3.459 \(\int \frac {\csc (x)}{a+b \cot (x)+c \csc (x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac {2 \tanh ^{-1}\left (\frac {a-(b-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3166, 3124, 618, 206} \[ -\frac {2 \tanh ^{-1}\left (\frac {a-(b-c) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + b*Cot[x] + c*Csc[x]),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 3166

Int[csc[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + csc[(d_.) + (e_.)*(x_)]*(b_.) + cot[(d_.) + (e_.)*(x_)]*(c_.))^(m_)
, x_Symbol] :> Int[1/(b + a*Sin[d + e*x] + c*Cos[d + e*x])^n, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + n, 0]
&& IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.98 \[ -\frac {2 \tanh ^{-1}\left (\frac {a+(c-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + b*Cot[x] + c*Csc[x]),x]

[Out]

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

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fricas [B]  time = 0.81, size = 349, normalized size = 6.84 \[ \left [\frac {\log \left (-\frac {a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} + {\left (a^{2} - b^{2}\right )} c^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} c \cos \relax (x) + {\left (a^{4} - b^{4} - 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \relax (x)^{2} + 2 \, {\left ({\left (a^{3} + a b^{2}\right )} c - {\left (a^{3} b + a b^{3} - 2 \, a b c^{2}\right )} \cos \relax (x)\right )} \sin \relax (x) - 2 \, {\left (2 \, a b c \cos \relax (x)^{2} - a b c + {\left (a^{3} + a b^{2}\right )} \cos \relax (x) - {\left (a^{2} b + b^{3} - {\left (a^{2} - b^{2}\right )} c \cos \relax (x)\right )} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2} - c^{2}}}{2 \, b c \cos \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (a b \cos \relax (x) + a c\right )} \sin \relax (x)}\right )}{2 \, \sqrt {a^{2} + b^{2} - c^{2}}}, \frac {\sqrt {-a^{2} - b^{2} + c^{2}} \arctan \left (\frac {{\left (b c \cos \relax (x) + a c \sin \relax (x) + a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}}{{\left (a^{3} + a b^{2} - a c^{2}\right )} \cos \relax (x) - {\left (a^{2} b + b^{3} - b c^{2}\right )} \sin \relax (x)}\right )}{a^{2} + b^{2} - c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 73, normalized size = 1.43 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, b - 2 \, c\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) - c \tan \left (\frac {1}{2} \, x\right ) - a}{\sqrt {-a^{2} - b^{2} + c^{2}}}\right )\right )}}{\sqrt {-a^{2} - b^{2} + c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cot(x)+c*csc(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 53, normalized size = 1.04 \[ -\frac {2 \arctan \left (\frac {2 \left (b -c \right ) \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}-b^{2}+c^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a+b*cot(x)+c*csc(x)),x)

[Out]

-2/(-a^2-b^2+c^2)^(1/2)*arctan(1/2*(2*(b-c)*tan(1/2*x)-2*a)/(-a^2-b^2+c^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*cot(x)+c*csc(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(c^2-b^2-a^2>0)', see `assume?`
for more details)Is c^2-b^2-a^2 positive or negative?

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mupad [B]  time = 2.79, size = 47, normalized size = 0.92 \[ -\frac {2\,\mathrm {atanh}\left (\frac {a-\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,b-2\,c\right )}{2}}{\sqrt {a^2+b^2-c^2}}\right )}{\sqrt {a^2+b^2-c^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + c/sin(x) + b*cot(x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cot(x)+c*csc(x)),x)

[Out]

Integral(csc(x)/(a + b*cot(x) + c*csc(x)), x)

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