3.56 \(\int \frac{\cot (x)}{(a+b \cot ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=78 \[ -\frac{1}{(a-b)^2 \sqrt{a+b \cot ^2(x)}}-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2) - 1/(3*(a - b)*(a + b*Cot[x]^2)^(3/2)) - 1/((a - b)^2*
Sqrt[a + b*Cot[x]^2])

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Rubi [A]  time = 0.0915714, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{1}{(a-b)^2 \sqrt{a+b \cot ^2(x)}}-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2) - 1/(3*(a - b)*(a + b*Cot[x]^2)^(3/2)) - 1/((a - b)^2*
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 )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{x}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\cot (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{5/2}} \, dx,x,\cot ^2(x)\right )\right )\\ &=-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{3/2}} \, dx,x,\cot ^2(x)\right )}{2 (a-b)}\\ &=-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{1}{(a-b)^2 \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)^2}\\ &=-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{1}{(a-b)^2 \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)^2 b}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}}-\frac{1}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}-\frac{1}{(a-b)^2 \sqrt{a+b \cot ^2(x)}}\\ \end{align*}

Mathematica [C]  time = 0.0390666, size = 47, normalized size = 0.6 \[ -\frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \cot ^2(x)}{a-b}\right )}{3 (a-b) \left (a+b \cot ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]/(a + b*Cot[x]^2)^(5/2),x]

[Out]

-Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Cot[x]^2)/(a - b)]/(3*(a - b)*(a + b*Cot[x]^2)^(3/2))

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Maple [A]  time = 0.017, size = 75, normalized size = 1. \begin{align*} -{\frac{1}{ \left ( a-b \right ) ^{2}}\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}{ \left ( a-b \right ) ^{2}}{\frac{1}{\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}}}-{\frac{1}{3\,a-3\,b} \left ( a+b \left ( \cot \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a+b*cot(x)^2)^(5/2),x)

[Out]

-1/(a-b)^2/(-a+b)^(1/2)*arctan((a+b*cot(x)^2)^(1/2)/(-a+b)^(1/2))-1/(a-b)^2/(a+b*cot(x)^2)^(1/2)-1/3/(a-b)/(a+
b*cot(x)^2)^(3/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13093, size = 1416, normalized size = 18.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*((a^2 - 2*a*b + b^2)*cos(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - 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) - 4*(2*(a^2 - 2*a*
b + b^2)*cos(2*x)^2 + 2*a^2 - a*b - b^2 - (4*a^2 - 5*a*b + b^2)*cos(2*x))*sqrt(((a - b)*cos(2*x) - a - b)/(cos
(2*x) - 1)))/(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5
*a*b^4 - b^5)*cos(2*x)^2 - 2*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*cos(2*x)), 1/3*(3*((a^2 -
2*a*b + b^2)*cos(2*x)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - 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)) - 2*(2*(a^2 - 2*a*b + b^2)*cos(2*x)^2 + 2*a^2 - a*b - b^2
- (4*a^2 - 5*a*b + b^2)*cos(2*x))*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1)))/(a^5 - a^4*b - 2*a^3*b^2 +
2*a^2*b^3 + a*b^4 - b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*cos(2*x)^2 - 2*(a^5 - 3*a^
4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*cos(2*x))]

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Sympy [A]  time = 51.1062, size = 70, normalized size = 0.9 \begin{align*} - \frac{1}{3 \left (a - b\right ) \left (a + b \cot ^{2}{\left (x \right )}\right )^{\frac{3}{2}}} - \frac{1}{\left (a - b\right )^{2} \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 )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)**2)**(5/2),x)

[Out]

-1/(3*(a - b)*(a + b*cot(x)**2)**(3/2)) - 1/((a - b)**2*sqrt(a + b*cot(x)**2)) - atan(sqrt(a + b*cot(x)**2)/sq
rt(-a + b))/(sqrt(-a + b)*(a - b)**2)

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Giac [B]  time = 1.35537, size = 270, normalized size = 3.46 \begin{align*} \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 )}{16 \,{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )}} + \frac{{\left (\frac{4 \,{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (x\right )^{2}}{a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}} + \frac{3 \,{\left (a b^{2} - b^{3}\right )}}{a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}}\right )} \sin \left (x\right )}{48 \,{\left (a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*cot(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/16*sqrt(a - b)*log(abs(-sqrt(a - b)*sin(x) + sqrt(a*sin(x)^2 - b*sin(x)^2 + b)))/(a^4*b - 4*a^3*b^2 + 6*a^2*
b^3 - 4*a*b^4 + b^5) + 1/48*(4*(a^2*b - 2*a*b^2 + b^3)*sin(x)^2/(a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b
^6) + 3*(a*b^2 - b^3)/(a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6))*sin(x)/(a*sin(x)^2 - b*sin(x)^2 + b)^
(3/2)