3.256 \(\int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx\)

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

[Out]

4/5*b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(1/2)+2/5/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))/a
rctanh(tanh(b*x+a))^(1/2)+16/5*b^2/(b*x-arctanh(tanh(b*x+a)))^3/x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-32/5*b^3*x^
(1/2)/(b*x-arctanh(tanh(b*x+a)))^4/arctanh(tanh(b*x+a))^(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} \[ -\frac {32 b^3 \sqrt {x}}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-32*b^3*Sqrt[x])/(5*(b*x - ArcTanh[Tanh[a + b*x]])^4*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*b^2)/(5*Sqrt[x]*(b*x
- ArcTanh[Tanh[a + b*x]])^3*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (4*b)/(5*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2
*Sqrt[ArcTanh[Tanh[a + b*x]]]) + 2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])*Sqrt[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} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {(6 b) \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {\left (8 b^2\right ) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {\left (16 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {32 b^3 \sqrt {x}}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {4 b}{5 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 80, normalized size = 0.54 \[ -\frac {2 \left (15 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-5 b x \tanh ^{-1}(\tanh (a+b x))^2+\tanh ^{-1}(\tanh (a+b x))^3+5 b^3 x^3\right )}{5 x^{5/2} \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(3/2)),x]

[Out]

(-2*(5*b^3*x^3 + 15*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 5*b*x*ArcTanh[Tanh[a + b*x]]^2 + ArcTanh[Tanh[a + b*x]]^3
))/(5*x^(5/2)*Sqrt[ArcTanh[Tanh[a + b*x]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.58, size = 58, normalized size = 0.39 \[ -\frac {2 \, {\left (16 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt {b x + a} \sqrt {x}}{5 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 161, normalized size = 1.09 \[ -\frac {2 \, b^{3} \sqrt {x}}{\sqrt {b x + a} a^{4}} + \frac {4 \, {\left (5 \, b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} - 30 \, a b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 80 \, a^{2} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 50 \, a^{3} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 11 \, a^{4} b^{\frac {5}{2}}\right )}}{5 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

-2*b^3*sqrt(x)/(sqrt(b*x + a)*a^4) + 4/5*(5*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 - 30*a*b^(5/2)*(sqrt(b
)*sqrt(x) - sqrt(b*x + a))^6 + 80*a^2*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 - 50*a^3*b^(5/2)*(sqrt(b)*sq
rt(x) - sqrt(b*x + a))^2 + 11*a^4*b^(5/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^5*a^3)

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maple [A]  time = 0.24, size = 151, normalized size = 1.02 \[ -\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {12 b \left (-\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {4 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 b \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)/arctanh(tanh(b*x+a))^(1/2)-12/5*b/(arctanh(tanh(b*x+a))-b*x)*(-1/3/(ar
ctanh(tanh(b*x+a))-b*x)/x^(3/2)/arctanh(tanh(b*x+a))^(1/2)-4/3*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh(tanh(
b*x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-2*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 = 54, normalized size = 0.36 \[ -\frac {2 \, {\left (16 \, b^{4} x^{4} + 24 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - a^{3} b x + a^{4}\right )}}{5 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

-2/5*(16*b^4*x^4 + 24*a*b^3*x^3 + 6*a^2*b^2*x^2 - a^3*b*x + a^4)/((b*x + a)^(3/2)*a^4*x^(5/2))

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mupad [B]  time = 1.77, size = 346, normalized size = 2.34 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,x}{5\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {4}{5\,b\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {128\,b\,x^2}{5\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}-\frac {512\,b^2\,x^3}{5\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}\right )}{x^{7/2}-\frac {x^{5/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*atanh(tanh(a + b*x))^(3/2)),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*x)
/(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)^2) + 4
/(5*b*(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*x^2)/(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)^3) - (512*b^2*x^3)/(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)^4)))/(x^(7/2) - (x^(5/2)*(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*b))

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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**(7/2)/atanh(tanh(b*x+a))**(3/2),x)

[Out]

Timed out

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