3.183 \(\int x^{7/2} \tanh ^{-1}(\tanh (a+b x))^3 \, dx\)

Optimal. Leaf size=69 \[ \frac {16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32 b^3 x^{15/2}}{6435} \]

[Out]

-32/6435*b^3*x^(15/2)+16/429*b^2*x^(13/2)*arctanh(tanh(b*x+a))-4/33*b*x^(11/2)*arctanh(tanh(b*x+a))^2+2/9*x^(9
/2)*arctanh(tanh(b*x+a))^3

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Rubi [A]  time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2168, 30} \[ \frac {16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {32 b^3 x^{15/2}}{6435} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(-32*b^3*x^(15/2))/6435 + (16*b^2*x^(13/2)*ArcTanh[Tanh[a + b*x]])/429 - (4*b*x^(11/2)*ArcTanh[Tanh[a + b*x]]^
2)/33 + (2*x^(9/2)*ArcTanh[Tanh[a + b*x]]^3)/9

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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^{7/2} \tanh ^{-1}(\tanh (a+b x))^3 \, dx &=\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{3} (2 b) \int x^{9/2} \tanh ^{-1}(\tanh (a+b x))^2 \, dx\\ &=-\frac {4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3+\frac {1}{33} \left (8 b^2\right ) \int x^{11/2} \tanh ^{-1}(\tanh (a+b x)) \, dx\\ &=\frac {16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3-\frac {1}{429} \left (16 b^3\right ) \int x^{13/2} \, dx\\ &=-\frac {32 b^3 x^{15/2}}{6435}+\frac {16}{429} b^2 x^{13/2} \tanh ^{-1}(\tanh (a+b x))-\frac {4}{33} b x^{11/2} \tanh ^{-1}(\tanh (a+b x))^2+\frac {2}{9} x^{9/2} \tanh ^{-1}(\tanh (a+b x))^3\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 0.83 \[ -\frac {2 x^{9/2} \left (-120 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+390 b x \tanh ^{-1}(\tanh (a+b x))^2-715 \tanh ^{-1}(\tanh (a+b x))^3+16 b^3 x^3\right )}{6435} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(-2*x^(9/2)*(16*b^3*x^3 - 120*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 390*b*x*ArcTanh[Tanh[a + b*x]]^2 - 715*ArcTanh[
Tanh[a + b*x]]^3))/6435

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fricas [A]  time = 0.43, size = 40, normalized size = 0.58 \[ \frac {2}{6435} \, {\left (429 \, b^{3} x^{7} + 1485 \, a b^{2} x^{6} + 1755 \, a^{2} b x^{5} + 715 \, a^{3} x^{4}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="fricas")

[Out]

2/6435*(429*b^3*x^7 + 1485*a*b^2*x^6 + 1755*a^2*b*x^5 + 715*a^3*x^4)*sqrt(x)

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giac [A]  time = 0.42, size = 35, normalized size = 0.51 \[ \frac {2}{15} \, b^{3} x^{\frac {15}{2}} + \frac {6}{13} \, a b^{2} x^{\frac {13}{2}} + \frac {6}{11} \, a^{2} b x^{\frac {11}{2}} + \frac {2}{9} \, a^{3} x^{\frac {9}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="giac")

[Out]

2/15*b^3*x^(15/2) + 6/13*a*b^2*x^(13/2) + 6/11*a^2*b*x^(11/2) + 2/9*a^3*x^(9/2)

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maple [A]  time = 0.29, size = 56, normalized size = 0.81 \[ \frac {2 x^{\frac {9}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{9}-\frac {4 b \left (\frac {x^{\frac {11}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{11}-\frac {4 b \left (\frac {x^{\frac {13}{2}} \arctanh \left (\tanh \left (b x +a \right )\right )}{13}-\frac {2 x^{\frac {15}{2}} b}{195}\right )}{11}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*arctanh(tanh(b*x+a))^3,x)

[Out]

2/9*x^(9/2)*arctanh(tanh(b*x+a))^3-4/3*b*(1/11*x^(11/2)*arctanh(tanh(b*x+a))^2-4/11*b*(1/13*x^(13/2)*arctanh(t
anh(b*x+a))-2/195*x^(15/2)*b))

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maxima [A]  time = 0.36, size = 55, normalized size = 0.80 \[ -\frac {4}{33} \, b x^{\frac {11}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} + \frac {2}{9} \, x^{\frac {9}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3} - \frac {16}{6435} \, {\left (2 \, b^{2} x^{\frac {15}{2}} - 15 \, b x^{\frac {13}{2}} \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(tanh(b*x+a))^3,x, algorithm="maxima")

[Out]

-4/33*b*x^(11/2)*arctanh(tanh(b*x + a))^2 + 2/9*x^(9/2)*arctanh(tanh(b*x + a))^3 - 16/6435*(2*b^2*x^(15/2) - 1
5*b*x^(13/2)*arctanh(tanh(b*x + a)))*b

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mupad [B]  time = 1.17, size = 182, normalized size = 2.64 \[ \frac {2\,b^3\,x^{15/2}}{15}-\frac {x^{9/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}{36}+\frac {3\,b\,x^{11/2}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{22}-\frac {3\,b^2\,x^{13/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{13} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*atanh(tanh(a + b*x))^3,x)

[Out]

(2*b^3*x^(15/2))/15 - (x^(9/2)*(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)/36 + (3*b*x^(11/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/22 - (3*b^2*x^(13/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*e
xp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/13

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*atanh(tanh(b*x+a))**3,x)

[Out]

Timed out

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