3.2.60 \(\int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\) [160]

Optimal. Leaf size=59 \[ -\frac {2 x^2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {8 x}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^3} \]

[Out]

-2/3*x^2/b/arctanh(tanh(b*x+a))^(3/2)-8/3*x/b^2/arctanh(tanh(b*x+a))^(1/2)+16/3*arctanh(tanh(b*x+a))^(1/2)/b^3

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Rubi [A]
time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2199, 2188, 30} \begin {gather*} \frac {16 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{3 b^3}-\frac {8 x}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x^2}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

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

Rule 30

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

Rule 2188

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 2199

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

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Mathematica [A]
time = 0.03, size = 48, normalized size = 0.81 \begin {gather*} -\frac {2 \left (b^2 x^2+4 b x \tanh ^{-1}(\tanh (a+b x))-8 \tanh ^{-1}(\tanh (a+b x))^2\right )}{3 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

(-2*(b^2*x^2 + 4*b*x*ArcTanh[Tanh[a + b*x]] - 8*ArcTanh[Tanh[a + b*x]]^2))/(3*b^3*ArcTanh[Tanh[a + b*x]]^(3/2)
)

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Maple [A]
time = 0.07, size = 91, normalized size = 1.54

method	result	size
default	\(\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (a^{2}+2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{3}}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/b^3*(arctanh(tanh(b*x+a))^(1/2)-1/3*(a^2+2*a*(arctanh(tanh(b*x+a))-b*x-a)+(arctanh(tanh(b*x+a))-b*x-a)^2)/ar
ctanh(tanh(b*x+a))^(3/2)-(-2*arctanh(tanh(b*x+a))+2*b*x)/arctanh(tanh(b*x+a))^(1/2))

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Maxima [A]
time = 0.55, size = 42, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (3 \, b^{3} x^{3} + 15 \, a b^{2} x^{2} + 20 \, a^{2} b x + 8 \, a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/3*(3*b^3*x^3 + 15*a*b^2*x^2 + 20*a^2*b*x + 8*a^3)/((b*x + a)^(5/2)*b^3)

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Fricas [A]
time = 0.33, size = 52, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} x^{2} + 12 \, a b x + 8 \, a^{2}\right )} \sqrt {b x + a}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*b^2*x^2 + 12*a*b*x + 8*a^2)*sqrt(b*x + a)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

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Sympy [A]
time = 71.90, size = 71, normalized size = 1.20 \begin {gather*} \begin {cases} - \frac {2 x^{2}}{3 b \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}} - \frac {8 x}{3 b^{2} \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}} + \frac {16 \sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}{3 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Piecewise((-2*x**2/(3*b*atanh(tanh(a + b*x))**(3/2)) - 8*x/(3*b**2*sqrt(atanh(tanh(a + b*x)))) + 16*sqrt(atanh
(tanh(a + b*x)))/(3*b**3), Ne(b, 0)), (x**3/(3*atanh(tanh(a))**(5/2)), True))

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Giac [A]
time = 0.40, size = 39, normalized size = 0.66 \begin {gather*} \frac {2 \, \sqrt {b x + a}}{b^{3}} + \frac {2 \, {\left (6 \, {\left (b x + a\right )} a - a^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

2*sqrt(b*x + a)/b^3 + 2/3*(6*(b*x + a)*a - a^2)/((b*x + a)^(3/2)*b^3)

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Mupad [B]
time = 1.26, size = 259, normalized size = 4.39 \begin {gather*} \frac {8\,\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 (-b^2\,x^2-2\,b\,x\,\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\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2-4\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,{\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\right )}{3\,b^3\,{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/atanh(tanh(a + b*x))^(5/2),x)

[Out]

(8*(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)*(2*lo
g(2/(exp(2*a)*exp(2*b*x) + 1))^2 - 4*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(2/(exp(2*a)*ex
p(2*b*x) + 1)) + 2*log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2 - b^2*x^2 - 2*b*x*log((2*exp(2*a)*
exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x*log(2/(exp(2*a)*exp(2*b*x) + 1))))/(3*b^3*(log((2*exp(2*a)*exp(
2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))^2)

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