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

Optimal. Leaf size=55 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}}-\frac{1}{(a-b) \sqrt{a+b \cot ^2(x)}} \]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(3/2) - 1/((a - b)*Sqrt[a + b*Cot[x]^2])

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Rubi [A]  time = 0.0750258, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3670, 444, 51, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2}}-\frac{1}{(a-b) \sqrt{a+b \cot ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/(a + b*Cot[x]^2)^(3/2),x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(3/2) - 1/((a - 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 51

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

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

Mathematica [C]  time = 0.0411101, size = 44, normalized size = 0.8 \[ \frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \cot ^2(x)}{a-b}\right )}{(b-a) \sqrt{a+b \cot ^2(x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Cot[x]^2)/(a - b)]/((-a + b)*Sqrt[a + b*Cot[x]^2])

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Maple [A]  time = 0.016, size = 56, normalized size = 1. \begin{align*} -{\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)/(a+b*cot(x)^2)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.11315, size = 774, normalized size = 14.07 \begin{align*} \left [\frac{{\left ({\left (a - b\right )} \cos \left (2 \, x\right ) - a - b\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 ({\left (a - b\right )} \cos \left (2 \, x\right ) - a + b\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{2 \,{\left (a^{3} - a^{2} b - a b^{2} + b^{3} -{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (2 \, x\right )\right )}}, -\frac{{\left ({\left (a - b\right )} \cos \left (2 \, x\right ) - a - b\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 ({\left (a - b\right )} \cos \left (2 \, x\right ) - a + b\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}}}{a^{3} - a^{2} b - a b^{2} + b^{3} -{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (2 \, x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 15.2512, size = 48, normalized size = 0.87 \begin{align*} - \frac{1}{\left (a - b\right ) \sqrt{a + b \cot ^{2}{\left (x \right )}}} - \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \cot ^{2}{\left (x \right )}}}{\sqrt{- a + b}} \right )}}{\sqrt{- a + b} \left (a - b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((a - b)*sqrt(a + b*cot(x)**2)) - atan(sqrt(a + b*cot(x)**2)/sqrt(-a + b))/(sqrt(-a + b)*(a - b))

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Giac [B]  time = 1.28151, size = 165, normalized size = 3. \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{\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 )} - \frac{\sin \left (x\right )}{\sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}{\left (a \mathrm{sgn}\left (\sin \left (x\right )\right ) - b \mathrm{sgn}\left (\sin \left (x\right )\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(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) - 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))) - sin(x)/(sqrt(a*sin(x)^2 - b*sin(x)^2 + b)*(a
*sgn(sin(x)) - b*sgn(sin(x))))