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

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Rubi [A]  time = 0.10, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.22, 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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fricas [B]  time = 2.05, size = 555, normalized size = 5.66 \[ \left [\frac {\sqrt {a^{2} + b^{2} - c^{2}} a c \log \left (\frac {a^{4} + 3 \, a^{2} b^{2} + 2 \, b^{4} + {\left (a^{2} - b^{2}\right )} c^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} c \cos \relax (x) + {\left (a^{4} - b^{4} - 2 \, {\left (a^{2} - b^{2}\right )} c^{2}\right )} \cos \relax (x)^{2} + 2 \, {\left ({\left (a^{3} + a b^{2}\right )} c - {\left (a^{3} b + a b^{3} - 2 \, a b c^{2}\right )} \cos \relax (x)\right )} \sin \relax (x) + 2 \, {\left (2 \, a b c \cos \relax (x)^{2} - a b c + {\left (a^{3} + a b^{2}\right )} \cos \relax (x) - {\left (a^{2} b + b^{3} - {\left (a^{2} - b^{2}\right )} c \cos \relax (x)\right )} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2} - c^{2}}}{2 \, b c \cos \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (a b \cos \relax (x) + a c\right )} \sin \relax (x)}\right ) + 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x - {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, b c \cos \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (a b \cos \relax (x) + a c\right )} \sin \relax (x)\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}, -\frac {2 \, \sqrt {-a^{2} - b^{2} + c^{2}} a c \arctan \left (\frac {{\left (b c \cos \relax (x) + a c \sin \relax (x) + a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}}{{\left (a^{3} + a b^{2} - a c^{2}\right )} \cos \relax (x) - {\left (a^{2} b + b^{3} - b c^{2}\right )} \sin \relax (x)}\right ) - 2 \, {\left (a^{3} + a b^{2} - a c^{2}\right )} x + {\left (a^{2} b + b^{3} - b c^{2}\right )} \log \left (2 \, b c \cos \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2} + c^{2} + 2 \, {\left (a b \cos \relax (x) + a c\right )} \sin \relax (x)\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} - {\left (a^{2} + b^{2}\right )} c^{2}\right )}}\right ] \]

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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giac [A]  time = 0.14, size = 158, normalized size = 1.61 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, b + 2 \, c\right ) + \arctan \left (-\frac {b \tan \left (\frac {1}{2} \, x\right ) - c \tan \left (\frac {1}{2} \, x\right ) - a}{\sqrt {-a^{2} - b^{2} + c^{2}}}\right )\right )} a c}{{\left (a^{2} + b^{2}\right )} \sqrt {-a^{2} - b^{2} + c^{2}}} + \frac {a x}{a^{2} + b^{2}} - \frac {b \log \left (-b \tan \left (\frac {1}{2} \, x\right )^{2} + c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, x\right ) + b + c\right )}{a^{2} + b^{2}} + \frac {b \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{a^{2} + b^{2}} \]

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)

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maple [B]  time = 0.13, size = 446, normalized size = 4.55 \[ -\frac {2 \ln \left (b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-c \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 a \tan \left (\frac {x}{2}\right )-b -c \right ) b^{2}}{\left (2 a^{2}+2 b^{2}\right ) \left (b -c \right )}+\frac {2 \ln \left (b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-c \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 a \tan \left (\frac {x}{2}\right )-b -c \right ) c b}{\left (2 a^{2}+2 b^{2}\right ) \left (b -c \right )}+\frac {4 \arctan \left (\frac {2 \left (b -c \right ) \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a b}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {-a^{2}-b^{2}+c^{2}}}+\frac {4 \arctan \left (\frac {2 \left (b -c \right ) \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a c}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {-a^{2}-b^{2}+c^{2}}}-\frac {4 \arctan \left (\frac {2 \left (b -c \right ) \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a \,b^{2}}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {-a^{2}-b^{2}+c^{2}}\, \left (b -c \right )}+\frac {4 \arctan \left (\frac {2 \left (b -c \right ) \tan \left (\frac {x}{2}\right )-2 a}{2 \sqrt {-a^{2}-b^{2}+c^{2}}}\right ) a c b}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {-a^{2}-b^{2}+c^{2}}\, \left (b -c \right )}+\frac {2 b \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )}{2 a^{2}+2 b^{2}}+\frac {4 a \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}+2 b^{2}} \]

