3.238 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

4/63*b*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/9*arctanh(tanh(b*x+a))^(7/2)/x^(9/2)/
(b*x-arctanh(tanh(b*x+a)))

________________________________________________________________________________________

Rubi [A]  time = 0.03, 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 = {2171, 2167} \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(11/2),x]

[Out]

(4*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(
7/2))/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]]))

Rule 2167

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] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2171

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 {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(2 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{9 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 48, normalized size = 0.67 \[ \frac {2 \left (9 b x-7 \tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{7/2}}{63 x^{9/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(11/2),x]

[Out]

(2*(9*b*x - 7*ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(9/2)*(-(b*x) + ArcTanh[Tanh[a + b*x
]])^2)

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 56, normalized size = 0.78 \[ \frac {2 \, {\left (2 \, b^{4} x^{4} - a b^{3} x^{3} - 15 \, a^{2} b^{2} x^{2} - 19 \, a^{3} b x - 7 \, a^{4}\right )} \sqrt {b x + a}}{63 \, a^{2} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="fricas")

[Out]

2/63*(2*b^4*x^4 - a*b^3*x^3 - 15*a^2*b^2*x^2 - 19*a^3*b*x - 7*a^4)*sqrt(b*x + a)/(a^2*x^(9/2))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 59, normalized size = 0.82 \[ \frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{9}}{a^{2}} - \frac {9 \, \sqrt {2} b^{9}}{a}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{63 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="giac")

[Out]

1/63*sqrt(2)*(2*sqrt(2)*(b*x + a)*b^9/a^2 - 9*sqrt(2)*b^9/a)*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(9/2)*abs(
b))

________________________________________________________________________________________

maple [A]  time = 0.26, size = 59, normalized size = 0.82 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {4 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x)

[Out]

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(7/2)+4/63*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(7/2)
*arctanh(tanh(b*x+a))^(7/2)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 34, normalized size = 0.47 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} - 5 \, a b x - 7 \, a^{2}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{63 \, a^{2} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(11/2),x, algorithm="maxima")

[Out]

2/63*(2*b^2*x^2 - 5*a*b*x - 7*a^2)*(b*x + a)^(5/2)/(a^2*x^(9/2))

________________________________________________________________________________________

mupad [B]  time = 1.62, size = 293, normalized size = 4.07 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {19\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{63}-\frac {10\,b^2\,x^2}{21}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{18}+\frac {4\,b^3\,x^3}{63\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {16\,b^4\,x^4}{63\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(11/2),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)*((19*b*
x*(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))/63 - (1
0*b^2*x^2)/21 - (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)^2/18 + (4*b^3*x^3)/(63*(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)) + (16*b^4*x^4)/(63*(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)))/x^(9/2)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**(11/2),x)

[Out]

Timed out

________________________________________________________________________________________