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

Optimal. Leaf size=62 \[ \frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \sqrt{a^2-b^2}}+\frac{b \tanh ^{-1}(\cos (x))}{a^2}-\frac{\cot (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.111467, antiderivative size = 62, normalized size of antiderivative = 1., 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, 12, 2747, 3770, 2660, 618, 204} \[ \frac{2 b^2 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \sqrt{a^2-b^2}}+\frac{b \tanh ^{-1}(\cos (x))}{a^2}-\frac{\cot (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Sin[x]),x]

[Out]

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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)} \, dx &=-\frac{\cot (x)}{a}-\frac{\int \frac{b \csc (x)}{a+b \sin (x)} \, dx}{a}\\ &=-\frac{\cot (x)}{a}-\frac{b \int \frac{\csc (x)}{a+b \sin (x)} \, dx}{a}\\ &=-\frac{\cot (x)}{a}-\frac{b \int \csc (x) \, dx}{a^2}+\frac{b^2 \int \frac{1}{a+b \sin (x)} \, dx}{a^2}\\ &=\frac{b \tanh ^{-1}(\cos (x))}{a^2}-\frac{\cot (x)}{a}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^2}\\ &=\frac{b \tanh ^{-1}(\cos (x))}{a^2}-\frac{\cot (x)}{a}-\frac{\left (4 b^2\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^2}\\ &=\frac{2 b^2 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \sqrt{a^2-b^2}}+\frac{b \tanh ^{-1}(\cos (x))}{a^2}-\frac{\cot (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.238603, size = 91, normalized size = 1.47 \[ \frac{\csc \left (\frac{x}{2}\right ) \sec \left (\frac{x}{2}\right ) \left (\frac{2 b^2 \sin (x) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-a \cos (x)+b \sin (x) \left (\log \left (\cos \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )\right )\right )\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + b*Sin[x]),x]

[Out]

(Csc[x/2]*Sec[x/2]*(-(a*Cos[x]) + (2*b^2*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]]*Sin[x])/Sqrt[a^2 - b^2] + b*
(Log[Cos[x/2]] - Log[Sin[x/2]])*Sin[x]))/(2*a^2)

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Maple [A]  time = 0.043, size = 77, normalized size = 1.2 \begin{align*}{\frac{1}{2\,a}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{2\,a} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{b}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a*tan(1/2*x)+2/a^2*b^2/(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/tan(1/2*x)-1
/a^2*b*ln(tan(1/2*x))

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.18704, size = 752, normalized size = 12.13 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} b^{2} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) \sin \left (x\right ) -{\left (a^{2} b - b^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) +{\left (a^{2} b - b^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}, -\frac{2 \, \sqrt{a^{2} - b^{2}} b^{2} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) \sin \left (x\right ) -{\left (a^{2} b - b^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) +{\left (a^{2} b - b^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^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*co
s(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2))*sin(x) - (a^2*b - b^3)*log(1/2*cos(x) + 1/2
)*sin(x) + (a^2*b - b^3)*log(-1/2*cos(x) + 1/2)*sin(x) + 2*(a^3 - a*b^2)*cos(x))/((a^4 - a^2*b^2)*sin(x)), -1/
2*(2*sqrt(a^2 - b^2)*b^2*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x)))*sin(x) - (a^2*b - b^3)*log(1/2*cos(x
) + 1/2)*sin(x) + (a^2*b - b^3)*log(-1/2*cos(x) + 1/2)*sin(x) + 2*(a^3 - a*b^2)*cos(x))/((a^4 - a^2*b^2)*sin(x
))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**2/(a + b*sin(x)), x)

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Giac [A]  time = 1.2401, size = 132, normalized size = 2.13 \begin{align*} \frac{2 \,{\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 )} b^{2}}{\sqrt{a^{2} - b^{2}} a^{2}} - \frac{b \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac{\tan \left (\frac{1}{2} \, x\right )}{2 \, a} + \frac{2 \, b \tan \left (\frac{1}{2} \, x\right ) - a}{2 \, a^{2} \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)),x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*b^2/(sqrt(a^2 - b^2)*a^2) - b
*log(abs(tan(1/2*x)))/a^2 + 1/2*tan(1/2*x)/a + 1/2*(2*b*tan(1/2*x) - a)/(a^2*tan(1/2*x))