3.460 \(\int \frac{\csc ^2(x)}{a+b \cot (x)+c \csc (x)} \, dx\)

Optimal. Leaf size=120 \[ -\frac{2 a c \tanh ^{-1}\left (\frac{a-(b-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{b \log \left (2 a \tan \left (\frac{x}{2}\right )-(b-c) \tan ^2\left (\frac{x}{2}\right )+b+c\right )}{b^2-c^2}+\frac{\log \left (\tan \left (\frac{x}{2}\right )\right )}{b+c} \]

[Out]

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

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Rubi [A]  time = 0.533174, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {4397, 12, 1628, 634, 618, 206, 628} \[ -\frac{2 a c \tanh ^{-1}\left (\frac{a-(b-c) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\left (b^2-c^2\right ) \sqrt{a^2+b^2-c^2}}-\frac{b \log \left (2 a \tan \left (\frac{x}{2}\right )-(b-c) \tan ^2\left (\frac{x}{2}\right )+b+c\right )}{b^2-c^2}+\frac{\log \left (\tan \left (\frac{x}{2}\right )\right )}{b+c} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Cot[x] + c*Csc[x]),x]

[Out]

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

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.286086, size = 104, normalized size = 0.87 \[ \frac{\frac{2 a c \tanh ^{-1}\left (\frac{a+(c-b) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2-c^2}}\right )}{\sqrt{a^2+b^2-c^2}}+b \log (a \sin (x)+b \cos (x)+c)+(c-b) \log \left (\sin \left (\frac{x}{2}\right )\right )-(b+c) \log \left (\cos \left (\frac{x}{2}\right )\right )}{(c-b) (b+c)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + b*Cot[x] + c*Csc[x]),x]

[Out]

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

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Maple [A]  time = 0.045, size = 184, normalized size = 1.5 \begin{align*} -{\frac{b}{ \left ( b+c \right ) \left ( b-c \right ) }\ln \left ( b \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}c-2\,a\tan \left ( x/2 \right ) -b-c \right ) }+2\,{\frac{a}{ \left ( b+c \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 ) }-2\,{\frac{ab}{ \left ( b+c \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 ) }+{\frac{1}{b+c}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^2/(a+b*cot(x)+c*csc(x)),x)

[Out]

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

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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(csc(x)^2/(a+b*cot(x)+c*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 23.5262, size = 1571, normalized size = 13.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (x \right )}}{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(csc(x)**2/(a+b*cot(x)+c*csc(x)),x)

[Out]

Integral(csc(x)**2/(a + b*cot(x) + c*csc(x)), x)

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Giac [A]  time = 1.17926, size = 192, normalized size = 1.6 \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}{\sqrt{-a^{2} - b^{2} + c^{2}}{\left (b^{2} - c^{2}\right )}} - \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 )}{b^{2} - c^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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