}{3 \, a^{2} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(tan(1/2*x)))/a^2 + 2/3*(6*tan(1/2*x)^2 + 9*tan(1/2*x) + 5)/(a^2*(tan(1/2*x) + 1)^3)

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maple [A]  time = 0.12, size = 50, normalized size = 1.32 \[ \frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}+\frac {4}{3 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {4}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/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+4/a^2/(tan(1/2*x)+1)

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maxima [B]  time = 1.00, size = 89, normalized size = 2.34 \[ \frac {2 \, {\left (\frac {9 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {6 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 5\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, a^{2} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {a^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}} + \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(9*sin(x)/(cos(x) + 1) + 6*sin(x)^2/(cos(x) + 1)^2 + 5)/(a^2 + 3*a^2*sin(x)/(cos(x) + 1) + 3*a^2*sin(x)^2/
(cos(x) + 1)^2 + a^2*sin(x)^3/(cos(x) + 1)^3) + log(sin(x)/(cos(x) + 1))/a^2

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mupad [B]  time = 6.48, size = 38, normalized size = 1.00 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2}+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {10}{3}}{a^2\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + a*sin(x))^2),x)

[Out]

log(tan(x/2))/a^2 + (6*tan(x/2) + 4*tan(x/2)^2 + 10/3)/(a^2*(tan(x/2) + 1)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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