3.11 \(\int \sqrt [3]{b \coth (c+d x)} \, dx\)

Optimal. Leaf size=132 \[ -\frac {\sqrt [3]{b} \log \left (b^{2/3}-(b \coth (c+d x))^{2/3}\right )}{2 d}+\frac {\sqrt [3]{b} \log \left (b^{2/3} (b \coth (c+d x))^{2/3}+b^{4/3}+(b \coth (c+d x))^{4/3}\right )}{4 d}-\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {b^{2/3}+2 (b \coth (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3476, 329, 275, 292, 31, 634, 617, 204, 628} \[ -\frac {\sqrt [3]{b} \log \left (b^{2/3}-(b \coth (c+d x))^{2/3}\right )}{2 d}+\frac {\sqrt [3]{b} \log \left (b^{2/3} (b \coth (c+d x))^{2/3}+b^{4/3}+(b \coth (c+d x))^{4/3}\right )}{4 d}-\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {b^{2/3}+2 (b \coth (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(1/3),x]

[Out]

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

Rule 31

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

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 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 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 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 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]

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.04, size = 38, normalized size = 0.29 \[ \frac {3 (b \coth (c+d x))^{4/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\coth ^2(c+d x)\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(1/3),x]

[Out]

(3*(b*Coth[c + d*x])^(4/3)*Hypergeometric2F1[2/3, 1, 5/3, Coth[c + d*x]^2])/(4*b*d)

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fricas [B]  time = 0.41, size = 291, normalized size = 2.20 \[ -\frac {2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b - 2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}}{3 \, b}\right ) - 2 \, \left (-b\right )^{\frac {1}{3}} \log \left (-\left (-b\right )^{\frac {2}{3}} + \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}}\right ) + \left (-b\right )^{\frac {1}{3}} \log \left (\frac {{\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \left (-b\right )^{\frac {2}{3}} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {2}{3}} - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \left (-b\right )^{\frac {1}{3}} + {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + b\right )} \left (\frac {b \cosh \left (d x + c\right )}{\sinh \left (d x + c\right )}\right )^{\frac {1}{3}}}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(-1/3*(sqrt(3)*b - 2*sqrt(3)*(-b)^(1/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3)
)/b) - 2*(-b)^(1/3)*log(-(-b)^(2/3) + (b*cosh(d*x + c)/sinh(d*x + c))^(2/3)) + (-b)^(1/3)*log(((cosh(d*x + c)^
2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*(-b)^(2/3)*(b*cosh(d*x + c)/sinh(d*x + c))^(2/3) - (b
*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)*(-b)^(1/3) + (b*cosh(d*x + c)^2 +
2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)*(b*cosh(d*x + c)/sinh(d*x + c))^(1/3))/(cosh(d*x + c)
^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)))/d

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giac [B]  time = 0.37, size = 217, normalized size = 1.64 \[ -\frac {b {\left (\frac {2 \, \sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}}{3 \, {\left | b \right |}^{\frac {2}{3}}}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left (\left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} {\left | b \right |}^{\frac {2}{3}} + {\left | b \right |}^{\frac {4}{3}} + \frac {{\left (b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {1}{3}}}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )}{b^{2}} + \frac {2 \, {\left | b \right |}^{\frac {4}{3}} \log \left ({\left | \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + b}{e^{\left (2 \, d x + 2 \, c\right )} - 1}\right )^{\frac {2}{3}} - {\left | b \right |}^{\frac {2}{3}} \right |}\right )}{b^{2}}\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*(2*sqrt(3)*abs(b)^(4/3)*arctan(1/3*sqrt(3)*(2*((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(2/3) + a
bs(b)^(2/3))/abs(b)^(2/3))/b^2 - abs(b)^(4/3)*log(((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(2/3)*abs(b)
^(2/3) + abs(b)^(4/3) + (b*e^(2*d*x + 2*c) + b)*((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(1/3)/(e^(2*d*
x + 2*c) - 1))/b^2 + 2*abs(b)^(4/3)*log(abs(((b*e^(2*d*x + 2*c) + b)/(e^(2*d*x + 2*c) - 1))^(2/3) - abs(b)^(2/
3)))/b^2)/d

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maple [A]  time = 0.10, size = 115, normalized size = 0.87 \[ -\frac {b \ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{2 d \left (b^{2}\right )^{\frac {1}{3}}}+\frac {b \ln \left (\left (b \coth \left (d x +c \right )\right )^{\frac {4}{3}}+\left (b^{2}\right )^{\frac {1}{3}} \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{4 d \left (b^{2}\right )^{\frac {1}{3}}}-\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \coth \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 d \left (b^{2}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c))^(1/3),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \coth \left (d x + c\right )\right )^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*coth(d*x + c))^(1/3), x)

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mupad [B]  time = 1.52, size = 146, normalized size = 1.11 \[ \frac {{\left (-b\right )}^{1/3}\,\ln \left (81\,{\left (-b\right )}^{16/3}\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}-81\,b^6\right )}{2\,d}-\frac {{\left (-b\right )}^{1/3}\,\ln \left (-\frac {81\,b^6}{d^4}-\frac {81\,{\left (-b\right )}^{16/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {{\left (-b\right )}^{1/3}\,\ln \left (-\frac {81\,b^6}{d^4}+\frac {162\,{\left (-b\right )}^{16/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x))^(1/3),x)

[Out]

((-b)^(1/3)*log(81*(-b)^(16/3)*(b*coth(c + d*x))^(2/3) - 81*b^6))/(2*d) - ((-b)^(1/3)*log(- (81*b^6)/d^4 - (81
*(-b)^(16/3)*((3^(1/2)*1i)/2 + 1/2)*(b*coth(c + d*x))^(2/3))/d^4)*((3^(1/2)*1i)/2 + 1/2))/(2*d) + ((-b)^(1/3)*
log((162*(-b)^(16/3)*((3^(1/2)*1i)/4 - 1/4)*(b*coth(c + d*x))^(2/3))/d^4 - (81*b^6)/d^4)*((3^(1/2)*1i)/4 - 1/4
))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt [3]{b \coth {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))**(1/3),x)

[Out]

Integral((b*coth(c + d*x))**(1/3), x)

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