3.54 \(\int \frac{\cot ^3(x)}{(a+b \cot ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=82 \[ \frac{a}{3 b (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{1}{(a-b)^2 \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.126773, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3670, 446, 78, 51, 63, 208} \[ \frac{a}{3 b (a-b) \left (a+b \cot ^2(x)\right )^{3/2}}+\frac{1}{(a-b)^2 \sqrt{a+b \cot ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^3/(a + b*Cot[x]^2)^(5/2),x]

[Out]

-(ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]/(a - b)^(5/2)) + a/(3*(a - b)*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 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 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 ^3(x)}{\left (a+b \cot ^2(x)\right )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{x^3}{\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{x}{(1+x) (a+b x)^{5/2}} \, dx,x,\cot ^2(x)\right )\right )\\ &=\frac{a}{3 (a-b) 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{a}{3 (a-b) 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{a}{3 (a-b) 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{a}{3 (a-b) 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 = 0.109064, size = 69, normalized size = 0.84 \[ \frac{3 b \left (a+b \cot ^2(x)\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \cot ^2(x)}{a-b}\right )+a (a-b)}{3 b (a-b)^2 \left (a+b \cot ^2(x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^3/(a + b*Cot[x]^2)^(5/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 88, normalized size = 1.1 \begin{align*}{\frac{1}{3\,b} \left ( a+b \left ( \cot \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\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)^3/(a+b*cot(x)^2)^(5/2),x)

[Out]

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

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.21944, size = 1538, normalized size = 18.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cot(x)^2)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**3/(a+b*cot(x)**2)**(5/2),x)

[Out]

Integral(cot(x)**3/(a + b*cot(x)**2)**(5/2), x)

________________________________________________________________________________________

Giac [B]  time = 1.42389, size = 370, normalized size = 4.51 \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 ) \mathrm{sgn}\left (\sin \left (x\right )\right )}{8 \,{\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )}} + \frac{\sqrt{a - b} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (\sin \left (x\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{{\left (a^{3} \mathrm{sgn}\left (\sin \left (x\right )\right ) + a^{2} b \mathrm{sgn}\left (\sin \left (x\right )\right ) - 5 \, a b^{2} \mathrm{sgn}\left (\sin \left (x\right )\right ) + 3 \, b^{3} \mathrm{sgn}\left (\sin \left (x\right )\right )\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} \mathrm{sgn}\left (\sin \left (x\right )\right ) - b^{3} \mathrm{sgn}\left (\sin \left (x\right )\right )\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 )}{24 \,{\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)^3/(a+b*cot(x)^2)^(5/2),x, algorithm="giac")

[Out]

-1/8*sqrt(a - b)*log(abs(-sqrt(a - b)*sin(x) + sqrt(a*sin(x)^2 - b*sin(x)^2 + b)))*sgn(sin(x))/(a^4*b - 4*a^3*
b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5) + 1/16*sqrt(a - b)*log(abs(b))*sgn(sin(x))/(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4
*a*b^4 + b^5) - 1/24*((a^3*sgn(sin(x)) + a^2*b*sgn(sin(x)) - 5*a*b^2*sgn(sin(x)) + 3*b^3*sgn(sin(x)))*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*sgn(sin(x)) - b^3*sgn(sin(x)))/(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)