3.394 \(\int \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=27 \[ \frac {\log (\cosh (a+b x))}{b}-\frac {\tanh ^2(a+b x)}{2 b} \]

[Out]

ln(cosh(b*x+a))/b-1/2*tanh(b*x+a)^2/b

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3473, 3475} \[ \frac {\log (\cosh (a+b x))}{b}-\frac {\tanh ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[a + b*x]^3,x]

[Out]

Log[Cosh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b)

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \tanh ^3(a+b x) \, dx &=-\frac {\tanh ^2(a+b x)}{2 b}+\int \tanh (a+b x) \, dx\\ &=\frac {\log (\cosh (a+b x))}{b}-\frac {\tanh ^2(a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {\log (\cosh (a+b x))}{b}-\frac {\tanh ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + b*x]^3,x]

[Out]

Log[Cosh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b)

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fricas [B]  time = 0.59, size = 339, normalized size = 12.56 \[ -\frac {b x \cosh \left (b x + a\right )^{4} + 4 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b x \sinh \left (b x + a\right )^{4} + 2 \, {\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b x \cosh \left (b x + a\right )^{2} + b x - 1\right )} \sinh \left (b x + a\right )^{2} + b x - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 4 \, {\left (b x \cosh \left (b x + a\right )^{3} + {\left (b x - 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

-(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 + 2*(b*x - 1)*cosh(b*x + a)^
2 + 2*(3*b*x*cosh(b*x + a)^2 + b*x - 1)*sinh(b*x + a)^2 + b*x - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x +
a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 +
cosh(b*x + a))*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(b*x*cosh(b*x + a)^
3 + (b*x - 1)*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*
x + a)^4 + 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 + b*cosh(b
*x + a))*sinh(b*x + a) + b)

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giac [B]  time = 0.16, size = 65, normalized size = 2.41 \[ -\frac {2 \, b x + 2 \, a + \frac {3 \, e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 3}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} - 2 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

-1/2*(2*b*x + 2*a + (3*e^(4*b*x + 4*a) + 2*e^(2*b*x + 2*a) + 3)/(e^(2*b*x + 2*a) + 1)^2 - 2*log(e^(2*b*x + 2*a
) + 1))/b

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maple [A]  time = 0.13, size = 26, normalized size = 0.96 \[ \frac {\ln \left (\cosh \left (b x +a \right )\right )}{b}-\frac {\tanh ^{2}\left (b x +a \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^3*sinh(b*x+a)^3,x)

[Out]

ln(cosh(b*x+a))/b-1/2*tanh(b*x+a)^2/b

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maxima [B]  time = 0.79, size = 61, normalized size = 2.26 \[ x + \frac {a}{b} + \frac {\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac {2 \, e^{\left (-2 \, b x - 2 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

x + a/b + log(e^(-2*b*x - 2*a) + 1)/b + 2*e^(-2*b*x - 2*a)/(b*(2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + 1))

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mupad [B]  time = 1.51, size = 25, normalized size = 0.93 \[ \frac {1}{2\,b\,{\mathrm {cosh}\left (a+b\,x\right )}^2}+\frac {\ln \left (\mathrm {cosh}\left (a+b\,x\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^3/cosh(a + b*x)^3,x)

[Out]

1/(2*b*cosh(a + b*x)^2) + log(cosh(a + b*x))/b

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**3*sinh(b*x+a)**3,x)

[Out]

Integral(sinh(a + b*x)**3*sech(a + b*x)**3, x)

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