3.19 \(\int \sqrt [3]{c \cot (a+b x)} \, dx\)

Optimal. Leaf size=131 \[ \frac{\sqrt [3]{c} \log \left ((c \cot (a+b x))^{2/3}+c^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \log \left (-c^{2/3} (c \cot (a+b x))^{2/3}+(c \cot (a+b x))^{4/3}+c^{4/3}\right )}{4 b}+\frac{\sqrt{3} \sqrt [3]{c} \tan ^{-1}\left (\frac{c^{2/3}-2 (c \cot (a+b x))^{2/3}}{\sqrt{3} c^{2/3}}\right )}{2 b} \]

[Out]

(Sqrt[3]*c^(1/3)*ArcTan[(c^(2/3) - 2*(c*Cot[a + b*x])^(2/3))/(Sqrt[3]*c^(2/3))])/(2*b) + (c^(1/3)*Log[c^(2/3)
+ (c*Cot[a + b*x])^(2/3)])/(2*b) - (c^(1/3)*Log[c^(4/3) - c^(2/3)*(c*Cot[a + b*x])^(2/3) + (c*Cot[a + b*x])^(4
/3)])/(4*b)

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Rubi [A]  time = 0.101623, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3476, 329, 275, 292, 31, 634, 617, 204, 628} \[ \frac{\sqrt [3]{c} \log \left ((c \cot (a+b x))^{2/3}+c^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \log \left (-c^{2/3} (c \cot (a+b x))^{2/3}+(c \cot (a+b x))^{4/3}+c^{4/3}\right )}{4 b}+\frac{\sqrt{3} \sqrt [3]{c} \tan ^{-1}\left (\frac{c^{2/3}-2 (c \cot (a+b x))^{2/3}}{\sqrt{3} c^{2/3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cot[a + b*x])^(1/3),x]

[Out]

(Sqrt[3]*c^(1/3)*ArcTan[(c^(2/3) - 2*(c*Cot[a + b*x])^(2/3))/(Sqrt[3]*c^(2/3))])/(2*b) + (c^(1/3)*Log[c^(2/3)
+ (c*Cot[a + b*x])^(2/3)])/(2*b) - (c^(1/3)*Log[c^(4/3) - c^(2/3)*(c*Cot[a + b*x])^(2/3) + (c*Cot[a + b*x])^(4
/3)])/(4*b)

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \sqrt [3]{c \cot (a+b x)} \, dx &=-\frac{c \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{c^2+x^2} \, dx,x,c \cot (a+b x)\right )}{b}\\ &=-\frac{(3 c) \operatorname{Subst}\left (\int \frac{x^3}{c^2+x^6} \, dx,x,\sqrt [3]{c \cot (a+b x)}\right )}{b}\\ &=-\frac{(3 c) \operatorname{Subst}\left (\int \frac{x}{c^2+x^3} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{2 b}\\ &=\frac{\sqrt [3]{c} \operatorname{Subst}\left (\int \frac{1}{c^{2/3}+x} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \operatorname{Subst}\left (\int \frac{c^{2/3}+x}{c^{4/3}-c^{2/3} x+x^2} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{2 b}\\ &=\frac{\sqrt [3]{c} \log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \operatorname{Subst}\left (\int \frac{-c^{2/3}+2 x}{c^{4/3}-c^{2/3} x+x^2} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{4 b}-\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{c^{4/3}-c^{2/3} x+x^2} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{4 b}\\ &=\frac{\sqrt [3]{c} \log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \log \left (c^{4/3}-c^{2/3} (c \cot (a+b x))^{2/3}+(c \cot (a+b x))^{4/3}\right )}{4 b}-\frac{\left (3 \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 (c \cot (a+b x))^{2/3}}{c^{2/3}}\right )}{2 b}\\ &=\frac{\sqrt{3} \sqrt [3]{c} \tan ^{-1}\left (\frac{1-\frac{2 (c \cot (a+b x))^{2/3}}{c^{2/3}}}{\sqrt{3}}\right )}{2 b}+\frac{\sqrt [3]{c} \log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b}-\frac{\sqrt [3]{c} \log \left (c^{4/3}-c^{2/3} (c \cot (a+b x))^{2/3}+(c \cot (a+b x))^{4/3}\right )}{4 b}\\ \end{align*}

Mathematica [C]  time = 0.0395061, size = 40, normalized size = 0.31 \[ -\frac{3 (c \cot (a+b x))^{4/3} \text{Hypergeometric2F1}\left (\frac{2}{3},1,\frac{5}{3},-\cot ^2(a+b x)\right )}{4 b c} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cot[a + b*x])^(1/3),x]

[Out]

(-3*(c*Cot[a + b*x])^(4/3)*Hypergeometric2F1[2/3, 1, 5/3, -Cot[a + b*x]^2])/(4*b*c)