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-c*tan(1/2*x)^2-2*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-c*tan(1/2*x)^2-2*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 = 13.76, size = 965, normalized size = 9.85 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\mathrm {i}\right )}{b+a\,1{}\mathrm {i}}-\frac {\ln \left (-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (b-c\right )}^2-\frac {\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (32\,a\,b^2+32\,a\,c^2-64\,a\,b\,c-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b-c\right )\,\left (a^2-c^2+b\,c\right )+\frac {\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (64\,a\,b^3-32\,a^3\,b+32\,a^3\,c+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b-c\right )\,\left (a^2\,b-2\,c\,a^2-2\,b^3+2\,c\,b^2\right )+64\,a\,b\,c^2-128\,a\,b^2\,c-\frac {32\,\left (b-c\right )\,\left (a^2\,b-b\,c^2+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^4+3\,a^3\,b+a^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2+3\,a\,b^3+a\,b^2\,c-4\,a\,b\,c^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3\,c+2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (b\,\left (a^2-c^2\right )+b^3+a\,c\,\sqrt {a^2+b^2-c^2}\right )}{c^2\,\left (a^2+b^2-c^2\right )+{\left (a^2+b^2-c^2\right )}^2}-\frac {\ln \left (-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (b-c\right )}^2-\frac {\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (32\,a\,b^2+32\,a\,c^2-64\,a\,b\,c-64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b-c\right )\,\left (a^2-c^2+b\,c\right )+\frac {\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (64\,a\,b^3-32\,a^3\,b+32\,a^3\,c+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b-c\right )\,\left (a^2\,b-2\,c\,a^2-2\,b^3+2\,c\,b^2\right )+64\,a\,b\,c^2-128\,a\,b^2\,c-\frac {32\,\left (b-c\right )\,\left (a^2\,b-b\,c^2+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )\,\left (3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^4+3\,a^3\,b+a^3\,c+3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b\,c-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,c^2+3\,a\,b^3+a\,b^2\,c-4\,a\,b\,c^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3\,c+2\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^2\,c^2\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )}{\left (a^2+b^2\right )\,\left (a^2+b^2-c^2\right )}\right )\,\left (b\,\left (a^2-c^2\right )+b^3-a\,c\,\sqrt {a^2+b^2-c^2}\right )}{c^2\,\left (a^2+b^2-c^2\right )+{\left (a^2+b^2-c^2\right )}^2}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a+b\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + c/sin(x) + b*cot(x)),x)

[Out]

log(tan(x/2) - 1i)/(a*1i + b) + (log(tan(x/2) + 1i)*1i)/(a + b*1i) - (log(- 64*tan(x/2)*(b - c)^2 - ((a^2*b -
b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a*b^2 + 32*a*c^2 - 64*a*b*c - 64*tan(x/2)*(b - c)*(b*c + a^2 -
c^2) + ((a^2*b - b*c^2 + b^3 + a*c*(a^2 + b^2 - c^2)^(1/2))*(64*a*b^3 - 32*a^3*b + 32*a^3*c + 32*tan(x/2)*(b -
c)*(a^2*b - 2*a^2*c + 2*b^2*c - 2*b^3) + 64*a*b*c^2 - 128*a*b^2*c - (32*(b - c)*(a^2*b - b*c^2 + b^3 + a*c*(a
^2 + b^2 - c^2)^(1/2))*(3*a^4*tan(x/2) + 3*a*b^3 + 3*a^3*b + a^3*c + 3*a^2*b^2*tan(x/2) - 2*a^2*c^2*tan(x/2) +
2*b^2*c^2*tan(x/2) - 2*b^3*c*tan(x/2) - 4*a*b*c^2 + a*b^2*c - 2*a^2*b*c*tan(x/2)))/((a^2 + b^2)*(a^2 + b^2 -
c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^2)))*(b*(a^2 - c^2) + b^3 + a*c*(a^2 + b
^2 - c^2)^(1/2)))/(c^2*(a^2 + b^2 - c^2) + (a^2 + b^2 - c^2)^2) - (log(- 64*tan(x/2)*(b - c)^2 - ((a^2*b - b*c
^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(32*a*b^2 + 32*a*c^2 - 64*a*b*c - 64*tan(x/2)*(b - c)*(b*c + a^2 - c^2
) + ((a^2*b - b*c^2 + b^3 - a*c*(a^2 + b^2 - c^2)^(1/2))*(64*a*b^3 - 32*a^3*b + 32*a^3*c + 32*tan(x/2)*(b - c)
*(a^2*b - 2*a^2*c + 2*b^2*c - 2*b^3) + 64*a*b*c^2 - 128*a*b^2*c - (32*(b - c)*(a^2*b - b*c^2 + b^3 - a*c*(a^2
+ b^2 - c^2)^(1/2))*(3*a^4*tan(x/2) + 3*a*b^3 + 3*a^3*b + a^3*c + 3*a^2*b^2*tan(x/2) - 2*a^2*c^2*tan(x/2) + 2*
b^2*c^2*tan(x/2) - 2*b^3*c*tan(x/2) - 4*a*b*c^2 + a*b^2*c - 2*a^2*b*c*tan(x/2)))/((a^2 + b^2)*(a^2 + b^2 - c^2
))))/((a^2 + b^2)*(a^2 + b^2 - c^2))))/((a^2 + b^2)*(a^2 + b^2 - c^2)))*(b*(a^2 - c^2) + b^3 - a*c*(a^2 + b^2
- c^2)^(1/2)))/(c^2*(a^2 + b^2 - c^2) + (a^2 + b^2 - c^2)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \cot {\relax (x )} + c \csc {\relax (x )}}\, dx \]

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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