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

Optimal. Leaf size=86 \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{5/2}}+\frac {3 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x^{3/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2168, 2169, 2165} \[ \frac {3 \sqrt {x} \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}}\right ) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^{5/2}}-\frac {2 x^{3/2}}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.09, size = 81, normalized size = 0.94 \[ \frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\sqrt {b} \sqrt {\tanh ^{-1}(\tanh (a+b x))}+b \sqrt {x}\right )}{b^{5/2}}+\frac {\sqrt {x} \left (3 \tanh ^{-1}(\tanh (a+b x))-2 b x\right )}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 145, normalized size = 1.69 \[ \left [\frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{b^{4} x + a b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 48, normalized size = 0.56 \[ \frac {\sqrt {x} {\left (\frac {x}{b} + \frac {3 \, a}{b^{2}}\right )}}{\sqrt {b x + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 130, normalized size = 1.51 \[ \frac {x^{\frac {3}{2}}}{b \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}+\frac {3 a \sqrt {x}}{b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {3 a \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {5}{2}}}+\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}}{b^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(3/2)/b/arctanh(tanh(b*x+a))^(1/2)+3/b^2*a*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-3/b^(5/2)*a*ln(b^(1/2)*x^(1/2)
+arctanh(tanh(b*x+a))^(1/2))+3/b^2*(arctanh(tanh(b*x+a))-b*x-a)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2)-3/b^(5/2)*(
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 \[ \int \frac {x^{\frac {3}{2}}}{\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(3/2)/arctanh(tanh(b*x + a))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{3/2}}{{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {3}{2}}}{\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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