3.2.33 \(\int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx\) [133]

Optimal. Leaf size=18 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2188, 30} \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

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

Rule 30

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

Rule 2188

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]

Rubi steps

\begin {align*} \int \tanh ^{-1}(\tanh (a+b x))^{5/2} \, dx &=\frac {\text {Subst}\left (\int x^{5/2} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b}\\ &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2),x]

[Out]

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

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Maple [A]
time = 0.06, size = 15, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7 b}\)	\(15\)
default	\(\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7 b}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 12, normalized size = 0.67 \begin {gather*} \frac {2 \, {\left (b x + a\right )}^{\frac {7}{2}}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

2/7*(b*x + a)^(7/2)/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (14) = 28\).
time = 0.34, size = 39, normalized size = 2.17 \begin {gather*} \frac {2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt {b x + a}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

2/7*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)/b

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Sympy [A]
time = 17.68, size = 26, normalized size = 1.44 \begin {gather*} \begin {cases} \frac {2 \operatorname {atanh}^{\frac {7}{2}}{\left (\tanh {\left (a + b x \right )} \right )}}{7 b} & \text {for}\: b \neq 0 \\x \operatorname {atanh}^{\frac {5}{2}}{\left (\tanh {\left (a \right )} \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2),x)

[Out]

Piecewise((2*atanh(tanh(a + b*x))**(7/2)/(7*b), Ne(b, 0)), (x*atanh(tanh(a))**(5/2), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (14) = 28\).
time = 0.41, size = 136, normalized size = 7.56 \begin {gather*} \frac {\sqrt {2} {\left (35 \, \sqrt {2} \sqrt {b x + a} a^{3} + 35 \, \sqrt {2} {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} + 7 \, \sqrt {2} {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a + \sqrt {2} {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )}\right )}}{35 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")

[Out]

1/35*sqrt(2)*(35*sqrt(2)*sqrt(b*x + a)*a^3 + 35*sqrt(2)*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2 + 7*sqrt(2)*
(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a + sqrt(2)*(5*(b*x + a)^(7/2) - 21*(b*x + a
)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3))/b

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Mupad [B]
time = 1.12, size = 337, normalized size = 18.72 \begin {gather*} \frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3\,\sqrt {\frac {\ln \left (\frac {{\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 {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{28\,b}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3\,\sqrt {\frac {\ln \left (\frac {{\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 {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{28\,b}-\frac {3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\sqrt {\frac {\ln \left (\frac {{\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 {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{28\,b}+\frac {3\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\sqrt {\frac {\ln \left (\frac {{\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 {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}}{28\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2),x)

[Out]

(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/
2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b) - (log(1/(exp(2*a)*exp(2*b*x) + 1))^3*(log((exp(2*a)*exp
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b) - (3*log(1/(exp(2*a)
*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*
exp(2*b*x) + 1))/2 - log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b) + (3*log(1/(exp(2*a)*exp(2*b*x) + 1))^2
*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2
- log(1/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2))/(28*b)

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