3.17 \(\int \frac{\csc ^5(x)}{a+b \cot (x)} \, dx\)

Optimal. Leaf size=101 \[ -\frac{\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{b^4}+\frac{\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sin (x) (b-a \cot (x))}{\sqrt{a^2+b^2}}\right )}{b^4}+\frac{a \tanh ^{-1}(\cos (x))}{2 b^2}+\frac{a \cot (x) \csc (x)}{2 b^2}-\frac{\csc ^3(x)}{3 b} \]

[Out]

(a*ArcTanh[Cos[x]])/(2*b^2) + (a*(a^2 + b^2)*ArcTanh[Cos[x]])/b^4 + ((a^2 + b^2)^(3/2)*ArcTanh[((b - a*Cot[x])
*Sin[x])/Sqrt[a^2 + b^2]])/b^4 - ((a^2 + b^2)*Csc[x])/b^3 + (a*Cot[x]*Csc[x])/(2*b^2) - Csc[x]^3/(3*b)

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Rubi [A]  time = 0.166973, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3510, 3486, 3768, 3770, 3509, 206} \[ -\frac{\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{b^4}+\frac{\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{\sin (x) (b-a \cot (x))}{\sqrt{a^2+b^2}}\right )}{b^4}+\frac{a \tanh ^{-1}(\cos (x))}{2 b^2}+\frac{a \cot (x) \csc (x)}{2 b^2}-\frac{\csc ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(a + b*Cot[x]),x]

[Out]

(a*ArcTanh[Cos[x]])/(2*b^2) + (a*(a^2 + b^2)*ArcTanh[Cos[x]])/b^4 + ((a^2 + b^2)^(3/2)*ArcTanh[((b - a*Cot[x])
*Sin[x])/Sqrt[a^2 + b^2]])/b^4 - ((a^2 + b^2)*Csc[x])/b^3 + (a*Cot[x]*Csc[x])/(2*b^2) - Csc[x]^3/(3*b)

Rule 3510

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[d^2/b^2, I
nt[(d*Sec[e + f*x])^(m - 2)*(a - b*Tan[e + f*x]), x], x] + Dist[(d^2*(a^2 + b^2))/b^2, Int[(d*Sec[e + f*x])^(m
- 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 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 ^5(x)}{a+b \cot (x)} \, dx &=-\frac{\int (a-b \cot (x)) \csc ^3(x) \, dx}{b^2}+\frac{\left (a^2+b^2\right ) \int \frac{\csc ^3(x)}{a+b \cot (x)} \, dx}{b^2}\\ &=-\frac{\csc ^3(x)}{3 b}-\frac{a \int \csc ^3(x) \, dx}{b^2}-\frac{\left (a^2+b^2\right ) \int (a-b \cot (x)) \csc (x) \, dx}{b^4}+\frac{\left (a^2+b^2\right )^2 \int \frac{\csc (x)}{a+b \cot (x)} \, dx}{b^4}\\ &=-\frac{\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac{a \cot (x) \csc (x)}{2 b^2}-\frac{\csc ^3(x)}{3 b}-\frac{a \int \csc (x) \, dx}{2 b^2}-\frac{\left (a \left (a^2+b^2\right )\right ) \int \csc (x) \, dx}{b^4}-\frac{\left (a^2+b^2\right )^2 \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,(-b+a \cot (x)) \sin (x)\right )}{b^4}\\ &=\frac{a \tanh ^{-1}(\cos (x))}{2 b^2}+\frac{a \left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{b^4}+\frac{\left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{(b-a \cot (x)) \sin (x)}{\sqrt{a^2+b^2}}\right )}{b^4}-\frac{\left (a^2+b^2\right ) \csc (x)}{b^3}+\frac{a \cot (x) \csc (x)}{2 b^2}-\frac{\csc ^3(x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 1.18382, size = 198, normalized size = 1.96 \[ -\frac{4 b \left (6 a^2+7 b^2\right ) \cot \left (\frac{x}{2}\right )-96 \left (a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac{b \tan \left (\frac{x}{2}\right )-a}{\sqrt{a^2+b^2}}\right )+24 a^2 b \tan \left (\frac{x}{2}\right )+48 a^3 \log \left (\sin \left (\frac{x}{2}\right )\right )-48 a^3 \log \left (\cos \left (\frac{x}{2}\right )\right )-6 a b^2 \csc ^2\left (\frac{x}{2}\right )+6 a b^2 \sec ^2\left (\frac{x}{2}\right )+72 a b^2 \log \left (\sin \left (\frac{x}{2}\right )\right )-72 a b^2 \log \left (\cos \left (\frac{x}{2}\right )\right )+28 b^3 \tan \left (\frac{x}{2}\right )+16 b^3 \sin ^4\left (\frac{x}{2}\right ) \csc ^3(x)+b^3 \sin (x) \csc ^4\left (\frac{x}{2}\right )}{48 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(a + b*Cot[x]),x]

[Out]

