3.264 \(\int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=186 \[ \frac {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

16/15*b/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^2/arctanh(tanh(b*x+a))^(3/2)+2/5/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))
/arctanh(tanh(b*x+a))^(3/2)+32/5*b^2/(b*x-arctanh(tanh(b*x+a)))^3/arctanh(tanh(b*x+a))^(3/2)/x^(1/2)-128/15*b^
3*x^(1/2)/(b*x-arctanh(tanh(b*x+a)))^4/arctanh(tanh(b*x+a))^(3/2)+256/15*b^3*x^(1/2)/(b*x-arctanh(tanh(b*x+a))
)^5/arctanh(tanh(b*x+a))^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(-128*b^3*Sqrt[x])/(15*(b*x - ArcTanh[Tanh[a + b*x]])^4*ArcTanh[Tanh[a + b*x]]^(3/2)) + (32*b^2)/(5*Sqrt[x]*(b
*x - ArcTanh[Tanh[a + b*x]])^3*ArcTanh[Tanh[a + b*x]]^(3/2)) + (16*b)/(15*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]
])^2*ArcTanh[Tanh[a + b*x]]^(3/2)) + 2/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])*ArcTanh[Tanh[a + b*x]]^(3/2))
+ (256*b^3*Sqrt[x])/(15*(b*x - ArcTanh[Tanh[a + b*x]])^5*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))^{5/2}} \, dx &=\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {(8 b) \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {\left (16 b^2\right ) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (64 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx}{5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {\left (128 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {128 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {32 b^2}{5 \sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {16 b}{15 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {256 b^3 \sqrt {x}}{15 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 100, normalized size = 0.54 \[ \frac {2 \left (60 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+90 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-20 b x \tanh ^{-1}(\tanh (a+b x))^3+3 \tanh ^{-1}(\tanh (a+b x))^4-5 b^4 x^4\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^5 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*ArcTanh[Tanh[a + b*x]]^(5/2)),x]

[Out]

(2*(-5*b^4*x^4 + 60*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 90*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 20*b*x*ArcTanh[Tanh
[a + b*x]]^3 + 3*ArcTanh[Tanh[a + b*x]]^4))/(15*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^5*ArcTanh[Tanh[a + b*x]
]^(3/2))

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fricas [A]  time = 0.73, size = 82, normalized size = 0.44 \[ -\frac {2 \, {\left (128 \, b^{4} x^{4} + 192 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x + 3 \, a^{4}\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(128*b^4*x^4 + 192*a*b^3*x^3 + 48*a^2*b^2*x^2 - 8*a^3*b*x + 3*a^4)*sqrt(b*x + a)*sqrt(x)/(a^5*b^2*x^5 +
2*a^6*b*x^4 + a^7*x^3)

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giac [A]  time = 0.24, size = 173, normalized size = 0.93 \[ -\frac {2 \, \sqrt {x} {\left (\frac {11 \, b^{4} x}{a^{5}} + \frac {12 \, b^{3}}{a^{4}}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {4 \, {\left (45 \, b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} - 240 \, a b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} + 490 \, a^{2} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} - 320 \, a^{3} b^{\frac {5}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} + 73 \, a^{4} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{5} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/3*sqrt(x)*(11*b^4*x/a^5 + 12*b^3/a^4)/(b*x + a)^(3/2) + 4/15*(45*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^
8 - 240*a*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 + 490*a^2*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 -
320*a^3*b^(5/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 + 73*a^4*b^(5/2))/(((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 -
a)^5*a^4)

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maple [A]  time = 0.26, size = 196, normalized size = 1.05 \[ -\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {16 b \left (-\frac {1}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {2 b \left (-\frac {1}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}\, \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {\sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {2 \sqrt {x}}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{\arctanh \left (\tanh \left (b x +a \right )\right )-b x}\right )}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)/arctanh(tanh(b*x+a))^(3/2)-16/5*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))^(3/2)-2*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh(tanh(b*
x+a))-b*x)/x^(1/2)/arctanh(tanh(b*x+a))^(3/2)-4*b/(arctanh(tanh(b*x+a))-b*x)*(1/3*x^(1/2)/(arctanh(tanh(b*x+a)
)-b*x)/arctanh(tanh(b*x+a))^(3/2)+2/3/(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 = 0.44, size = 67, normalized size = 0.36 \[ -\frac {2 \, {\left (128 \, b^{5} x^{5} + 320 \, a b^{4} x^{4} + 240 \, a^{2} b^{3} x^{3} + 40 \, a^{3} b^{2} x^{2} - 5 \, a^{4} b x + 3 \, a^{5}\right )}}{15 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/15*(128*b^5*x^5 + 320*a*b^4*x^4 + 240*a^2*b^3*x^3 + 40*a^3*b^2*x^2 - 5*a^4*b*x + 3*a^5)/((b*x + a)^(5/2)*a^
5*x^(5/2))

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mupad [B]  time = 1.86, size = 466, normalized size = 2.51 \[ \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\,b^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 )}+\frac {256\,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 {64\,x}{15\,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 )}^2}-\frac {2048\,b\,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}+\frac {8192\,b^2\,x^4}{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 )}^5}\right )}{x^{9/2}-\frac {x^{7/2}\,\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 )}{b}+\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}{4\,b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*atanh(tanh(a + b*x))^(5/2)),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*b
^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)) + (256
*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
) + (64*x)/(15*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)^2) - (2048*b*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) + (8192*b^2*x^4)/(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)^5)))/(x^(9/2) - (x^(7/2)*(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))/b + (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)/(4*b^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**(7/2)/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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