3.140 \(\int \frac {x^4}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx\)

Optimal. Leaf size=99 \[ \frac {256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \]

[Out]

-16/3*x^3*arctanh(tanh(b*x+a))^(3/2)/b^2+32/5*x^2*arctanh(tanh(b*x+a))^(5/2)/b^3-128/35*x*arctanh(tanh(b*x+a))
^(7/2)/b^4+256/315*arctanh(tanh(b*x+a))^(9/2)/b^5+2*x^4*arctanh(tanh(b*x+a))^(1/2)/b

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Rubi [A]  time = 0.07, antiderivative size = 99, 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, 2157, 30} \[ \frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*x^4*Sqrt[ArcTanh[Tanh[a + b*x]]])/b - (16*x^3*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b^2) + (32*x^2*ArcTanh[Tanh[
a + b*x]]^(5/2))/(5*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^4) + (256*ArcTanh[Tanh[a + b*x]]^(9/2))/
(315*b^5)

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^4}{\sqrt {\tanh ^{-1}(\tanh (a+b x))}} \, dx &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {8 \int x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {16 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{b^2}\\ &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {64 \int x \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^3}\\ &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^4}\\ &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {128 \operatorname {Subst}\left (\int x^{7/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{35 b^5}\\ &=\frac {2 x^4 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 83, normalized size = 0.84 \[ \frac {2 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \left (-840 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+1008 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-576 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+315 b^4 x^4\right )}{315 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[ArcTanh[Tanh[a + b*x]]],x]

[Out]

(2*Sqrt[ArcTanh[Tanh[a + b*x]]]*(315*b^4*x^4 - 840*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 1008*b^2*x^2*ArcTanh[Tanh[
a + b*x]]^2 - 576*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(315*b^5)

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fricas [A]  time = 0.65, size = 53, normalized size = 0.54 \[ \frac {2 \, {\left (35 \, b^{4} x^{4} - 40 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x + 128 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^4*x^4 - 40*a*b^3*x^3 + 48*a^2*b^2*x^2 - 64*a^3*b*x + 128*a^4)*sqrt(b*x + a)/b^5

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giac [A]  time = 0.16, size = 61, normalized size = 0.62 \[ \frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )}}{315 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(1/2),x, algorithm="giac")

[Out]

2/315*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sq
rt(b*x + a)*a^4)/b^5

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maple [A]  time = 0.17, size = 153, normalized size = 1.55 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {4 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}+2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{4} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^(1/2),x)

[Out]

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

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maxima [A]  time = 0.53, size = 64, normalized size = 0.65 \[ \frac {2 \, {\left (35 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 64 \, a^{4} b x + 128 \, a^{5}\right )}}{315 \, \sqrt {b x + a} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*b^5*x^5 - 5*a*b^4*x^4 + 8*a^2*b^3*x^3 - 16*a^3*b^2*x^2 + 64*a^4*b*x + 128*a^5)/(sqrt(b*x + a)*b^5)

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mupad [B]  time = 1.09, size = 496, normalized size = 5.01 \[ \frac {2\,x^4\,\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}}}{9\,b}+\frac {256\,\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 {\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 )}^4}{315\,b^5}+\frac {16\,x^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 (\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 )}{63\,b^2}+\frac {128\,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}}\,{\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}{315\,b^4}+\frac {32\,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}}\,{\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 )}^2}{105\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/atanh(tanh(a + b*x))^(1/2),x)

[Out]

(2*x^4*(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))/
(9*b) + (256*(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)^4)/
(315*b^5) + (16*x^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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b
*x))/(63*b^2) + (128*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)*(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)/(315*b^4) + (32*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)*(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)^2)/(105*b^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {\operatorname {atanh}{\left (\tanh {\left (a + b x \right )} \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/atanh(tanh(b*x+a))**(1/2),x)

[Out]

Integral(x**4/sqrt(atanh(tanh(a + b*x))), x)

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