-(-96*(a^2 + b^2)^(3/2)*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]] + 4*b*(6*a^2 + 7*b^2)*Cot[x/2] - 6*a*b^2*Cs
c[x/2]^2 - 48*a^3*Log[Cos[x/2]] - 72*a*b^2*Log[Cos[x/2]] + 48*a^3*Log[Sin[x/2]] + 72*a*b^2*Log[Sin[x/2]] + 6*a
*b^2*Sec[x/2]^2 + 16*b^3*Csc[x]^3*Sin[x/2]^4 + b^3*Csc[x/2]^4*Sin[x] + 24*a^2*b*Tan[x/2] + 28*b^3*Tan[x/2])/(4
8*b^4)

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Maple [B]  time = 0.121, size = 232, normalized size = 2.3 \begin{align*} -{\frac{1}{24\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{a}{8\,{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{{a}^{2}}{2\,{b}^{3}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{5}{8\,b}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{{a}^{4}}{\sqrt{{a}^{2}+{b}^{2}}{b}^{4}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tan \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+4\,{\frac{{a}^{2}}{\sqrt{{a}^{2}+{b}^{2}}{b}^{2}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tan \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }+2\,{\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,\tan \left ( x/2 \right ) b-2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{24\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{{a}^{2}}{2\,{b}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{5}{8\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{a}{8\,{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{{a}^{3}}{{b}^{4}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{3\,a}{2\,{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)^5/(a+b*cot(x)),x)

[Out]

-1/24/b*tan(1/2*x)^3-1/8/b^2*a*tan(1/2*x)^2-1/2/b^3*a^2*tan(1/2*x)-5/8/b*tan(1/2*x)+2/b^4/(a^2+b^2)^(1/2)*arct
anh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))*a^4+4/b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+
b^2)^(1/2))*a^2+2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))-1/24/b/tan(1/2*x)^3-1/2/b^
3/tan(1/2*x)*a^2-5/8/b/tan(1/2*x)+1/8*a/b^2/tan(1/2*x)^2-1/b^4*a^3*ln(tan(1/2*x))-3/2/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)^5/(a+b*cot(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.9923, size = 662, normalized size = 6.55 \begin{align*} -\frac{6 \, a b^{2} \cos \left (x\right ) \sin \left (x\right ) - 6 \,{\left ({\left (a^{2} + b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - b^{2}\right )} \sqrt{a^{2} + b^{2}} \log \left (-\frac{2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) \sin \left (x\right ) - 12 \, a^{2} b - 16 \, b^{3} + 12 \,{\left (a^{2} b + b^{3}\right )} \cos \left (x\right )^{2} + 3 \,{\left (2 \, a^{3} + 3 \, a b^{2} -{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right ) - 3 \,{\left (2 \, a^{3} + 3 \, a b^{2} -{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) \sin \left (x\right )}{12 \,{\left (b^{4} \cos \left (x\right )^{2} - b^{4}\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+b*cot(x)),x, algorithm="fricas")

[Out]

-1/12*(6*a*b^2*cos(x)*sin(x) - 6*((a^2 + b^2)*cos(x)^2 - a^2 - b^2)*sqrt(a^2 + b^2)*log(-(2*a*b*cos(x)*sin(x)
- (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 + 2*sqrt(a^2 + b^2)*(a*cos(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 -
b^2)*cos(x)^2 + a^2))*sin(x) - 12*a^2*b - 16*b^3 + 12*(a^2*b + b^3)*cos(x)^2 + 3*(2*a^3 + 3*a*b^2 - (2*a^3 + 3
*a*b^2)*cos(x)^2)*log(1/2*cos(x) + 1/2)*sin(x) - 3*(2*a^3 + 3*a*b^2 - (2*a^3 + 3*a*b^2)*cos(x)^2)*log(-1/2*cos
(x) + 1/2)*sin(x))/((b^4*cos(x)^2 - b^4)*sin(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{5}{\left (x \right )}}{a + b \cot{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**5/(a+b*cot(x)),x)

[Out]

Integral(csc(x)**5/(a + b*cot(x)), x)

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Giac [B]  time = 1.3842, size = 297, normalized size = 2.94 \begin{align*} -\frac{b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 3 \, a b \tan \left (\frac{1}{2} \, x\right )^{2} + 12 \, a^{2} \tan \left (\frac{1}{2} \, x\right ) + 15 \, b^{2} \tan \left (\frac{1}{2} \, x\right )}{24 \, b^{3}} - \frac{{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{2 \, b^{4}} - \frac{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\frac{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} b^{4}} + \frac{44 \, a^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 66 \, a b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 12 \, a^{2} b \tan \left (\frac{1}{2} \, x\right )^{2} - 15 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, x\right ) - b^{3}}{24 \, b^{4} \tan \left (\frac{1}{2} \, x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+b*cot(x)),x, algorithm="giac")

[Out]

-1/24*(b^2*tan(1/2*x)^3 + 3*a*b*tan(1/2*x)^2 + 12*a^2*tan(1/2*x) + 15*b^2*tan(1/2*x))/b^3 - 1/2*(2*a^3 + 3*a*b
^2)*log(abs(tan(1/2*x)))/b^4 - (a^4 + 2*a^2*b^2 + b^4)*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2
*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b^4) + 1/24*(44*a^3*tan(1/2*x)^3 + 66*a*b^2*tan(1/2
*x)^3 - 12*a^2*b*tan(1/2*x)^2 - 15*b^3*tan(1/2*x)^2 + 3*a*b^2*tan(1/2*x) - b^3)/(b^4*tan(1/2*x)^3)