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

Optimal. Leaf size=111 \[ \frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {10 x^{3/2}}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

5*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))*(b*x-arctanh(tanh(b*x+a)))/b^(7/2)-2/3*x^(5/2)/b/arctanh
(tanh(b*x+a))^(3/2)-10/3*x^(3/2)/b^2/arctanh(tanh(b*x+a))^(1/2)+5*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)/b^3

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Rubi [A]  time = 0.07, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ -\frac {10 x^{3/2}}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]]*(b*x - ArcTanh[Tanh[a + b*x]]))/b^(7/2) - (2*x^(5/2
))/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (5*Sqrt[x]*Sqrt[Ar
cTanh[Tanh[a + b*x]]])/b^3

Rule 2165

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

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]))

Rule 2169

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 + n + 1)), x] - Dist[(n*(b*u - a*v))/(a*(m + n + 1)), Int[u^m*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]]
/; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !In
tegerQ[n] || LtQ[0, m, n])) &&  !ILtQ[m + n, -2]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx &=-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {5 \int \frac {x^{3/2}}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{3 b}\\ &=-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {\left (5 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{2 b^3}\\ &=\frac {5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{7/2}}-\frac {2 x^{5/2}}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {5 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 101, normalized size = 0.91 \[ \frac {5 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{b^{7/2}}-\frac {\sqrt {x} \left (10 b x \tanh ^{-1}(\tanh (a+b x))-15 \tanh ^{-1}(\tanh (a+b x))^2+2 b^2 x^2\right )}{3 b^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[x]*(2*b^2*x^2 + 10*b*x*ArcTanh[Tanh[a + b*x]] - 15*ArcTanh[Tanh[a + b*x]]^2))/(b^3*ArcTanh[Tanh[a +
b*x]]^(3/2)) + (5*(b*x - ArcTanh[Tanh[a + b*x]])*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]])/b^(7/
2)

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fricas [A]  time = 0.80, size = 214, normalized size = 1.93 \[ \left [\frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^3*x^2
+ 20*a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(15*(a*b^2*x^2 + 2*a^2*b
*x + a^3)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (3*b^3*x^2 + 20*a*b^2*x + 15*a^2*b)*sqrt(b*x +
a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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giac [A]  time = 0.15, size = 61, normalized size = 0.55 \[ \frac {{\left (x {\left (\frac {3 \, x}{b} + \frac {20 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} \sqrt {x}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}}} + \frac {5 \, a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x*(3*x/b + 20*a/b^2) + 15*a^2/b^3)*sqrt(x)/(b*x + a)^(3/2) + 5*a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)
))/b^(7/2)

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maple [B]  time = 0.25, size = 180, normalized size = 1.62 \[ \frac {x^{\frac {5}{2}}}{b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {5 a \,x^{\frac {3}{2}}}{3 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {5 a \sqrt {x}}{b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {5 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {7}{2}}}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x^{\frac {3}{2}}}{3 b^{2} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}+\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {5 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {5}{2}}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(5/2)/arctanh(tanh(b*x + a))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/atanh(tanh(a + b*x))^(5/2),x)

[Out]

int(x^(5/2)/atanh(tanh(a + b*x))^(5/2), x)

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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(x**(5/2)/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Timed out

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