3.49 \(\int \frac{\cot ^3(x)}{(a+b \cot ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=59 \[ \frac{a}{b (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.110584, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {3670, 446, 78, 63, 208} \[ \frac{a}{b (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^3/(a + b*Cot[x]^2)^(3/2),x]

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot ^3(x)}{\left (a+b \cot ^2(x)\right )^{3/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\cot (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{(1+x) (a+b x)^{3/2}} \, dx,x,\cot ^2(x)\right )\right )\\ &=\frac{a}{(a-b) b \sqrt{a+b \cot ^2(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)}\\ &=\frac{a}{(a-b) b \sqrt{a+b \cot ^2(x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \cot ^2(x)}\right )}{(a-b) b}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}}+\frac{a}{(a-b) b \sqrt{a+b \cot ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.217555, size = 59, normalized size = 1. \[ \frac{a}{b (a-b) \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^3/(a + b*Cot[x]^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.025, size = 68, normalized size = 1.2 \begin{align*}{\frac{1}{b}{\frac{1}{\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}}}+{\frac{1}{a-b}\arctan \left ({\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+{\frac{1}{a-b}{\frac{1}{\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^3/(a+b*cot(x)^2)^(3/2),x)

[Out]

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

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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(cot(x)^3/(a+b*cot(x)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.11659, size = 841, normalized size = 14.25 \begin{align*} \left [-\frac{{\left (a b + b^{2} -{\left (a b - b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt{a - b} \log \left (-\sqrt{a - b} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}{\left (\cos \left (2 \, x\right ) - 1\right )} -{\left (a - b\right )} \cos \left (2 \, x\right ) + a\right ) - 2 \,{\left (a^{2} - a b -{\left (a^{2} - a b\right )} \cos \left (2 \, x\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{2 \,{\left (a^{3} b - a^{2} b^{2} - a b^{3} + b^{4} -{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \cos \left (2 \, x\right )\right )}}, -\frac{{\left (a b + b^{2} -{\left (a b - b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt{-a + b} \arctan \left (-\frac{\sqrt{-a + b} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{a - b}\right ) -{\left (a^{2} - a b -{\left (a^{2} - a b\right )} \cos \left (2 \, x\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{a^{3} b - a^{2} b^{2} - a b^{3} + b^{4} -{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \cos \left (2 \, x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cot(x)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (x \right )}}{\left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**3/(a+b*cot(x)**2)**(3/2),x)

[Out]

Integral(cot(x)**3/(a + b*cot(x)**2)**(3/2), x)

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Giac [B]  time = 1.40318, size = 163, normalized size = 2.76 \begin{align*} -\frac{\sqrt{a - b} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (\sin \left (x\right )\right )}{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{a \mathrm{sgn}\left (\sin \left (x\right )\right ) \sin \left (x\right )}{\sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}{\left (a b - b^{2}\right )}} + \frac{\sqrt{a - b} \log \left ({\left | -\sqrt{a - b} \sin \left (x\right ) + \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b} \right |}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \mathrm{sgn}\left (\sin \left (x\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cot(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/2*sqrt(a - b)*log(abs(b))*sgn(sin(x))/(a^2 - 2*a*b + b^2) + a*sgn(sin(x))*sin(x)/(sqrt(a*sin(x)^2 - b*sin(x
)^2 + b)*(a*b - b^2)) + sqrt(a - b)*log(abs(-sqrt(a - b)*sin(x) + sqrt(a*sin(x)^2 - b*sin(x)^2 + b)))/((a^2 -
2*a*b + b^2)*sgn(sin(x)))