3.248 \(\int \frac {1}{x^{9/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\)

Optimal. Leaf size=148 \[ \frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

16/35*b^2*arctanh(tanh(b*x+a))^(1/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^3+12/35*b*arctanh(tanh(b*x+a))^(1/2)/x
^(5/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/7*arctanh(tanh(b*x+a))^(1/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))+32/35*b^
3*arctanh(tanh(b*x+a))^(1/2)/(b*x-arctanh(tanh(b*x+a)))^4/x^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(9/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(32*b^3*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*Sqrt[ArcTanh[Tan
h[a + b*x]]])/(35*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (12*b*Sqrt[ArcTanh[Tanh[a + b*x]]])/(35*x^(5/2)*
(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*Sqrt[ArcTanh[Tanh[a + b*x]]])/(7*x^(7/2)*(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^{9/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(6 b) \int \frac {1}{x^{7/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{7 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (24 b^2\right ) \int \frac {1}{x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (16 b^3\right ) \int \frac {1}{x^{3/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{35 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {32 b^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{35 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{7 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.55 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-35 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+21 b x \tanh ^{-1}(\tanh (a+b x))^2-5 \tanh ^{-1}(\tanh (a+b x))^3+35 b^3 x^3\right )}{35 x^{7/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(9/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]),x]

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(35*b^3*x^3 - 35*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 21*b*x*ArcTanh[Tanh[a + b*x]
]^2 - 5*ArcTanh[Tanh[a + b*x]]^3))/(35*x^(7/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.71, size = 45, normalized size = 0.30 \[ \frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{4} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x + a)/(a^4*x^(7/2))

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giac [A]  time = 0.15, size = 103, normalized size = 0.70 \[ \frac {64 \, {\left (35 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 21 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 7 \, a^{2} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3}\right )} b^{\frac {7}{2}}}{35 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

64/35*(35*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 - 21*a*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 7*a^2*(sqrt(b)*sqrt
(x) - sqrt(b*x + a))^2 - a^3)*b^(7/2)/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7

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maple [A]  time = 0.30, size = 151, normalized size = 1.02 \[ -\frac {2 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {12 b \left (-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {2 b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {x}}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(9/2)/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

-2/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(1/2)-12/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(ar
ctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(1/2)-4/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(arctanh(tan
h(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(tanh(b*x+
a))^(1/2)))

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maxima [A]  time = 0.43, size = 55, normalized size = 0.37 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 2 \, a^{2} b^{2} x^{2} + a^{3} b x - 5 \, a^{4}\right )}}{35 \, \sqrt {b x + a} a^{4} x^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(9/2)/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

2/35*(16*b^4*x^4 + 8*a*b^3*x^3 - 2*a^2*b^2*x^2 + a^3*b*x - 5*a^4)/(sqrt(b*x + a)*a^4*x^(7/2))

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mupad [B]  time = 1.64, size = 287, normalized size = 1.94 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {4}{7\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,b^2\,x^2}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^3\,x^3}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}+\frac {48\,b\,x}{35\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}\right )}{x^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(9/2)*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/(7*(
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)/(35*(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) + (512*b^3*x^3)/(35*(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)^4) + (48*b*x)/(35*(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^(7/2)

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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(1/x**(9/2)/atanh(tanh(b*x+a))**(1/2),x)

[Out]

Timed out

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