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

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Rubi [A]  time = 0.15, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 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 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 3770

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

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*}

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Mathematica [A]  time = 0.20, 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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fricas [B]  time = 0.67, size = 308, normalized size = 4.97 \[ \left [\frac {\sqrt {a^{2} - b^{2}} a^{2} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) \sin \relax (x) + {\left (a^{3} - a b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - {\left (a^{3} - a b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) - 2 \, {\left (a^{2} b - b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \relax (x)}, -\frac {2 \, \sqrt {-a^{2} + b^{2}} a^{2} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) \sin \relax (x) - {\left (a^{3} - a b^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + {\left (a^{3} - a b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \relax (x)}\right ] \]

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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giac [A]  time = 0.35, size = 98, normalized size = 1.58 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{2}}{\sqrt {-a^{2} + b^{2}} b^{2}} - \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{b^{2}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, b} + \frac {2 \, a \tan \left (\frac {1}{2} \, x\right ) - b}{2 \, b^{2} \tan \left (\frac {1}{2} \, x\right )} \]

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))

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maple [A]  time = 0.21, size = 77, normalized size = 1.24 \[ \frac {\tan \left (\frac {x}{2}\right )}{2 b}+\frac {2 a^{2} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{2} \sqrt {-a^{2}+b^{2}}}-\frac {1}{2 b \tan \left (\frac {x}{2}\right )}-\frac {a \ln \left (\tan \left (\frac {x}{2}\right )\right )}{b^{2}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*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(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 0.49, size = 135, normalized size = 2.18 \[ -\frac {1}{b\,\mathrm {tan}\relax (x)}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{b^2}-\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,4{}\mathrm {i}-b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}\,1{}\mathrm {i}+a\,b\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{4\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3+2\,a^2\,b-3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^2-b^3}\right )\,2{}\mathrm {i}}{b^2\,\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^3*(a + b/sin(x))),x)

[Out]

- 1/(b*tan(x)) - (a*log(tan(x/2)))/b^2 - (a^2*atan((a^2*tan(x/2)*(a^2 - b^2)^(1/2)*4i - b^2*tan(x/2)*(a^2 - b^
2)^(1/2)*1i + a*b*(a^2 - b^2)^(1/2)*2i)/(4*a^3*tan(x/2) + 2*a^2*b - b^3 - 3*a*b^2*tan(x/2)))*2i)/(b^2*(a^2 - b
^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

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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