3.20 \(\int \frac{1}{\sqrt [3]{c \cot (a+b x)}} \, dx\)

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

[Out]

(Sqrt[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*c
^(1/3))

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Rubi [A]  time = 0.0963444, 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, 200, 31, 634, 617, 204, 628} \[ -\frac{\log \left ((c \cot (a+b x))^{2/3}+c^{2/3}\right )}{2 b \sqrt [3]{c}}+\frac{\log \left (-c^{2/3} (c \cot (a+b x))^{2/3}+(c \cot (a+b x))^{4/3}+c^{4/3}\right )}{4 b \sqrt [3]{c}}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{c^{2/3}-2 (c \cot (a+b x))^{2/3}}{\sqrt{3} c^{2/3}}\right )}{2 b \sqrt [3]{c}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cot[a + b*x])^(-1/3),x]

[Out]

(Sqrt[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*c
^(1/3))

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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 \frac{1}{\sqrt [3]{c \cot (a+b x)}} \, dx &=-\frac{c \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{x} \left (c^2+x^2\right )} \, dx,x,c \cot (a+b x)\right )}{b}\\ &=-\frac{(3 c) \operatorname{Subst}\left (\int \frac{x}{c^2+x^6} \, dx,x,\sqrt [3]{c \cot (a+b x)}\right )}{b}\\ &=-\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{c^2+x^3} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{c^{2/3}+x} \, dx,x,(c \cot (a+b x))^{2/3}\right )}{2 b \sqrt [3]{c}}-\frac{\operatorname{Subst}\left (\int \frac{2 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 \sqrt [3]{c}}\\ &=-\frac{\log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b \sqrt [3]{c}}+\frac{\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 \sqrt [3]{c}}-\frac{\left (3 \sqrt [3]{c}\right ) \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{\log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b \sqrt [3]{c}}+\frac{\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 \sqrt [3]{c}}-\frac{3 \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 \sqrt [3]{c}}\\ &=\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 (c \cot (a+b x))^{2/3}}{c^{2/3}}}{\sqrt{3}}\right )}{2 b \sqrt [3]{c}}-\frac{\log \left (c^{2/3}+(c \cot (a+b x))^{2/3}\right )}{2 b \sqrt [3]{c}}+\frac{\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 \sqrt [3]{c}}\\ \end{align*}

Mathematica [A]  time = 0.150585, size = 98, normalized size = 0.75 \[ \frac{\sqrt [3]{\cot (a+b x)} \left (-2 \log \left (\cot ^{\frac{2}{3}}(a+b x)+1\right )+\log \left (\cot ^{\frac{4}{3}}(a+b x)-\cot ^{\frac{2}{3}}(a+b x)+1\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 \cot ^{\frac{2}{3}}(a+b x)-1}{\sqrt{3}}\right )\right )}{4 b \sqrt [3]{c \cot (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cot[a + b*x])^(-1/3),x]

[Out]

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

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Maple [A]  time = 0.024, 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 ) \left ({c}^{2} \right ) ^{-{\frac{2}{3}}}}+{\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 ) \left ({c}^{2} \right ) ^{-{\frac{2}{3}}}}-{\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 ) \left ({c}^{2} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*c/(c^2)^(2/3)*ln((c*cot(b*x+a))^(2/3)+(c^2)^(1/3))+1/4/b*c/(c^2)^(2/3)*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/(c^2)^(2/3)*3^(1/2)*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 = 1.57281, size = 163, normalized size = 1.24 \begin{align*} -\frac{c{\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{2}{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{2}{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{2}{3}}}\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.77121, size = 1639, normalized size = 12.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*cot(b*x + a))^(-1/3), x)