3.239 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

16/693*b^2*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^3+8/99*b*arctanh(tanh(b*x+a))^(7/2)/x
^(9/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/11*arctanh(tanh(b*x+a))^(7/2)/x^(11/2)/(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]  time = 0.06, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(13/2),x]

[Out]

(16*b^2*ArcTanh[Tanh[a + b*x]]^(7/2))/(693*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*ArcTanh[Tanh[a + b
*x]]^(7/2))/(99*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(7/2))/(11*x^(11/2)*(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^{13/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(4 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx}{11 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{99 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{693 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{99 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{11 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 66, normalized size = 0.60 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-154 b x \tanh ^{-1}(\tanh (a+b x))+63 \tanh ^{-1}(\tanh (a+b x))^2+99 b^2 x^2\right )}{693 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(13/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(99*b^2*x^2 - 154*b*x*ArcTanh[Tanh[a + b*x]] + 63*ArcTanh[Tanh[a + b*x]]^2))/(
693*x^(11/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3)

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fricas [A]  time = 0.57, size = 67, normalized size = 0.61 \[ -\frac {2 \, {\left (8 \, b^{5} x^{5} - 4 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{3} + 113 \, a^{3} b^{2} x^{2} + 161 \, a^{4} b x + 63 \, a^{5}\right )} \sqrt {b x + a}}{693 \, a^{3} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(13/2),x, algorithm="fricas")

[Out]

-2/693*(8*b^5*x^5 - 4*a*b^4*x^4 + 3*a^2*b^3*x^3 + 113*a^3*b^2*x^2 + 161*a^4*b*x + 63*a^5)*sqrt(b*x + a)/(a^3*x
^(11/2))

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giac [A]  time = 0.20, size = 78, normalized size = 0.71 \[ -\frac {\sqrt {2} {\left (\frac {99 \, \sqrt {2} b^{11}}{a} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{11}}{a^{3}} - \frac {11 \, \sqrt {2} b^{11}}{a^{2}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{693 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {11}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(13/2),x, algorithm="giac")

[Out]

-1/693*sqrt(2)*(99*sqrt(2)*b^11/a + 4*(2*sqrt(2)*(b*x + a)*b^11/a^3 - 11*sqrt(2)*b^11/a^2)*(b*x + a))*(b*x + a
)^(7/2)*b/(((b*x + a)*b - a*b)^(11/2)*abs(b))

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maple [A]  time = 0.30, size = 105, normalized size = 0.95 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {8 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(13/2),x)

[Out]

-2/11/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*arctanh(tanh(b*x+a))^(7/2)-8/11*b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/(
arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(7/2)+2/63*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(7/2)*arcta
nh(tanh(b*x+a))^(7/2))

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maxima [A]  time = 0.42, size = 45, normalized size = 0.41 \[ -\frac {2 \, {\left (8 \, b^{3} x^{3} - 20 \, a b^{2} x^{2} + 35 \, a^{2} b x + 63 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{693 \, a^{3} x^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(13/2),x, algorithm="maxima")

[Out]

-2/693*(8*b^3*x^3 - 20*a*b^2*x^2 + 35*a^2*b*x + 63*a^3)*(b*x + a)^(5/2)/(a^3*x^(11/2))

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mupad [B]  time = 1.66, size = 353, normalized size = 3.21 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {23\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{99}-\frac {226\,b^2\,x^2}{693}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{22}+\frac {4\,b^3\,x^3}{231\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {32\,b^4\,x^4}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^5\,x^5}{693\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(13/2),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)*((23*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))/99 - (2
26*b^2*x^2)/693 - (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)^2/22 + (4*b^3*x^3)/(231*(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)) + (32*b^4*x^4)/(693*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2) + (128*b^5*x^5)/(693*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*e
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3)))/x^(11/2)

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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**(13/2),x)

[Out]

Timed out

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