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

Optimal. Leaf size=88 \[ -\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}+(a-b) \sqrt{a+b \cot ^2(x)}-(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.136862, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {3670, 446, 80, 50, 63, 208} \[ -\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}+(a-b) \sqrt{a+b \cot ^2(x)}-(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \cot ^3(x) \left (a+b \cot ^2(x)\right )^{3/2} \, dx &=-\operatorname{Subst}\left (\int \frac{x^3 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\cot (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (a+b x)^{3/2}}{1+x} \, dx,x,\cot ^2(x)\right )\right )\\ &=-\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1+x} \, dx,x,\cot ^2(x)\right )\\ &=\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}-\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{1}{2} (a-b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\cot ^2(x)\right )\\ &=(a-b) \sqrt{a+b \cot ^2(x)}+\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}-\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{1}{2} (a-b)^2 \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\cot ^2(x)\right )\\ &=(a-b) \sqrt{a+b \cot ^2(x)}+\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}-\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}+\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \cot ^2(x)}\right )}{b}\\ &=-(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )+(a-b) \sqrt{a+b \cot ^2(x)}+\frac{1}{3} \left (a+b \cot ^2(x)\right )^{3/2}-\frac{\left (a+b \cot ^2(x)\right )^{5/2}}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.497448, size = 91, normalized size = 1.03 \[ (a-b)^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right )\right )-\frac{\sqrt{a+b \cot ^2(x)} \left (3 a^2+b (6 a-5 b) \cot ^2(x)-20 a b+3 b^2 \cot ^4(x)+15 b^2\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - b)^(3/2)*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]) - (Sqrt[a + b*Cot[x]^2]*(3*a^2 - 20*a*b + 15*b^2 +
(6*a - 5*b)*b*Cot[x]^2 + 3*b^2*Cot[x]^4))/(15*b)

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Maple [B]  time = 0.023, size = 150, normalized size = 1.7 \begin{align*} -{\frac{1}{5\,b} \left ( a+b \left ( \cot \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{b \left ( \cot \left ( x \right ) \right ) ^{2}}{3}\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}+{\frac{4\,a}{3}\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}-b\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}+{{b}^{2}\arctan \left ({\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-2\,{\frac{ab}{\sqrt{-a+b}}\arctan \left ({\frac{\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}}{\sqrt{-a+b}}} \right ) }+{{a}^{2}\arctan \left ({\sqrt{a+b \left ( \cot \left ( x \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}} \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]

-1/5*(a+b*cot(x)^2)^(5/2)/b+1/3*b*cot(x)^2*(a+b*cot(x)^2)^(1/2)+4/3*a*(a+b*cot(x)^2)^(1/2)-b*(a+b*cot(x)^2)^(1
/2)+b^2/(-a+b)^(1/2)*arctan((a+b*cot(x)^2)^(1/2)/(-a+b)^(1/2))-2*a*b/(-a+b)^(1/2)*arctan((a+b*cot(x)^2)^(1/2)/
(-a+b)^(1/2))+a^2/(-a+b)^(1/2)*arctan((a+b*cot(x)^2)^(1/2)/(-a+b)^(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 = 2.32319, size = 1152, normalized size = 13.09 \begin{align*} \left [-\frac{15 \,{\left ({\left (a b - b^{2}\right )} \cos \left (2 \, x\right )^{2} + a b - b^{2} - 2 \,{\left (a b - b^{2}\right )} \cos \left (2 \, x\right )\right )} \sqrt{a - b} \log \left (-2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, x\right )^{2} - 2 \, a^{2} + b^{2} - 2 \,{\left ({\left (a - b\right )} \cos \left (2 \, x\right )^{2} -{\left (2 \, a - b\right )} \cos \left (2 \, x\right ) + a\right )} \sqrt{a - b} \sqrt{\frac{{\left (a - b\right )} \cos \left (2 \, x\right ) - a - b}{\cos \left (2 \, x\right ) - 1}} + 4 \,{\left (a^{2} - a b\right )} \cos \left (2 \, x\right )\right ) + 4 \,{\left ({\left (3 \, a^{2} - 26 \, a b + 23 \, b^{2}\right )} \cos \left (2 \, x\right )^{2} + 3 \, a^{2} - 14 \, a b + 13 \, b^{2} - 2 \,{\left (3 \, a^{2} - 20 \, a b + 12 \, b^{2}\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}}}{60 \,{\left (b \cos \left (2 \, x\right )^{2} - 2 \, b \cos \left (2 \, x\right ) + b\right )}}, -\frac{15 \,{\left ({\left (a b - b^{2}\right )} \cos \left (2 \, x\right )^{2} + a b - b^{2} - 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}}{\left (\cos \left (2 \, x\right ) - 1\right )}}{{\left (a - b\right )} \cos \left (2 \, x\right ) - a}\right ) + 2 \,{\left ({\left (3 \, a^{2} - 26 \, a b + 23 \, b^{2}\right )} \cos \left (2 \, x\right )^{2} + 3 \, a^{2} - 14 \, a b + 13 \, b^{2} - 2 \,{\left (3 \, a^{2} - 20 \, a b + 12 \, b^{2}\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}}}{30 \,{\left (b \cos \left (2 \, x\right )^{2} - 2 \, b \cos \left (2 \, x\right ) + b\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]

[-1/60*(15*((a*b - b^2)*cos(2*x)^2 + a*b - b^2 - 2*(a*b - b^2)*cos(2*x))*sqrt(a - b)*log(-2*(a^2 - 2*a*b + b^2
)*cos(2*x)^2 - 2*a^2 + b^2 - 2*((a - b)*cos(2*x)^2 - (2*a - b)*cos(2*x) + a)*sqrt(a - b)*sqrt(((a - b)*cos(2*x
) - a - b)/(cos(2*x) - 1)) + 4*(a^2 - a*b)*cos(2*x)) + 4*((3*a^2 - 26*a*b + 23*b^2)*cos(2*x)^2 + 3*a^2 - 14*a*
b + 13*b^2 - 2*(3*a^2 - 20*a*b + 12*b^2)*cos(2*x))*sqrt(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1)))/(b*cos(2*x
)^2 - 2*b*cos(2*x) + b), -1/30*(15*((a*b - b^2)*cos(2*x)^2 + a*b - b^2 - 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))*(cos(2*x) - 1)/((a - b)*cos(2*x) - a)) +
2*((3*a^2 - 26*a*b + 23*b^2)*cos(2*x)^2 + 3*a^2 - 14*a*b + 13*b^2 - 2*(3*a^2 - 20*a*b + 12*b^2)*cos(2*x))*sqrt
(((a - b)*cos(2*x) - a - b)/(cos(2*x) - 1)))/(b*cos(2*x)^2 - 2*b*cos(2*x) + b)]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out