3.73 \(\int \sinh (a+b x) \tanh ^4(a+b x) \, dx\)

Optimal. Leaf size=37 \[ \frac{\cosh (a+b x)}{b}-\frac{\text{sech}^3(a+b x)}{3 b}+\frac{2 \text{sech}(a+b x)}{b} \]

[Out]

Cosh[a + b*x]/b + (2*Sech[a + b*x])/b - Sech[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0329711, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2590, 270} \[ \frac{\cosh (a+b x)}{b}-\frac{\text{sech}^3(a+b x)}{3 b}+\frac{2 \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]*Tanh[a + b*x]^4,x]

[Out]

Cosh[a + b*x]/b + (2*Sech[a + b*x])/b - Sech[a + b*x]^3/(3*b)

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \sinh (a+b x) \tanh ^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}-\frac{2}{x^2}\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\cosh (a+b x)}{b}+\frac{2 \text{sech}(a+b x)}{b}-\frac{\text{sech}^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0348394, size = 37, normalized size = 1. \[ \frac{\cosh (a+b x)}{b}-\frac{\text{sech}^3(a+b x)}{3 b}+\frac{2 \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]*Tanh[a + b*x]^4,x]

[Out]

Cosh[a + b*x]/b + (2*Sech[a + b*x])/b - Sech[a + b*x]^3/(3*b)

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Maple [A]  time = 0.017, size = 67, normalized size = 1.8 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}-{\frac{8\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,\cosh \left ( bx+a \right ) }}+{\frac{8\,\cosh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)*tanh(b*x+a)^4,x)

[Out]

1/b*(sinh(b*x+a)^4/cosh(b*x+a)^3+4/3*sinh(b*x+a)^2/cosh(b*x+a)^3-8/3*sinh(b*x+a)^2/cosh(b*x+a)+8/3*cosh(b*x+a)
)

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Maxima [B]  time = 1.02946, size = 132, normalized size = 3.57 \begin{align*} \frac{e^{\left (-b x - a\right )}}{2 \, b} + \frac{33 \, e^{\left (-2 \, b x - 2 \, a\right )} + 41 \, e^{\left (-4 \, b x - 4 \, a\right )} + 27 \, e^{\left (-6 \, b x - 6 \, a\right )} + 3}{6 \, b{\left (e^{\left (-b x - a\right )} + 3 \, e^{\left (-3 \, b x - 3 \, a\right )} + 3 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^4,x, algorithm="maxima")

[Out]

1/2*e^(-b*x - a)/b + 1/6*(33*e^(-2*b*x - 2*a) + 41*e^(-4*b*x - 4*a) + 27*e^(-6*b*x - 6*a) + 3)/(b*(e^(-b*x - a
) + 3*e^(-3*b*x - 3*a) + 3*e^(-5*b*x - 5*a) + e^(-7*b*x - 7*a)))

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Fricas [B]  time = 1.75935, size = 257, normalized size = 6.95 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{4} + 3 \, \sinh \left (b x + a\right )^{4} + 18 \,{\left (\cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 36 \, \cosh \left (b x + a\right )^{2} + 25}{6 \,{\left (b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 3 \, b \cosh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^4,x, algorithm="fricas")

[Out]

1/6*(3*cosh(b*x + a)^4 + 3*sinh(b*x + a)^4 + 18*(cosh(b*x + a)^2 + 2)*sinh(b*x + a)^2 + 36*cosh(b*x + a)^2 + 2
5)/(b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a)*sinh(b*x + a)^2 + 3*b*cosh(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (a + b x \right )} \tanh ^{4}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)**4,x)

[Out]

Integral(sinh(a + b*x)*tanh(a + b*x)**4, x)

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Giac [B]  time = 1.27727, size = 96, normalized size = 2.59 \begin{align*} \frac{\frac{8 \,{\left (3 \, e^{\left (5 \, b x + 5 \, a\right )} + 4 \, e^{\left (3 \, b x + 3 \, a\right )} + 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} + 3 \, e^{\left (b x + a\right )} + 3 \, e^{\left (-b x - a\right )}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)*tanh(b*x+a)^4,x, algorithm="giac")

[Out]

1/6*(8*(3*e^(5*b*x + 5*a) + 4*e^(3*b*x + 3*a) + 3*e^(b*x + a))/(e^(2*b*x + 2*a) + 1)^3 + 3*e^(b*x + a) + 3*e^(
-b*x - a))/b