3.190 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {32 b^3 \sqrt {x}}{5} \]

[Out]

-4/5*b*arctanh(tanh(b*x+a))^2/x^(3/2)-2/5*arctanh(tanh(b*x+a))^3/x^(5/2)-16/5*b^2*arctanh(tanh(b*x+a))/x^(1/2)
+32/5*b^3*x^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 30} \[ -\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {32 b^3 \sqrt {x}}{5} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]

[Out]

(32*b^3*Sqrt[x])/5 - (16*b^2*ArcTanh[Tanh[a + b*x]])/(5*Sqrt[x]) - (4*b*ArcTanh[Tanh[a + b*x]]^2)/(5*x^(3/2))
- (2*ArcTanh[Tanh[a + b*x]]^3)/(5*x^(5/2))

Rule 30

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

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 {\tanh ^{-1}(\tanh (a+b x))^3}{x^{7/2}} \, dx &=-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} (6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^2}{x^{5/2}} \, dx\\ &=-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} \left (8 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))}{x^{3/2}} \, dx\\ &=-\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}+\frac {1}{5} \left (16 b^3\right ) \int \frac {1}{\sqrt {x}} \, dx\\ &=\frac {32 b^3 \sqrt {x}}{5}-\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))}{5 \sqrt {x}}-\frac {4 b \tanh ^{-1}(\tanh (a+b x))^2}{5 x^{3/2}}-\frac {2 \tanh ^{-1}(\tanh (a+b x))^3}{5 x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.83 \[ \frac {2 \left (-8 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-2 b x \tanh ^{-1}(\tanh (a+b x))^2-\tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^3/x^(7/2),x]

[Out]

(2*(16*b^3*x^3 - 8*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 2*b*x*ArcTanh[Tanh[a + b*x]]^2 - ArcTanh[Tanh[a + b*x]]^3)
)/(5*x^(5/2))

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fricas [A]  time = 0.54, size = 35, normalized size = 0.51 \[ \frac {2 \, {\left (5 \, b^{3} x^{3} - 15 \, a b^{2} x^{2} - 5 \, a^{2} b x - a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="fricas")

[Out]

2/5*(5*b^3*x^3 - 15*a*b^2*x^2 - 5*a^2*b*x - a^3)/x^(5/2)

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giac [A]  time = 0.17, size = 34, normalized size = 0.49 \[ 2 \, b^{3} \sqrt {x} - \frac {2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x + a^{3}\right )}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="giac")

[Out]

2*b^3*sqrt(x) - 2/5*(15*a*b^2*x^2 + 5*a^2*b*x + a^3)/x^(5/2)

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maple [A]  time = 0.26, size = 56, normalized size = 0.81 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{5 x^{\frac {5}{2}}}+\frac {12 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{3 x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )}{\sqrt {x}}+2 b \sqrt {x}\right )}{3}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^3/x^(7/2),x)

[Out]

-2/5*arctanh(tanh(b*x+a))^3/x^(5/2)+12/5*b*(-1/3*arctanh(tanh(b*x+a))^2/x^(3/2)+4/3*b*(-arctanh(tanh(b*x+a))/x
^(1/2)+2*b*x^(1/2)))

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maxima [A]  time = 0.36, size = 55, normalized size = 0.80 \[ \frac {16}{5} \, {\left (2 \, b^{2} \sqrt {x} - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )}{\sqrt {x}}\right )} b - \frac {4 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{5 \, x^{\frac {3}{2}}} - \frac {2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^3/x^(7/2),x, algorithm="maxima")

[Out]

16/5*(2*b^2*sqrt(x) - b*arctanh(tanh(b*x + a))/sqrt(x))*b - 4/5*b*arctanh(tanh(b*x + a))^2/x^(3/2) - 2/5*arcta
nh(tanh(b*x + a))^3/x^(5/2)

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mupad [B]  time = 1.18, size = 182, normalized size = 2.64 \[ \frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{20\,x^{5/2}}+2\,b^3\,\sqrt {x}+\frac {3\,b^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{\sqrt {x}}-\frac {b\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^3/x^(7/2),x)

[Out]

(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/(20*x^(5
/2)) + 2*b^3*x^(1/2) + (3*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*
b*x) + 1)) + 2*b*x))/x^(1/2) - (b*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(2*b*x) + 1)) + 2*b*x)^2)/(2*x^(3/2))

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sympy [A]  time = 155.33, size = 70, normalized size = 1.01 \[ \frac {32 b^{3} \sqrt {x}}{5} - \frac {16 b^{2} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}{5 \sqrt {x}} - \frac {4 b \operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {3}{2}}} - \frac {2 \operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}{5 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**3/x**(7/2),x)

[Out]

32*b**3*sqrt(x)/5 - 16*b**2*atanh(tanh(a + b*x))/(5*sqrt(x)) - 4*b*atanh(tanh(a + b*x))**2/(5*x**(3/2)) - 2*at
anh(tanh(a + b*x))**3/(5*x**(5/2))

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