3.41 \(\int \frac{\csc ^3(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=62 \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}+\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{\cot (x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.154549, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3790, 3789, 3770, 3831, 2660, 618, 206} \[ -\frac{2 a^2 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}+\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{\cot (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a + b*Csc[x]),x]

[Out]

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

Rule 3790

Int[csc[(e_.) + (f_.)*(x_)]^3/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(b*f), x
] - Dist[a/b, Int[Csc[e + f*x]^2/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3789

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 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 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{\csc ^3(x)}{a+b \csc (x)} \, dx &=-\frac{\cot (x)}{b}-\frac{a \int \frac{\csc ^2(x)}{a+b \csc (x)} \, dx}{b}\\ &=-\frac{\cot (x)}{b}-\frac{a \int \csc (x) \, dx}{b^2}+\frac{a^2 \int \frac{\csc (x)}{a+b \csc (x)} \, dx}{b^2}\\ &=\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{\cot (x)}{b}+\frac{a^2 \int \frac{1}{1+\frac{a \sin (x)}{b}} \, dx}{b^3}\\ &=\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{\cot (x)}{b}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{2 a x}{b}+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^3}\\ &=\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{\cot (x)}{b}-\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1-\frac{a^2}{b^2}\right )-x^2} \, dx,x,\frac{2 a}{b}+2 \tan \left (\frac{x}{2}\right )\right )}{b^3}\\ &=\frac{a \tanh ^{-1}(\cos (x))}{b^2}-\frac{2 a^2 \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}+\tan \left (\frac{x}{2}\right )\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \sqrt{a^2-b^2}}-\frac{\cot (x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.203814, size = 106, normalized size = 1.71 \[ \frac{\csc \left (\frac{x}{2}\right ) \sec \left (\frac{x}{2}\right ) \left (2 a^2 \sin (x) \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+\sqrt{b^2-a^2} \left (a \sin (x) \left (\log \left (\cos \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )\right )\right )-b \cos (x)\right )\right )}{2 b^2 \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a + b*Csc[x]),x]

[Out]

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

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Maple [A]  time = 0.049, size = 77, normalized size = 1.2 \begin{align*}{\frac{1}{2\,b}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{{a}^{2}}{{b}^{2}\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{2\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{a}{{b}^{2}}\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)^3/(a+b*csc(x)),x)

[Out]

1/2/b*tan(1/2*x)+2*a^2/b^2/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))-1/2/b/tan(1/2*x)
-1/b^2*a*ln(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(csc(x)^3/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.623509, size = 768, normalized size = 12.39 \begin{align*} \left [\frac{\sqrt{a^{2} - b^{2}} a^{2} \log \left (-\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} - 2 \,{\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) \sin \left (x\right ) +{\left (a^{3} - a b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) -{\left (a^{3} - a b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) - 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )}, -\frac{2 \, \sqrt{-a^{2} + b^{2}} a^{2} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \sin \left (x\right ) -{\left (a^{3} - a b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) +{\left (a^{3} - a b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*csc(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (x \right )}}{a + b \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a+b*csc(x)),x)

[Out]

Integral(csc(x)**3/(a + b*csc(x)), x)

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Giac [A]  time = 1.28532, size = 132, normalized size = 2.13 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )} a^{2}}{\sqrt{-a^{2} + b^{2}} b^{2}} - \frac{a \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{b^{2}} + \frac{\tan \left (\frac{1}{2} \, x\right )}{2 \, b} + \frac{2 \, a \tan \left (\frac{1}{2} \, x\right ) - b}{2 \, b^{2} \tan \left (\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*csc(x)),x, algorithm="giac")

[Out]

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