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

Optimal. Leaf size=110 \[ \frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

8/15*b*arctanh(tanh(b*x+a))^(1/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/5*arctanh(tanh(b*x+a))^(1/2)/x^(5/2)/
(b*x-arctanh(tanh(b*x+a)))+16/15*b^2*arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^3/x^(1/2)

________________________________________________________________________________________

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} \[ \frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(16*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(15*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (8*b*Sqrt[ArcTanh[Tanh[a
+ b*x]]])/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(5*x^(5/2)*(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 {1}{x^{7/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(4 b) \int \frac {1}{x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {8 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {1}{x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {8 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 66, normalized size = 0.60 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-10 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+15 b^2 x^2\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(15*b^2*x^2 - 10*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15
*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 34, normalized size = 0.31 \[ -\frac {2 \, {\left (8 \, b^{2} x^{2} - 4 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 77, normalized size = 0.70 \[ \frac {32 \, {\left (10 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 5 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + a^{2}\right )} b^{\frac {5}{2}}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/15*(10*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 5*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + a^2)*b^(5/2)/((sqrt(
b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5

________________________________________________________________________________________

maple [A]  time = 0.30, size = 105, normalized size = 0.95 \[ -\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {8 b \left (-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-8/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(arc
tanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh(tanh(b*x+a))^(1/2)+2/3*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(1/2)*arctanh(t
anh(b*x+a))^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 45, normalized size = 0.41 \[ -\frac {2 \, {\left (8 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - a^{2} b x + 3 \, a^{3}\right )}}{15 \, \sqrt {b x + a} a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.53, size = 227, normalized size = 2.06 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{5\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,b^2\,x^2}{15\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {32\,b\,x}{15\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*atanh(tanh(a + b*x))^(1/2)),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)*(4/(5*(
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)) + (128*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)^
3) + (32*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)))/x^(5/2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________