3.179 \(\int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\)

Optimal. Leaf size=34 \[ -\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

[Out]

-1/2*x/b/arccoth(tanh(b*x+a))^2-1/2/b^2/arccoth(tanh(b*x+a))

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2168, 2157, 30} \[ -\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-x/(2*b*ArcCoth[Tanh[a + b*x]]^2) - 1/(2*b^2*ArcCoth[Tanh[a + b*x]])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {x}{\coth ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {\int \frac {1}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 b}\\ &=-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{2 b^2}\\ &=-\frac {x}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {1}{2 b^2 \coth ^{-1}(\tanh (a+b x))}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 0.79 \[ -\frac {\coth ^{-1}(\tanh (a+b x))+b x}{2 b^2 \coth ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

-1/2*(b*x + ArcCoth[Tanh[a + b*x]])/(b^2*ArcCoth[Tanh[a + b*x]]^2)

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fricas [B]  time = 0.84, size = 124, normalized size = 3.65 \[ -\frac {2 \, {\left (8 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 16 \, a^{2} b x + \pi ^{2} a + 4 \, a^{3}\right )}}{16 \, b^{6} x^{4} + 64 \, a b^{5} x^{3} + \pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} + 16 \, a^{4} b^{2} + 8 \, {\left (\pi ^{2} b^{4} + 12 \, a^{2} b^{4}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b^{3} + 4 \, a^{3} b^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

-2*(8*b^3*x^3 + 20*a*b^2*x^2 + 16*a^2*b*x + pi^2*a + 4*a^3)/(16*b^6*x^4 + 64*a*b^5*x^3 + pi^4*b^2 + 8*pi^2*a^2
*b^2 + 16*a^4*b^2 + 8*(pi^2*b^4 + 12*a^2*b^4)*x^2 + 16*(pi^2*a*b^3 + 4*a^3*b^3)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

integrate(x/arccoth(tanh(b*x + a))^3, x)

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maple [C]  time = 0.27, size = 634, normalized size = 18.65 \[ -\frac {2 i \left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2 \pi +4 i \ln \left ({\mathrm e}^{b x +a}\right )+4 i b x \right )}{b^{2} \left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{b x +a}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 b x +2 a}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 b x +2 a}}{{\mathrm e}^{2 b x +2 a}+1}\right )^{3}+2 \pi +4 i \ln \left ({\mathrm e}^{b x +a}\right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/arccoth(tanh(b*x+a))^3,x)

[Out]

-2*I*(-2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x
+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*Pi*csgn(
I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*
x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi
*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+2*Pi+4*I*ln(exp(b*x+a))+4*I*b*x)/b^2/(-2*Pi*csgn(I/(exp(2*b*x+2*a
)+1))^2+Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-Pi*csgn
(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+2*Pi*csgn(I/(exp(2*b*x+2*a)+1))^3+Pi*csgn(I
*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-2*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+Pi*csgn(I*exp(2*b*x+2*a
))^3-Pi*csgn(I*exp(2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x
+2*a)+1))^3+2*Pi+4*I*ln(exp(b*x+a)))^2

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maxima [C]  time = 1.11, size = 62, normalized size = 1.82 \[ -\frac {8 \, {\left (-i \, \pi + 4 \, b x + 2 \, a\right )}}{32 \, b^{4} x^{2} - 8 \, \pi ^{2} b^{2} - 32 i \, \pi a b^{2} + 32 \, a^{2} b^{2} + {\left (-32 i \, \pi b^{3} + 64 \, a b^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arccoth(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-8*(-I*pi + 4*b*x + 2*a)/(32*b^4*x^2 - 8*pi^2*b^2 - 32*I*pi*a*b^2 + 32*a^2*b^2 + (-32*I*pi*b^3 + 64*a*b^3)*x)

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mupad [B]  time = 0.09, size = 25, normalized size = 0.74 \[ -\frac {\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )+b\,x}{2\,b^2\,{\mathrm {acoth}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/acoth(tanh(a + b*x))^3,x)

[Out]

-(acoth(tanh(a + b*x)) + b*x)/(2*b^2*acoth(tanh(a + b*x))^2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/acoth(tanh(b*x+a))**3,x)

[Out]

Exception raised: TypeError

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