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

Optimal. Leaf size=74 \[ \frac {32 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

-16*x*arctanh(tanh(b*x+a))^(3/2)/b^3+32/5*arctanh(tanh(b*x+a))^(5/2)/b^4-2*x^3/b/arctanh(tanh(b*x+a))^(1/2)+12
*x^2*arctanh(tanh(b*x+a))^(1/2)/b^2

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Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2168, 2157, 30} \[ \frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {32 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^3)/(b*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (12*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]])/b^2 - (16*x*ArcTanh[Tanh[a +
b*x]]^(3/2))/b^3 + (32*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b^4)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, 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]))

Rubi steps

\begin {align*} \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx &=-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {6 \int \frac {x^2}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b}\\ &=-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {24 \int x \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {16 \int \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b^3}\\ &=-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {16 \operatorname {Subst}\left (\int x^{3/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=-\frac {2 x^3}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {12 x^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {16 x \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {32 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 66, normalized size = 0.89 \[ \frac {2 \left (30 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-40 b x \tanh ^{-1}(\tanh (a+b x))^2+16 \tanh ^{-1}(\tanh (a+b x))^3-5 b^3 x^3\right )}{5 b^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-5*b^3*x^3 + 30*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 40*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x
]]^3))/(5*b^4*Sqrt[ArcTanh[Tanh[a + b*x]]])

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fricas [A]  time = 0.52, size = 51, normalized size = 0.69 \[ \frac {2 \, {\left (b^{3} x^{3} - 2 \, a b^{2} x^{2} + 8 \, a^{2} b x + 16 \, a^{3}\right )} \sqrt {b x + a}}{5 \, {\left (b^{5} x + a b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^3*x^3 - 2*a*b^2*x^2 + 8*a^2*b*x + 16*a^3)*sqrt(b*x + a)/(b^5*x + a*b^4)

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giac [A]  time = 0.18, size = 61, normalized size = 0.82 \[ \frac {2 \, a^{3}}{\sqrt {b x + a} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} b^{16} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{16} + 15 \, \sqrt {b x + a} a^{2} b^{16}\right )}}{5 \, b^{20}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3/(sqrt(b*x + a)*b^4) + 2/5*((b*x + a)^(5/2)*b^16 - 5*(b*x + a)^(3/2)*a*b^16 + 15*sqrt(b*x + a)*a^2*b^16)/
b^20

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maple [B]  time = 0.15, size = 201, normalized size = 2.72 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a -2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+12 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}+6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (-a^{3}-3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 52, normalized size = 0.70 \[ \frac {2 \, {\left (b^{4} x^{4} - a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 24 \, a^{3} b x + 16 \, a^{4}\right )}}{5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^4*x^4 - a*b^3*x^3 + 6*a^2*b^2*x^2 + 24*a^3*b*x + 16*a^4)/((b*x + a)^(3/2)*b^4)

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mupad [B]  time = 1.26, size = 660, normalized size = 8.92 \[ \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 {{\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}{2\,b^3}+\frac {2\,\left (\frac {\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}{b^2}+\frac {8\,\left (\frac {\ln \left (\frac {2}{{\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}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{5\,b^2}\right )\,\left (\frac {\ln \left (\frac {2}{{\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}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{3\,b}\right )}{b}+\frac {2\,x^2\,\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}}}{5\,b^2}+\frac {x\,\left (\frac {\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}{b^2}+\frac {8\,\left (\frac {\ln \left (\frac {2}{{\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}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}+b\,x\right )}{5\,b^2}\right )\,\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}}}{3\,b}-\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 (\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}{2\,b^4\,\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/atanh(tanh(a + b*x))^(3/2),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)*((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/(2*b^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)/b^2 + (8*(
log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(5*b^2))
*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/(3*b))
)/b + (2*x^2*(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))/(5*b^2) + (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)/b^2 + (8*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
)/2 + b*x))/(5*b^2))*(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))/(3*b) - ((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)*(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)/(2*b^4*(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))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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