3.120 \(\int x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=101 \[ \frac {256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

[Out]

2/5*x^4*arctanh(tanh(b*x+a))^(5/2)/b-16/35*x^3*arctanh(tanh(b*x+a))^(7/2)/b^2+32/105*x^2*arctanh(tanh(b*x+a))^
(9/2)/b^3-128/1155*x*arctanh(tanh(b*x+a))^(11/2)/b^4+256/15015*arctanh(tanh(b*x+a))^(13/2)/b^5

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Rubi [A]  time = 0.07, antiderivative size = 101, 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} \[ \frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*x^4*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*b) - (16*x^3*ArcTanh[Tanh[a + b*x]]^(7/2))/(35*b^2) + (32*x^2*ArcTanh[
Tanh[a + b*x]]^(9/2))/(105*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(11/2))/(1155*b^4) + (256*ArcTanh[Tanh[a + b*x
]]^(13/2))/(15015*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 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {8 \int x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {48 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^2}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {64 \int x \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{105 b^3}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{11/2} \, dx}{1155 b^4}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {128 \operatorname {Subst}\left (\int x^{11/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{1155 b^5}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{9/2}}{105 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{11/2}}{1155 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{13/2}}{15015 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 83, normalized size = 0.82 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \left (-3432 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+2288 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-832 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+3003 b^4 x^4\right )}{15015 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcTanh[Tanh[a + b*x]]^(3/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(3003*b^4*x^4 - 3432*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 2288*b^2*x^2*ArcTanh[Tan
h[a + b*x]]^2 - 832*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(15015*b^5)

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fricas [A]  time = 0.48, size = 75, normalized size = 0.74 \[ \frac {2 \, {\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt {b x + a}}{15015 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a
^6)*sqrt(b*x + a)/b^5

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giac [B]  time = 0.19, size = 241, normalized size = 2.39 \[ \frac {\sqrt {2} {\left (\frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2}}{b^{4}} + \frac {130 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a}{b^{4}} + \frac {15 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )}}{b^{4}}\right )}}{45045 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="giac")

[Out]

1/45045*sqrt(2)*(143*sqrt(2)*(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*sqrt(b*x + a)*a^4)*a^2/b^4 + 130*sqrt(2)*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 9
90*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x + a)*a^5)*a/b^4 +
15*sqrt(2)*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a
^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)/b^4)/b

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maple [A]  time = 0.14, size = 154, normalized size = 1.52 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {4 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctanh(tanh(b*x+a))^(3/2),x)

[Out]

2/b^5*(1/13*arctanh(tanh(b*x+a))^(13/2)+1/11*(-4*arctanh(tanh(b*x+a))+4*b*x)*arctanh(tanh(b*x+a))^(11/2)+1/9*(
2*(b*x-arctanh(tanh(b*x+a)))^2+(-2*arctanh(tanh(b*x+a))+2*b*x)^2)*arctanh(tanh(b*x+a))^(9/2)+2/7*(b*x-arctanh(
tanh(b*x+a)))^2*(-2*arctanh(tanh(b*x+a))+2*b*x)*arctanh(tanh(b*x+a))^(7/2)+1/5*(b*x-arctanh(tanh(b*x+a)))^4*ar
ctanh(tanh(b*x+a))^(5/2))

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maxima [A]  time = 0.53, size = 64, normalized size = 0.63 \[ \frac {2 \, {\left (1155 \, b^{5} x^{5} + 315 \, a b^{4} x^{4} - 280 \, a^{2} b^{3} x^{3} + 240 \, a^{3} b^{2} x^{2} - 192 \, a^{4} b x + 128 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{15015 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(tanh(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/15015*(1155*b^5*x^5 + 315*a*b^4*x^4 - 280*a^2*b^3*x^3 + 240*a^3*b^2*x^2 - 192*a^4*b*x + 128*a^5)*(b*x + a)^(
3/2)/b^5

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mupad [B]  time = 1.19, size = 1813, normalized size = 17.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(tanh(a + b*x))^(3/2),x)

[Out]

(2*b*x^6*(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)
)/13 + (x^5*(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)*((24*b*(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))/13 - 2*b*(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)))/(11*b) + (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)*((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 + (10*((24*b*(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))/13 - 2*b*(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))*(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))/(11*b)))/(9*b) + (128*(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 + (10*((24*b*(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))/13 - 2*b*(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))*(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))/(11*b))*(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) + (8*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)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*
x) + 1)) + 2*b*x)^2/2 + (10*((24*b*(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))/13 - 2*b*(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))*(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))/(11*b))*(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) + (64*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)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(
2*b*x) + 1)) + 2*b*x)^2/2 + (10*((24*b*(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))/13 - 2*b*(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))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex
p(2*b*x) + 1))/2 + b*x))/(11*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1))/2 + b*x)^3)/(315*b^4) + (16*x^2*(log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - lo
g(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 + (10*((24*b*(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))/13 - 2*b*(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))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1))/2 + b*x))/(11*b))*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(e
xp(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 \[ \int x^{4} \operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*atanh(tanh(b*x+a))**(3/2),x)

[Out]

Integral(x**4*atanh(tanh(a + b*x))**(3/2), x)

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