3.3.20 \(\int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx\) [220]

Optimal. Leaf size=72 \[ \frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

4/15*b*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/5*arctanh(tanh(b*x+a))^(3/2)/x^(5/2)/
(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]
time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2202, 2198} \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
3/2))/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

Rule 2198

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-u^(m + 1))*(
v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] &&
EqQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2202

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-u^(m + 1))*(
v^(n + 1)/((m + 1)*(b*u - a*v))), x] + Dist[b*((m + n + 2)/((m + 1)*(b*u - a*v))), Int[u^(m + 1)*v^n, x], x] /
; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(2 b) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 48, normalized size = 0.67 \begin {gather*} \frac {2 \left (5 b x-3 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}{15 x^{5/2} \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*b*x - 3*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(5/2)*(-(b*x) + ArcTanh[Tanh[a + b*x
]])^2)

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Maple [A]
time = 0.12, size = 59, normalized size = 0.82

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\)	\(59\)
default	\(-\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+4/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)
*arctanh(tanh(b*x+a))^(3/2)

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Maxima [A]
time = 0.47, size = 34, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 34, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(1/2)/x**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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Giac [A]
time = 0.39, size = 112, normalized size = 1.56 \begin {gather*} \frac {8 \, {\left (15 \, b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 5 \, a b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 5 \, a^{2} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

8/15*(15*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 + 5*a*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 5*a^2
*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^3*b^(5/2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5

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Mupad [B]
time = 1.41, size = 174, normalized size = 2.42 \begin {gather*} \frac {\sqrt {\frac {\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}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,b^2\,x^2}{15\,{\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}+\frac {4\,b\,x}{15\,\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 )}-\frac {2}{5}\right )}{x^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2)/x^(7/2),x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((16*b^
2*x^2)/(15*(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)
^2) + (4*b*x)/(15*(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)) - 2/5))/x^(5/2)

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