3.2 \(\int \tanh ^5(a+b x) \, dx\)

Optimal. Leaf size=42 \[ -\frac{\tanh ^4(a+b x)}{4 b}-\frac{\tanh ^2(a+b x)}{2 b}+\frac{\log (\cosh (a+b x))}{b} \]

[Out]

Log[Cosh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b) - Tanh[a + b*x]^4/(4*b)

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Rubi [A]  time = 0.0333051, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3473, 3475} \[ -\frac{\tanh ^4(a+b x)}{4 b}-\frac{\tanh ^2(a+b x)}{2 b}+\frac{\log (\cosh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[a + b*x]^5,x]

[Out]

Log[Cosh[a + b*x]]/b - Tanh[a + b*x]^2/(2*b) - Tanh[a + b*x]^4/(4*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 ^5(a+b x) \, dx &=-\frac{\tanh ^4(a+b x)}{4 b}+\int \tanh ^3(a+b x) \, dx\\ &=-\frac{\tanh ^2(a+b x)}{2 b}-\frac{\tanh ^4(a+b x)}{4 b}+\int \tanh (a+b x) \, dx\\ &=\frac{\log (\cosh (a+b x))}{b}-\frac{\tanh ^2(a+b x)}{2 b}-\frac{\tanh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0864425, size = 37, normalized size = 0.88 \[ \frac{-\tanh ^4(a+b x)-2 \tanh ^2(a+b x)+4 \log (\cosh (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + b*x]^5,x]

[Out]

(4*Log[Cosh[a + b*x]] - 2*Tanh[a + b*x]^2 - Tanh[a + b*x]^4)/(4*b)

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Maple [A]  time = 0.004, size = 56, normalized size = 1.3 \begin{align*} -{\frac{ \left ( \tanh \left ( bx+a \right ) \right ) ^{4}}{4\,b}}-{\frac{ \left ( \tanh \left ( bx+a \right ) \right ) ^{2}}{2\,b}}-{\frac{\ln \left ( -1+\tanh \left ( bx+a \right ) \right ) }{2\,b}}-{\frac{\ln \left ( 1+\tanh \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(b*x+a)^5,x)

[Out]

-1/4*tanh(b*x+a)^4/b-1/2*tanh(b*x+a)^2/b-1/2/b*ln(-1+tanh(b*x+a))-1/2*ln(1+tanh(b*x+a))/b

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Maxima [B]  time = 1.54729, size = 138, normalized size = 3.29 \begin{align*} x + \frac{a}{b} + \frac{\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} + \frac{4 \,{\left (e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}}{b{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} + 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + a/b + log(e^(-2*b*x - 2*a) + 1)/b + 4*(e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a) + e^(-6*b*x - 6*a))/(b*(4*e^(-2
*b*x - 2*a) + 6*e^(-4*b*x - 4*a) + 4*e^(-6*b*x - 6*a) + e^(-8*b*x - 8*a) + 1))

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Fricas [B]  time = 2.37979, size = 2642, normalized size = 62.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x*cosh(b*x + a)^8 + 8*b*x*cosh(b*x + a)*sinh(b*x + a)^7 + b*x*sinh(b*x + a)^8 + 4*(b*x - 1)*cosh(b*x + a)^
6 + 4*(7*b*x*cosh(b*x + a)^2 + b*x - 1)*sinh(b*x + a)^6 + 8*(7*b*x*cosh(b*x + a)^3 + 3*(b*x - 1)*cosh(b*x + a)
)*sinh(b*x + a)^5 + 2*(3*b*x - 2)*cosh(b*x + a)^4 + 2*(35*b*x*cosh(b*x + a)^4 + 30*(b*x - 1)*cosh(b*x + a)^2 +
3*b*x - 2)*sinh(b*x + a)^4 + 8*(7*b*x*cosh(b*x + a)^5 + 10*(b*x - 1)*cosh(b*x + a)^3 + (3*b*x - 2)*cosh(b*x +
a))*sinh(b*x + a)^3 + 4*(b*x - 1)*cosh(b*x + a)^2 + 4*(7*b*x*cosh(b*x + a)^6 + 15*(b*x - 1)*cosh(b*x + a)^4 +
3*(3*b*x - 2)*cosh(b*x + a)^2 + b*x - 1)*sinh(b*x + a)^2 + b*x - (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x
+ a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^6 + 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^
3 + 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 + 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cos
h(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 + 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x +
a)^6 + 15*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 +
3*cosh(b*x + a)^5 + 3*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))) + 8*(b*x*cosh(b*x + a)^7 + 3*(b*x - 1)*cosh(b*x + a)^5 + (3*b*x - 2)*cosh(b*x + a)^3 + (b*x -
1)*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*x + a)^8 +
4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^6 + 8*(7*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a
))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 + 30*b*cosh(b*x + a)^2 + 3*b)*sinh(b*x + a)
^4 + 8*(7*b*cosh(b*x + a)^5 + 10*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 4*b*cosh(b*x + a)^2
+ 4*(7*b*cosh(b*x + a)^6 + 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 8*(b*cosh(b*x + a
)^7 + 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) + b)

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Sympy [A]  time = 0.47104, size = 42, normalized size = 1. \begin{align*} \begin{cases} x - \frac{\log{\left (\tanh{\left (a + b x \right )} + 1 \right )}}{b} - \frac{\tanh ^{4}{\left (a + b x \right )}}{4 b} - \frac{\tanh ^{2}{\left (a + b x \right )}}{2 b} & \text{for}\: b \neq 0 \\x \tanh ^{5}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(b*x+a)**5,x)

[Out]

Piecewise((x - log(tanh(a + b*x) + 1)/b - tanh(a + b*x)**4/(4*b) - tanh(a + b*x)**2/(2*b), Ne(b, 0)), (x*tanh(
a)**5, True))

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Giac [A]  time = 1.18176, size = 99, normalized size = 2.36 \begin{align*} -\frac{b x + a}{b} + \frac{\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b} + \frac{4 \,{\left (e^{\left (6 \, b x + 6 \, a\right )} + e^{\left (4 \, b x + 4 \, a\right )} + e^{\left (2 \, b x + 2 \, a\right )}\right )}}{b{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + a)/b + log(e^(2*b*x + 2*a) + 1)/b + 4*(e^(6*b*x + 6*a) + e^(4*b*x + 4*a) + e^(2*b*x + 2*a))/(b*(e^(2*b
*x + 2*a) + 1)^4)