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

Optimal. Leaf size=128 \[ -\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2163, 2162} \[ -\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2162

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-2*ArcTanh[Sqrt
[v]/Rt[-((b*u - a*v)/a), 2]])/(a*Rt[-((b*u - a*v)/a), 2]), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /;
PiecewiseLinearQ[u, v, x]

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {b \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{-b x+\tanh ^{-1}(\tanh (a+b x))}\\ &=\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b^2 \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {b^3 \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{7/2}}+\frac {2 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 107, normalized size = 0.84 \[ \frac {2 \left (-11 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+23 b^2 x^2\right )}{15 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {2 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])
^(7/2) + (2*(23*b^2*x^2 - 11*b*x*ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(b*x - ArcT
anh[Tanh[a + b*x]])^3)

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fricas [A]  time = 0.57, size = 144, normalized size = 1.12 \[ \left [\frac {15 \, b^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, \frac {2 \, {\left (15 \, b^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*b^2*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^2*x^2 - 5*a*b*x + 3*a
^2)*sqrt(x))/(a^3*x^3), 2/15*(15*b^2*x^3*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*b^2*x^2 - 5*a*b*x + 3
*a^2)*sqrt(x))/(a^3*x^3)]

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giac [A]  time = 0.17, size = 52, normalized size = 0.41 \[ -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 120, normalized size = 0.94 \[ -\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}-\frac {2 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {x}}+\frac {2 b}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 52, normalized size = 0.41 \[ -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.63, size = 822, normalized size = 6.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/(5*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)) + (8*b)/(3*x^(3/2)*(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) + (16*b^2)/(x^(1/2)*(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)^3) + (8*2^(1/2)*b^(5/2)*log((b^(1/2)*(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)^(1/2)*(2^(1/2)*(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) - 4*b^(1/2)*x^(1/2)*(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)^(1/2) + 2*2^(1/2)*b*x)*((2*a - log((2*exp(
2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^6 + 60*a^2*(2*a - log(
(2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4 - 160*a^3*(2*
a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 + 240
*a^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)
^2 + 64*a^6 - 12*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x)
+ 1)) + 2*b*x)^5 - 192*a^5*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp
(2*b*x) + 1)) + 2*b*x)))/(2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*
x) + 1))))))/(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)^(7/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{\frac {7}{2}} \operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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