3.181 \(\int \frac {\csc (x)}{a+b \sin (x)} \, dx\)

Optimal. Leaf size=53 \[ -\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {\tanh ^{-1}(\cos (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {2747, 3770, 2660, 618, 204} \[ -\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {\tanh ^{-1}(\cos (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + b*Sin[x]),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 62, normalized size = 1.17 \[ \frac {-\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\log \left (\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + b*Sin[x]),x]

[Out]

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

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fricas [A]  time = 0.58, size = 239, normalized size = 4.51 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} b \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} - 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, \frac {2 \, \sqrt {a^{2} - b^{2}} b \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 63, normalized size = 1.19 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(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/(sqrt(a^2 - b^2)*a) + log(
abs(tan(1/2*x)))/a

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maple [A]  time = 0.10, size = 53, normalized size = 1.00 \[ \frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*ln(tan(1/2*x))-2*b/a/(a^2-b^2)^(1/2)*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)/(a+b*sin(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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 6.86, size = 122, normalized size = 2.30 \[ \frac {\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{a}+\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {b^2-a^2}\,\left (-1{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,a^2+2{}\mathrm {i}\,\cos \left (\frac {x}{2}\right )\,a\,b+4{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,b^2\right )}{1{}\mathrm {i}\,\cos \left (\frac {x}{2}\right )\,a^3+3{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,a^2\,b-2{}\mathrm {i}\,\cos \left (\frac {x}{2}\right )\,a\,b^2-4{}\mathrm {i}\,\sin \left (\frac {x}{2}\right )\,b^3}\right )}{a\,\sqrt {b^2-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + b*sin(x))),x)

[Out]

log(sin(x/2)/cos(x/2))/a + (2*b*atanh(((b^2 - a^2)^(1/2)*(b^2*sin(x/2)*4i - a^2*sin(x/2)*1i + a*b*cos(x/2)*2i)
)/(a^3*cos(x/2)*1i - b^3*sin(x/2)*4i - a*b^2*cos(x/2)*2i + a^2*b*sin(x/2)*3i)))/(a*(b^2 - a^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)/(a + b*sin(x)), x)

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