3.159 \(\int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 76, 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} \[ -\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^3)/(3*b*ArcTanh[Tanh[a + b*x]]^(3/2)) - (4*x^2)/(b^2*Sqrt[ArcTanh[Tanh[a + b*x]]]) + (16*x*Sqrt[ArcTanh[
Tanh[a + b*x]]])/b^3 - (32*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*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))^{5/2}} \, dx &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}+\frac {2 \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^{3/2}} \, dx}{b}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {8 \int \frac {x}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx}{b^2}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {16 \int \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {16 \operatorname {Subst}\left (\int \sqrt {x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=-\frac {2 x^3}{3 b \tanh ^{-1}(\tanh (a+b x))^{3/2}}-\frac {4 x^2}{b^2 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}+\frac {16 x \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^3}-\frac {32 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 0.86 \[ -\frac {2 \left (6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-24 b x \tanh ^{-1}(\tanh (a+b x))^2+16 \tanh ^{-1}(\tanh (a+b x))^3+b^3 x^3\right )}{3 b^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^3*x^3 + 6*b^2*x^2*ArcTanh[Tanh[a + b*x]] - 24*b*x*ArcTanh[Tanh[a + b*x]]^2 + 16*ArcTanh[Tanh[a + b*x]]^
3))/(3*b^4*ArcTanh[Tanh[a + b*x]]^(3/2))

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fricas [A]  time = 0.60, size = 62, normalized size = 0.82 \[ \frac {2 \, {\left (b^{3} x^{3} - 6 \, a b^{2} x^{2} - 24 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 59, normalized size = 0.78 \[ -\frac {2 \, {\left (9 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} b^{8} - 9 \, \sqrt {b x + a} a b^{8}\right )}}{3 \, b^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

-2/3*(9*(b*x + a)*a^2 - a^3)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)*b^8 - 9*sqrt(b*x + a)*a*b^8)/b^12

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maple [B]  time = 0.15, size = 186, normalized size = 2.45 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}-6 a \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-6 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (3 a^{2}+6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}-\frac {2 \left (-a^{3}-3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right )}{3 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arctanh(tanh(b*x+a))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.53, size = 52, normalized size = 0.68 \[ \frac {2 \, {\left (b^{4} x^{4} - 5 \, a b^{3} x^{3} - 30 \, a^{2} b^{2} x^{2} - 40 \, a^{3} b x - 16 \, a^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^4*x^4 - 5*a*b^3*x^3 - 30*a^2*b^2*x^2 - 40*a^3*b*x - 16*a^4)/((b*x + a)^(5/2)*b^4)

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mupad [B]  time = 1.30, size = 533, normalized size = 7.01 \[ \frac {2\,x\,\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^3}+\frac {\left (\frac {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 )}{b^3}+\frac {4\,\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^3}\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}}}{b}-\frac {3\,\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 )}^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 )}-\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}{3\,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 )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/atanh(tanh(a + b*x))^(5/2),x)

[Out]

(2*x*(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^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^3 + (4*(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^3))*(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))/b - (3*(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)/
(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)))) - ((log((2*e
xp(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)/(3*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)))^2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/atanh(tanh(b*x+a))**(5/2),x)

[Out]

Exception raised: TypeError

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