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

Optimal. Leaf size=101 \[ \frac {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

[Out]

2/3*x^4*arctanh(tanh(b*x+a))^(3/2)/b-16/15*x^3*arctanh(tanh(b*x+a))^(5/2)/b^2+32/35*x^2*arctanh(tanh(b*x+a))^(
7/2)/b^3-128/315*x*arctanh(tanh(b*x+a))^(9/2)/b^4+256/3465*arctanh(tanh(b*x+a))^(11/2)/b^5

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Rubi [A]  time = 0.06, 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} \[ \frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^4*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*b) - (16*x^3*ArcTanh[Tanh[a + b*x]]^(5/2))/(15*b^2) + (32*x^2*ArcTanh[
Tanh[a + b*x]]^(7/2))/(35*b^3) - (128*x*ArcTanh[Tanh[a + b*x]]^(9/2))/(315*b^4) + (256*ArcTanh[Tanh[a + b*x]]^
(11/2))/(3465*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 \sqrt {\tanh ^{-1}(\tanh (a+b x))} \, dx &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {8 \int x^3 \tanh ^{-1}(\tanh (a+b x))^{3/2} \, dx}{3 b}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {16 \int x^2 \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx}{5 b^2}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {64 \int x \tanh ^{-1}(\tanh (a+b x))^{7/2} \, dx}{35 b^3}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {128 \int \tanh ^{-1}(\tanh (a+b x))^{9/2} \, dx}{315 b^4}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {128 \operatorname {Subst}\left (\int x^{9/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{315 b^5}\\ &=\frac {2 x^4 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{3 b}-\frac {16 x^3 \tanh ^{-1}(\tanh (a+b x))^{5/2}}{15 b^2}+\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^3}-\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{9/2}}{315 b^4}+\frac {256 \tanh ^{-1}(\tanh (a+b x))^{11/2}}{3465 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.82 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{3/2} \left (-1848 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+1584 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-704 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+1155 b^4 x^4\right )}{3465 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(1155*b^4*x^4 - 1848*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 1584*b^2*x^2*ArcTanh[Tan
h[a + b*x]]^2 - 704*b*x*ArcTanh[Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(3465*b^5)

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fricas [A]  time = 0.52, size = 64, normalized size = 0.63 \[ \frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \]

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]

2/3465*(315*b^5*x^5 + 35*a*b^4*x^4 - 40*a^2*b^3*x^3 + 48*a^3*b^2*x^2 - 64*a^4*b*x + 128*a^5)*sqrt(b*x + a)/b^5

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giac [A]  time = 0.62, size = 150, normalized size = 1.49 \[ \frac {\sqrt {2} {\left (\frac {11 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a}{b^{4}} + \frac {5 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )}}{b^{4}}\right )}}{3465 \, b} \]

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]

1/3465*sqrt(2)*(11*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/b^4 + 5*sqrt(2)*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990*(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)/b^4)/b

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maple [A]  time = 0.17, size = 154, normalized size = 1.52 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {2 \left (-4 \arctanh \left (\tanh \left (b x +a \right )\right )+4 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right )^{2}\right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}+\frac {4 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{2} \left (-2 \arctanh \left (\tanh \left (b x +a \right )\right )+2 b x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (b x -\arctanh \left (\tanh \left (b x +a \right )\right )\right )^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{3}}{b^{5}} \]

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*(1/11*arctanh(tanh(b*x+a))^(11/2)+1/9*(-4*arctanh(tanh(b*x+a))+4*b*x)*arctanh(tanh(b*x+a))^(9/2)+1/7*(2*
(b*x-arctanh(tanh(b*x+a)))^2+(-2*arctanh(tanh(b*x+a))+2*b*x)^2)*arctanh(tanh(b*x+a))^(7/2)+2/5*(b*x-arctanh(ta
nh(b*x+a)))^2*(-2*arctanh(tanh(b*x+a))+2*b*x)*arctanh(tanh(b*x+a))^(5/2)+1/3*(b*x-arctanh(tanh(b*x+a)))^4*arct
anh(tanh(b*x+a))^(3/2))

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maxima [A]  time = 1.84, size = 64, normalized size = 0.63 \[ \frac {2 \, {\left (315 \, b^{5} x^{5} + 35 \, a b^{4} x^{4} - 40 \, a^{2} b^{3} x^{3} + 48 \, a^{3} b^{2} x^{2} - 64 \, a^{4} b x + 128 \, a^{5}\right )} \sqrt {b x + a}}{3465 \, b^{5}} \]

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]

2/3465*(315*b^5*x^5 + 35*a*b^4*x^4 - 40*a^2*b^3*x^3 + 48*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 = 1.05, size = 811, normalized size = 8.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(tanh(a + b*x))^(1/2),x)

[Out]

(2*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))/
11 - (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))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 + (2*b*x)/1
1))/(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x)
^4*(log(2/(exp(2*a)*exp(2*b*x) + 1))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 + (2*b*x)/
11))/(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))/2 - 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))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11 + (2*
b*x)/11))/(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))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)
)/2 + b*x)^3*(log(2/(exp(2*a)*exp(2*b*x) + 1))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/11
+ (2*b*x)/11))/(315*b^4) - (16*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*(log(2/(exp(2*a)*exp(2*b*x) + 1))/11 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) +
1))/11 + (2*b*x)/11))/(105*b^3)

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

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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