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

Optimal. Leaf size=95 \[ -\frac {256 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^5}+\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {16 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {2 x^4}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 95, 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} \[ \frac {16 x^3 \sqrt {\tanh ^{-1}(\tanh (a+b x))}}{b^2}-\frac {32 x^2 \tanh ^{-1}(\tanh (a+b x))^{3/2}}{b^3}+\frac {128 x \tanh ^{-1}(\tanh (a+b x))^{5/2}}{5 b^4}-\frac {256 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{35 b^5}-\frac {2 x^4}{b \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.05, size = 83, normalized size = 0.87 \[ -\frac {2 \left (-280 b^3 x^3 \tanh ^{-1}(\tanh (a+b x))+560 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2-448 b x \tanh ^{-1}(\tanh (a+b x))^3+128 \tanh ^{-1}(\tanh (a+b x))^4+35 b^4 x^4\right )}{35 b^5 \sqrt {\tanh ^{-1}(\tanh (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(35*b^4*x^4 - 280*b^3*x^3*ArcTanh[Tanh[a + b*x]] + 560*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 448*b*x*ArcTanh[
Tanh[a + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(35*b^5*Sqrt[ArcTanh[Tanh[a + b*x]]])

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fricas [A]  time = 0.52, size = 63, normalized size = 0.66 \[ \frac {2 \, {\left (5 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 16 \, a^{2} b^{2} x^{2} - 64 \, a^{3} b x - 128 \, a^{4}\right )} \sqrt {b x + a}}{35 \, {\left (b^{6} x + a b^{5}\right )}} \]

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]

2/35*(5*b^4*x^4 - 8*a*b^3*x^3 + 16*a^2*b^2*x^2 - 64*a^3*b*x - 128*a^4)*sqrt(b*x + a)/(b^6*x + a*b^5)

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giac [A]  time = 0.19, size = 77, normalized size = 0.81 \[ -\frac {2 \, a^{4}}{\sqrt {b x + a} b^{5}} + \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{30} - 28 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{30} + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{30} - 140 \, \sqrt {b x + a} a^{3} b^{30}\right )}}{35 \, b^{35}} \]

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]

-2*a^4/(sqrt(b*x + a)*b^5) + 2/35*(5*(b*x + a)^(7/2)*b^30 - 28*(b*x + a)^(5/2)*a*b^30 + 70*(b*x + a)^(3/2)*a^2
*b^30 - 140*sqrt(b*x + a)*a^3*b^30)/b^35

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maple [B]  time = 0.15, size = 319, normalized size = 3.36 \[ \frac {\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}-\frac {8 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}} a}{5}-\frac {8 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{5}+4 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a^{2}+8 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+4 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-8 \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}\, a^{3}-24 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-24 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-8 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3} \sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {2 \left (a^{4}+4 a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )+6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}+4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}+\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}\right )}{\sqrt {\arctanh \left (\tanh \left (b x +a \right )\right )}}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^5*(1/7*arctanh(tanh(b*x+a))^(7/2)-4/5*arctanh(tanh(b*x+a))^(5/2)*a-4/5*arctanh(tanh(b*x+a))^(5/2)*(arctanh
(tanh(b*x+a))-b*x-a)+2*arctanh(tanh(b*x+a))^(3/2)*a^2+4*arctanh(tanh(b*x+a))^(3/2)*a*(arctanh(tanh(b*x+a))-b*x
-a)+2*arctanh(tanh(b*x+a))^(3/2)*(arctanh(tanh(b*x+a))-b*x-a)^2-4*arctanh(tanh(b*x+a))^(1/2)*a^3-12*a^2*(arcta
nh(tanh(b*x+a))-b*x-a)*arctanh(tanh(b*x+a))^(1/2)-12*a*(arctanh(tanh(b*x+a))-b*x-a)^2*arctanh(tanh(b*x+a))^(1/
2)-4*(arctanh(tanh(b*x+a))-b*x-a)^3*arctanh(tanh(b*x+a))^(1/2)-(a^4+4*a^3*(arctanh(tanh(b*x+a))-b*x-a)+6*a^2*(
arctanh(tanh(b*x+a))-b*x-a)^2+4*a*(arctanh(tanh(b*x+a))-b*x-a)^3+(arctanh(tanh(b*x+a))-b*x-a)^4)/arctanh(tanh(
b*x+a))^(1/2))

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maxima [A]  time = 0.54, size = 64, normalized size = 0.67 \[ \frac {2 \, {\left (5 \, b^{5} x^{5} - 3 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 48 \, a^{3} b^{2} x^{2} - 192 \, a^{4} b x - 128 \, a^{5}\right )}}{35 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5}} \]

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]

2/35*(5*b^5*x^5 - 3*a*b^4*x^4 + 8*a^2*b^3*x^3 - 48*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 = 1.45, size = 1057, normalized size = 11.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/atanh(tanh(a + b*x))^(3/2),x)

[Out]

((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/(4*b^4) + (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)
+ (4*((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
+ (12*(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))/(
7*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))
/(5*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))
/(3*b)))/b + (2*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))/(7*b^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)*((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) + (4*((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 + (12*(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))/(7*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))/(5*b)))/(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)^4)/(4*b^5*(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)))) + (x^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 + (12*(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))/(7*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))/(5*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {atanh}^{\frac {3}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

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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