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

Optimal. Leaf size=148 \[ \frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{315 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{21 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

32/315*b^3*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^4+16/105*b^2*arctanh(tanh(b*x+a))^(3/
2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^3+4/21*b*arctanh(tanh(b*x+a))^(3/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^2
+2/9*arctanh(tanh(b*x+a))^(3/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))

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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} \[ \frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{315 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{21 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(315*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*ArcTanh[Tanh[a
+ b*x]]^(3/2))/(105*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(21*x^(7/2)
*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(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 {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{11/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(2 b) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{9/2}} \, dx}{3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{21 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{7/2}} \, dx}{21 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{21 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (16 b^3\right ) \int \frac {\sqrt {\tanh ^{-1}(\tanh (a+b x))}}{x^{5/2}} \, dx}{105 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{315 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{105 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {4 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}{21 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{9 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 82, normalized size = 0.55 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-189 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+135 b x \tanh ^{-1}(\tanh (a+b x))^2-35 \tanh ^{-1}(\tanh (a+b x))^3+105 b^3 x^3\right )}{315 x^{9/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(105*b^3*x^3 - 189*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 135*b*x*ArcTanh[Tanh[a + b
*x]]^2 - 35*ArcTanh[Tanh[a + b*x]]^3))/(315*x^(9/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.75, size = 56, normalized size = 0.38 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="fricas")

[Out]

2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt(b*x + a)/(a^4*x^(9/2))

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giac [A]  time = 0.20, size = 166, normalized size = 1.12 \[ \frac {64 \, {\left (315 \, b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{10} + 189 \, a b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} + 84 \, a^{2} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 36 \, a^{3} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 9 \, a^{4} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{5} b^{\frac {9}{2}}\right )}}{315 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="giac")

[Out]

64/315*(315*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^10 + 189*a*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 +
84*a^2*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 - 36*a^3*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 9*a
^4*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^5*b^(9/2))/((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^9

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maple [A]  time = 0.30, size = 151, normalized size = 1.02 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x)

[Out]

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(3/2)-4/3*b/(arctanh(tanh(b*x+a))-b*x)*(-1/7/(arc
tanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh(tanh(b*x+a))^(3/2)-4/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(arctanh(tanh
(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+2/15*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(3/2)*arctanh(tanh(b*x+
a))^(3/2)))

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maxima [A]  time = 0.43, size = 56, normalized size = 0.38 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="maxima")

[Out]

2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt(b*x + a)/(a^4*x^(9/2))

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mupad [B]  time = 1.52, size = 294, normalized size = 1.99 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,b^2\,x^2}{105\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^3\,x^3}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^4\,x^4}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}+\frac {4\,b\,x}{63\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {2}{9}\right )}{x^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(1/2)/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)*((16*b^
2*x^2)/(105*(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) + (128*b^3*x^3)/(315*(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^4*x^4)/(315*(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) + (4*b*x)/(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)) - 2/9))/x^(9/2)

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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))**(1/2)/x**(11/2),x)

[Out]

Timed out

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