3.458 \(\int \frac{1}{a+b \cot (x)+c \csc (x)} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 a c \tanh ^{-1}\left (\frac{a-(b-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{b \log (a \sin (x)+b \cos (x)+c)}{a^2+b^2}+\frac{a x}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.102696, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3160, 3137, 3124, 618, 206} \[ \frac{2 a c \tanh ^{-1}\left (\frac{a-(b-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (a^2+b^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{b \log (a \sin (x)+b \cos (x)+c)}{a^2+b^2}+\frac{a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[x] + c*Csc[x])^(-1),x]

[Out]

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

Rule 3160

Int[((a_.) + csc[(d_.) + (e_.)*(x_)]*(b_.) + cot[(d_.) + (e_.)*(x_)]*(c_.))^(-1), x_Symbol] :> Int[Sin[d + e*x
]/(b + a*Sin[d + e*x] + c*Cos[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rule 3137

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)]), x_Symbol] :> Simp[(c*C*(d + e*x))/(e*(b^2 + c^2)), x] + (Dist[(A*(b^2 + c^2) - a*c*C)/(b^2 + c^2), Int[
1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] - Simp[(b*C*Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2
+ c^2)), x]) /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*c*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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.224312, size = 80, normalized size = 0.82 \[ \frac{\frac{2 a c \tanh ^{-1}\left (\frac{a+(c-b) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}}-b \log (a \sin (x)+b \cos (x)+c)+a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cot[x] + c*Csc[x])^(-1),x]

[Out]

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

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Maple [B]  time = 0.051, size = 446, normalized size = 4.6 \begin{align*} -2\,{\frac{\ln \left ( b \left ( \tan \left ( x/2 \right ) \right ) ^{2}- \left ( \tan \left ( x/2 \right ) \right ) ^{2}c-2\,a\tan \left ( x/2 \right ) -b-c \right ){b}^{2}}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \left ( b-c \right ) }}+2\,{\frac{\ln \left ( b \left ( \tan \left ( x/2 \right ) \right ) ^{2}- \left ( \tan \left ( x/2 \right ) \right ) ^{2}c-2\,a\tan \left ( x/2 \right ) -b-c \right ) cb}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \left ( b-c \right ) }}+4\,{\frac{ab}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( b-c \right ) \tan \left ( x/2 \right ) -2\,a}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }+4\,{\frac{ac}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( b-c \right ) \tan \left ( x/2 \right ) -2\,a}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }-4\,{\frac{a{b}^{2}}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( b-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( b-c \right ) \tan \left ( x/2 \right ) -2\,a}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }+4\,{\frac{abc}{ \left ( 2\,{a}^{2}+2\,{b}^{2} \right ) \sqrt{-{a}^{2}-{b}^{2}+{c}^{2}} \left ( b-c \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( b-c \right ) \tan \left ( x/2 \right ) -2\,a}{\sqrt{-{a}^{2}-{b}^{2}+{c}^{2}}}} \right ) }+2\,{\frac{b\ln \left ( 1+ \left ( \tan \left ( x/2 \right ) \right ) ^{2} \right ) }{2\,{a}^{2}+2\,{b}^{2}}}+4\,{\frac{a\arctan \left ( \tan \left ( x/2 \right ) \right ) }{2\,{a}^{2}+2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(2*a^2+2*b^2)/(b-c)*ln(b*tan(1/2*x)^2-tan(1/2*x)^2*c-2*a*tan(1/2*x)-b-c)*b^2+2/(2*a^2+2*b^2)/(b-c)*ln(b*tan
(1/2*x)^2-tan(1/2*x)^2*c-2*a*tan(1/2*x)-b-c)*c*b+4/(2*a^2+2*b^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))*a*b+4/(2*a^2+2*b^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))*a*c-4/(2*a^2+2*b^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))*a/(b-c)*b^2+4/(2*a^2+2*b^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))*a/(b-c)*c*b+2/(2*a^2+2*b^2)*b*ln(1+tan(1/2*x)^2)+4/(2*a^2+2*b^2)*a*arctan(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(1/(a+b*cot(x)+c*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.17959, size = 1277, normalized size = 13.03 \begin{align*} \left [\frac{\sqrt{a^{2} + b^{2} - c^{2}} a c \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 \left (x\right ) +{\left (a^{4} - b^{4} - 2 \,{\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \left (x\right )^{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 \left (x\right )\right )} \sin \left (x\right ) + 2 \,{\left (2 \, a b c \cos \left (x\right )^{2} - a b c +{\left (a^{3} + a b^{2}\right )} \cos \left (x\right ) -{\left (a^{2} b + b^{3} -{\left (a^{2} - b^{2}\right )} c \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2} - c^{2}}}{2 \, b c \cos \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (a b \cos \left (x\right ) + a c\right )} \sin \left (x\right )}\right ) + 2 \,{\left (a^{3} + a b^{2} - a c^{2}\right )} x -{\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, b c \cos \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (a b \cos \left (x\right ) + a c\right )} \sin \left (x\right )\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} -{\left (a^{2} + b^{2}\right )} c^{2}\right )}}, -\frac{2 \, \sqrt{-a^{2} - b^{2} + c^{2}} a c \arctan \left (\frac{{\left (b c \cos \left (x\right ) + a c \sin \left (x\right ) + a^{2} + b^{2}\right )} \sqrt{-a^{2} - b^{2} + c^{2}}}{{\left (a^{3} + a b^{2} - a c^{2}\right )} \cos \left (x\right ) -{\left (a^{2} b + b^{3} - b c^{2}\right )} \sin \left (x\right )}\right ) - 2 \,{\left (a^{3} + a b^{2} - a c^{2}\right )} x +{\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, b c \cos \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + c^{2} + 2 \,{\left (a b \cos \left (x\right ) + a c\right )} \sin \left (x\right )\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} -{\left (a^{2} + b^{2}\right )} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(x)+c*csc(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \cot{\left (x \right )} + c \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(x)+c*csc(x)),x)

[Out]

Integral(1/(a + b*cot(x) + c*csc(x)), x)

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Giac [A]  time = 1.16176, size = 213, normalized size = 2.17 \begin{align*} -\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 )} a c}{{\left (a^{2} + b^{2}\right )} \sqrt{-a^{2} - b^{2} + c^{2}}} + \frac{a x}{a^{2} + b^{2}} - \frac{b \log \left (-b \tan \left (\frac{1}{2} \, x\right )^{2} + c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, x\right ) + b + c\right )}{a^{2} + b^{2}} + \frac{b \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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