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Maple [A]  time = 0.028, size = 114, normalized size = 0.9 \begin{align*}{\frac{c}{2\,b}\ln \left ( \left ( c\cot \left ( bx+a \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{{c}^{2}} \right ){\frac{1}{\sqrt [3]{{c}^{2}}}}}-{\frac{c}{4\,b}\ln \left ( \left ( c\cot \left ( bx+a \right ) \right ) ^{{\frac{4}{3}}}-\sqrt [3]{{c}^{2}} \left ( c\cot \left ( bx+a \right ) \right ) ^{{\frac{2}{3}}}+ \left ({c}^{2} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{c}^{2}}}}}-{\frac{c\sqrt{3}}{2\,b}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{ \left ( c\cot \left ( bx+a \right ) \right ) ^{2/3}}{\sqrt [3]{{c}^{2}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{c}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*cot(b*x+a))^(1/3),x)

[Out]

1/2/b*c/(c^2)^(1/3)*ln((c*cot(b*x+a))^(2/3)+(c^2)^(1/3))-1/4/b*c/(c^2)^(1/3)*ln((c*cot(b*x+a))^(4/3)-(c^2)^(1/
3)*(c*cot(b*x+a))^(2/3)+(c^2)^(2/3))-1/2/b*c*3^(1/2)/(c^2)^(1/3)*arctan(1/3*3^(1/2)*(2/(c^2)^(1/3)*(c*cot(b*x+
a))^(2/3)-1))

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Maxima [A]  time = 1.63484, size = 162, normalized size = 1.24 \begin{align*} -\frac{{\left (\frac{2 \, \sqrt{3} \arctan \left (\frac{\sqrt{3}{\left (2 \, \left (\frac{c}{\tan \left (b x + a\right )}\right )^{\frac{2}{3}} -{\left (c^{2}\right )}^{\frac{1}{3}}\right )}}{3 \,{\left (c^{2}\right )}^{\frac{1}{3}}}\right )}{{\left (c^{2}\right )}^{\frac{1}{3}}} + \frac{\log \left (\left (\frac{c}{\tan \left (b x + a\right )}\right )^{\frac{4}{3}} -{\left (c^{2}\right )}^{\frac{1}{3}} \left (\frac{c}{\tan \left (b x + a\right )}\right )^{\frac{2}{3}} +{\left (c^{2}\right )}^{\frac{2}{3}}\right )}{{\left (c^{2}\right )}^{\frac{1}{3}}} - \frac{2 \, \log \left (\left (\frac{c}{\tan \left (b x + a\right )}\right )^{\frac{2}{3}} +{\left (c^{2}\right )}^{\frac{1}{3}}\right )}{{\left (c^{2}\right )}^{\frac{1}{3}}}\right )} c}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cot(b*x+a))^(1/3),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(c/tan(b*x + a))^(2/3) - (c^2)^(1/3))/(c^2)^(1/3))/(c^2)^(1/3) + log((c/
tan(b*x + a))^(4/3) - (c^2)^(1/3)*(c/tan(b*x + a))^(2/3) + (c^2)^(2/3))/(c^2)^(1/3) - 2*log((c/tan(b*x + a))^(
2/3) + (c^2)^(1/3))/(c^2)^(1/3))*c/b

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Fricas [B]  time = 1.67033, size = 560, normalized size = 4.27 \begin{align*} -\frac{2 \, \sqrt{3} c^{\frac{1}{3}} \arctan \left (-\frac{\sqrt{3} c - 2 \, \sqrt{3} c^{\frac{1}{3}} \left (\frac{c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}\right )^{\frac{2}{3}}}{3 \, c}\right ) - 2 \, c^{\frac{1}{3}} \log \left (c^{\frac{2}{3}} + \left (\frac{c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}\right )^{\frac{2}{3}}\right ) + c^{\frac{1}{3}} \log \left (\frac{c^{\frac{4}{3}} \sin \left (2 \, b x + 2 \, a\right ) - c^{\frac{2}{3}} \left (\frac{c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}\right )^{\frac{2}{3}} \sin \left (2 \, b x + 2 \, a\right ) +{\left (c \cos \left (2 \, b x + 2 \, a\right ) + c\right )} \left (\frac{c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}\right )^{\frac{1}{3}}}{\sin \left (2 \, b x + 2 \, a\right )}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cot(b*x+a))^(1/3),x, algorithm="fricas")

[Out]

-1/4*(2*sqrt(3)*c^(1/3)*arctan(-1/3*(sqrt(3)*c - 2*sqrt(3)*c^(1/3)*((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))
^(2/3))/c) - 2*c^(1/3)*log(c^(2/3) + ((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))^(2/3)) + c^(1/3)*log((c^(4/3)
*sin(2*b*x + 2*a) - c^(2/3)*((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))^(2/3)*sin(2*b*x + 2*a) + (c*cos(2*b*x
+ 2*a) + c)*((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a))^(1/3))/sin(2*b*x + 2*a)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cot(b*x+a))**(1/3),x)

[Out]

Integral((c*cot(a + b*x))**(1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \cot \left (b x + a\right )\right )^{\frac{1}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cot(b*x+a))^(1/3),x, algorithm="giac")

[Out]

integrate((c*cot(b*x + a))^(1/3), x)