3.1.30 \(\int \frac {\csc ^2(x)}{a+b \cos (x)} \, dx\) [30]

Optimal. Leaf size=67 \[ -\frac {2 b^2 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}+\frac {(b-a \cos (x)) \csc (x)}{a^2-b^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2775, 12, 2738, 211} \begin {gather*} \frac {\csc (x) (b-a \cos (x))}{a^2-b^2}-\frac {2 b^2 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Cos[x]),x]

[Out]

(-2*b^2*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/((a - b)^(3/2)*(a + b)^(3/2)) + ((b - a*Cos[x])*Csc[x])/(a
^2 - b^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 211

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

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 66, normalized size = 0.99 \begin {gather*} -\frac {2 b^2 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {(b-a \cos (x)) \csc (x)}{a^2-b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + b*Cos[x]),x]

[Out]

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

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Maple [A]
time = 0.13, size = 78, normalized size = 1.16

method	result	size
default	\(\frac {\tan \left (\frac {x}{2}\right )}{2 a -2 b}-\frac {1}{2 \left (a +b \right ) \tan \left (\frac {x}{2}\right )}-\frac {2 b^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\)	\(78\)
risch	\(-\frac {2 i \left (-b \,{\mathrm e}^{i x}+a \right )}{\left ({\mathrm e}^{2 i x}-1\right ) \left (a^{2}-b^{2}\right )}+\frac {b^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {b^{2} \ln \left ({\mathrm e}^{i x}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right )}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^2/(a+b*cos(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2/(a+b*cos(x)),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 de

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Fricas [A]
time = 0.40, size = 230, normalized size = 3.43 \begin {gather*} \left [\frac {\sqrt {-a^{2} + b^{2}} b^{2} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) + 2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )}, -\frac {\sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{2}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**2/(a + b*cos(x)), x)

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Giac [A]
time = 0.55, size = 91, normalized size = 1.36 \begin {gather*} \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, {\left (a - b\right )}} - \frac {1}{2 \, {\left (a + b\right )} \tan \left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*b^2/(a^2
- b^2)^(3/2) + 1/2*tan(1/2*x)/(a - b) - 1/2/((a + b)*tan(1/2*x))

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Mupad [B]
time = 0.47, size = 86, normalized size = 1.28 \begin {gather*} \frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a-2\,b}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,\sqrt {a-b}}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {a-b}{\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a+b\right )\,\left (2\,a-2\,b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a + b*cos(x))),x)

[Out]

tan(x/2)/(2*a - 2*b) - (2*b^2*atan((tan(x/2)*(a^2 - b^2))/((a + b)^(3/2)*(a - b)^(1/2))))/((a + b)^(3/2)*(a -
b)^(3/2)) - (a - b)/(tan(x/2)*(a + b)*(2*a - 2*b))

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