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

Optimal. Leaf size=172 \[ \frac {7 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {7 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac {7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2163, 2162} \[ \frac {7 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\tanh ^{-1}(\tanh (a+b x))}}\right )}{\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^{9/2}}-\frac {7 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac {7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]])/(b*x - ArcTanh[Tanh[a + b*x]])^(9/2)
- (7*b^2)/(Sqrt[x]*(b*x - ArcTanh[Tanh[a + b*x]])^4) - (7*b)/(3*x^(3/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) - 7
/(5*x^(5/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) - 1/(b*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])) - 1/(b*x^(7/2)*Ar
cTanh[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]

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {7 \int \frac {1}{x^{9/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 b}\\ &=-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {7 \int \frac {1}{x^{7/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}-\frac {(7 b) \int \frac {1}{x^{5/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=-\frac {7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (7 b^2\right ) \int \frac {1}{x^{3/2} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=-\frac {7 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac {7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (7 b^3\right ) \int \frac {1}{\sqrt {x} \tanh ^{-1}(\tanh (a+b x))} \, dx}{2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {7 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 )^{9/2}}-\frac {7 b^2}{\sqrt {x} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}-\frac {7 b}{3 x^{3/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}-\frac {7}{5 x^{5/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}-\frac {1}{b x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}-\frac {1}{b x^{7/2} \tanh ^{-1}(\tanh (a+b x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.28, size = 139, normalized size = 0.81 \[ -\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))-b x}}\right )}{\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^{9/2}}-\frac {b^3 \sqrt {x}}{\tanh ^{-1}(\tanh (a+b x)) \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}-\frac {2 \left (-16 b x \tanh ^{-1}(\tanh (a+b x))+3 \tanh ^{-1}(\tanh (a+b x))^2+58 b^2 x^2\right )}{15 x^{5/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[-(b*x) + ArcTanh[Tanh[a + b*x]]]])/(-(b*x) + ArcTanh[Tanh[a + b*x]])
^(9/2) - (b^3*Sqrt[x])/(ArcTanh[Tanh[a + b*x]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4) - (2*(58*b^2*x^2 - 16*b*x*
ArcTanh[Tanh[a + b*x]] + 3*ArcTanh[Tanh[a + b*x]]^2))/(15*x^(5/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 210, normalized size = 1.22 \[ \left [\frac {105 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (105 \, b^{3} x^{3} + 70 \, a b^{2} x^{2} - 14 \, a^{2} b x + 6 \, a^{3}\right )} \sqrt {x}}{30 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac {105 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (105 \, b^{3} x^{3} + 70 \, a b^{2} x^{2} - 14 \, a^{2} b x + 6 \, a^{3}\right )} \sqrt {x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(105*(b^3*x^4 + a*b^2*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(105*b^3*x^3
+ 70*a*b^2*x^2 - 14*a^2*b*x + 6*a^3)*sqrt(x))/(a^4*b*x^4 + a^5*x^3), 1/15*(105*(b^3*x^4 + a*b^2*x^3)*sqrt(b/a
)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (105*b^3*x^3 + 70*a*b^2*x^2 - 14*a^2*b*x + 6*a^3)*sqrt(x))/(a^4*b*x^4 + a^
5*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.51, size = 70, normalized size = 0.41 \[ -\frac {7 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} - \frac {b^{3} \sqrt {x}}{{\left (b x + a\right )} a^{4}} - \frac {2 \, {\left (45 \, b^{2} x^{2} - 10 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{4} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - b^3*sqrt(x)/((b*x + a)*a^4) - 2/15*(45*b^2*x^2 - 10*a*b*x
+ 3*a^2)/(a^4*x^(5/2))

________________________________________________________________________________________

maple [A]  time = 0.27, size = 151, normalized size = 0.88 \[ -\frac {2}{5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}-\frac {6 b^{2}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {x}}+\frac {4 b}{3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{3} x^{\frac {3}{2}}}-\frac {b^{3} \sqrt {x}}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {7 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{4} \sqrt {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)-6/(arctanh(tanh(b*x+a))-b*x)^4*b^2/x^(1/2)+4/3/(arctanh(tanh(b*x+a))
-b*x)^3*b/x^(3/2)-1/(arctanh(tanh(b*x+a))-b*x)^4*b^3*x^(1/2)/arctanh(tanh(b*x+a))-7/(arctanh(tanh(b*x+a))-b*x)
^4*b^3/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2)*arctan(b*x^(1/2)/((arctanh(tanh(b*x+a))-b*x)*b)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 75, normalized size = 0.44 \[ -\frac {105 \, b^{3} x^{3} + 70 \, a b^{2} x^{2} - 14 \, a^{2} b x + 6 \, a^{3}}{15 \, {\left (a^{4} b x^{\frac {7}{2}} + a^{5} x^{\frac {5}{2}}\right )}} - \frac {7 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(105*b^3*x^3 + 70*a*b^2*x^2 - 14*a^2*b*x + 6*a^3)/(a^4*b*x^(7/2) + a^5*x^(5/2)) - 7*b^3*arctan(b*sqrt(x)
/sqrt(a*b))/(sqrt(a*b)*a^4)

________________________________________________________________________________________

mupad [B]  time = 2.14, size = 1051, normalized size = 6.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((96*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)^
3 - (224*b^3*x)/(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^(1/2)*(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))))
- (32*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)^3) - 8/(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)^2) + (56*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)) - l
og((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)^8 + 112*a^2*(2*a - log((2*e
xp(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 - 448*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)^5 + 1120*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)^4
- 1792*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)^3 + 1792*a^6*(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 + 256*a^8 - 16*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)^7 - 1024*a^7*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + lo
g(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)^(9/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________