3.40 \(\int \frac{\csc ^4(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=84 \[ -\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \sqrt{a^2-b^2}}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b} \]

[Out]

-((2*a^2 + b^2)*ArcTanh[Cos[x]])/(2*b^3) + (2*a^3*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^3*Sqrt[a^2 - b
^2]) + (a*Cot[x])/b^2 - (Cot[x]*Csc[x])/(2*b)

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Rubi [A]  time = 0.252198, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {3851, 4082, 3998, 3770, 3831, 2660, 618, 206} \[ -\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \sqrt{a^2-b^2}}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + b*Csc[x]),x]

[Out]

-((2*a^2 + b^2)*ArcTanh[Cos[x]])/(2*b^3) + (2*a^3*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^3*Sqrt[a^2 - b
^2]) + (a*Cot[x])/b^2 - (Cot[x]*Csc[x])/(2*b)

Rule 3851

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(d^3*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 3))/(b*f*(n - 2)), x] + Dist[d^3/(b*(n - 2)), Int[((d*Csc[e + f*x])^(n - 3)*Simp
[a*(n - 3) + b*(n - 3)*Csc[e + f*x] - a*(n - 2)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a,
b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 3]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

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 ^4(x)}{a+b \csc (x)} \, dx &=-\frac{\cot (x) \csc (x)}{2 b}+\frac{\int \frac{\csc (x) \left (a+b \csc (x)-2 a \csc ^2(x)\right )}{a+b \csc (x)} \, dx}{2 b}\\ &=\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}+\frac{\int \frac{\csc (x) \left (a b+\left (2 a^2+b^2\right ) \csc (x)\right )}{a+b \csc (x)} \, dx}{2 b^2}\\ &=\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}-\frac{a^3 \int \frac{\csc (x)}{a+b \csc (x)} \, dx}{b^3}+\frac{\left (2 a^2+b^2\right ) \int \csc (x) \, dx}{2 b^3}\\ &=-\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}-\frac{a^3 \int \frac{1}{1+\frac{a \sin (x)}{b}} \, dx}{b^4}\\ &=-\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}-\frac{\left (2 a^3\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^4}\\ &=-\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}+\frac{\left (4 a^3\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^4}\\ &=-\frac{\left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{2 a^3 \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}+\tan \left (\frac{x}{2}\right )\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \sqrt{a^2-b^2}}+\frac{a \cot (x)}{b^2}-\frac{\cot (x) \csc (x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.486741, size = 144, normalized size = 1.71 \[ \frac{-\frac{16 a^3 \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+8 a^2 \log \left (\sin \left (\frac{x}{2}\right )\right )-8 a^2 \log \left (\cos \left (\frac{x}{2}\right )\right )-4 a b \tan \left (\frac{x}{2}\right )+4 a b \cot \left (\frac{x}{2}\right )-b^2 \csc ^2\left (\frac{x}{2}\right )+b^2 \sec ^2\left (\frac{x}{2}\right )+4 b^2 \log \left (\sin \left (\frac{x}{2}\right )\right )-4 b^2 \log \left (\cos \left (\frac{x}{2}\right )\right )}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + b*Csc[x]),x]

[Out]

((-16*a^3*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 4*a*b*Cot[x/2] - b^2*Csc[x/2]^2 - 8*a^
2*Log[Cos[x/2]] - 4*b^2*Log[Cos[x/2]] + 8*a^2*Log[Sin[x/2]] + 4*b^2*Log[Sin[x/2]] + b^2*Sec[x/2]^2 - 4*a*b*Tan
[x/2])/(8*b^3)

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Maple [A]  time = 0.045, size = 112, normalized size = 1.3 \begin{align*}{\frac{1}{8\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{a}{2\,{b}^{2}}\tan \left ({\frac{x}{2}} \right ) }-2\,{\frac{{a}^{3}}{{b}^{3}\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}{8\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{1}{2\,b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{a}{2\,{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^4/(a+b*csc(x)),x)

[Out]

1/8/b*tan(1/2*x)^2-1/2/b^2*a*tan(1/2*x)-2*a^3/b^3/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^
(1/2))-1/8/b/tan(1/2*x)^2+1/b^3*ln(tan(1/2*x))*a^2+1/2/b*ln(tan(1/2*x))+1/2*a/b^2/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)^4/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.878674, size = 1146, normalized size = 13.64 \begin{align*} \left [\frac{4 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) - 2 \,{\left (a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \sqrt{a^{2} - b^{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 ) - 2 \,{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4} -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (2 \, a^{4} - a^{2} b^{2} - b^{4} -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{2} b^{3} - b^{5} -{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2}\right )}}, \frac{4 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) - 4 \,{\left (a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \sqrt{-a^{2} + b^{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 ) - 2 \,{\left (a^{2} b^{2} - b^{4}\right )} \cos \left (x\right ) -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4} -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (2 \, a^{4} - a^{2} b^{2} - b^{4} -{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{2} b^{3} - b^{5} -{\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/4*(4*(a^3*b - a*b^3)*cos(x)*sin(x) - 2*(a^3*cos(x)^2 - a^3)*sqrt(a^2 - b^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)) - 2*(a^2*b^2 - b^4)*cos(x) - (2*a^4 - a^2*b^2 - b^4 - (2*a^4 - a^2*b^2 - b^4)*cos(x)^2)*log(1/2*cos(x)
+ 1/2) + (2*a^4 - a^2*b^2 - b^4 - (2*a^4 - a^2*b^2 - b^4)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a^2*b^3 - b^5 - (
a^2*b^3 - b^5)*cos(x)^2), 1/4*(4*(a^3*b - a*b^3)*cos(x)*sin(x) - 4*(a^3*cos(x)^2 - a^3)*sqrt(-a^2 + b^2)*arcta
n(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) - 2*(a^2*b^2 - b^4)*cos(x) - (2*a^4 - a^2*b^2 - b^4 -
(2*a^4 - a^2*b^2 - b^4)*cos(x)^2)*log(1/2*cos(x) + 1/2) + (2*a^4 - a^2*b^2 - b^4 - (2*a^4 - a^2*b^2 - b^4)*co
s(x)^2)*log(-1/2*cos(x) + 1/2))/(a^2*b^3 - b^5 - (a^2*b^3 - b^5)*cos(x)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\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)**4/(a+b*csc(x)),x)

[Out]

Integral(csc(x)**4/(a + b*csc(x)), x)

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Giac [A]  time = 1.44523, size = 190, normalized size = 2.26 \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^{3}}{\sqrt{-a^{2} + b^{2}} b^{3}} + \frac{b \tan \left (\frac{1}{2} \, x\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, x\right )}{8 \, b^{2}} + \frac{{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{2 \, b^{3}} - \frac{12 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 6 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + b^{2}}{8 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(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^3/(sqrt(-a^2 + b^2)*b^3)
+ 1/8*(b*tan(1/2*x)^2 - 4*a*tan(1/2*x))/b^2 + 1/2*(2*a^2 + b^2)*log(abs(tan(1/2*x)))/b^3 - 1/8*(12*a^2*tan(1/2
*x)^2 + 6*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + b^2)/(b^3*tan(1/2*x)^2)