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

Optimal. Leaf size=38 \[ \frac{\sinh ^3(a+b x)}{3 b}-\frac{\sinh (a+b x)}{b}+\frac{\tan ^{-1}(\sinh (a+b x))}{b} \]

[Out]

ArcTan[Sinh[a + b*x]]/b - Sinh[a + b*x]/b + Sinh[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.0264162, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2592, 302, 203} \[ \frac{\sinh ^3(a+b x)}{3 b}-\frac{\sinh (a+b x)}{b}+\frac{\tan ^{-1}(\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^3*Tanh[a + b*x],x]

[Out]

ArcTan[Sinh[a + b*x]]/b - Sinh[a + b*x]/b + Sinh[a + b*x]^3/(3*b)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0144086, size = 38, normalized size = 1. \[ \frac{\sinh ^3(a+b x)}{3 b}-\frac{\sinh (a+b x)}{b}+\frac{\tan ^{-1}(\sinh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^3*Tanh[a + b*x],x]

[Out]

ArcTan[Sinh[a + b*x]]/b - Sinh[a + b*x]/b + Sinh[a + b*x]^3/(3*b)

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Maple [A]  time = 0.019, size = 38, normalized size = 1. \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{3\,b}}-{\frac{\sinh \left ( bx+a \right ) }{b}}+2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sinh(b*x+a)^3/b-sinh(b*x+a)/b+2*arctan(exp(b*x+a))/b

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Maxima [A]  time = 1.56535, size = 96, normalized size = 2.53 \begin{align*} -\frac{{\left (15 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{24 \, b} + \frac{15 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac{2 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*e^(-2*b*x - 2*a) - 1)*e^(3*b*x + 3*a)/b + 1/24*(15*e^(-b*x - a) - e^(-3*b*x - 3*a))/b - 2*arctan(e^(
-b*x - a))/b

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Fricas [B]  time = 1.87216, size = 836, normalized size = 22. \begin{align*} \frac{\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + 15 \,{\left (\cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{4} - 15 \, \cosh \left (b x + a\right )^{4} + 20 \,{\left (\cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 15 \,{\left (\cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 48 \,{\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3}\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 15 \, \cosh \left (b x + a\right )^{2} + 6 \,{\left (\cosh \left (b x + a\right )^{5} - 10 \, \cosh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1}{24 \,{\left (b \cosh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + b \sinh \left (b x + a\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 15*(cosh(b*x + a)^2 - 1)*sinh(b*x
+ a)^4 - 15*cosh(b*x + a)^4 + 20*(cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 15*(cosh(b*x + a)^4 - 6
*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 48*(cosh(b*x + a)^3 + 3*cosh(b*x + a)^2*sinh(b*x + a) + 3*cosh(b*x + a
)*sinh(b*x + a)^2 + sinh(b*x + a)^3)*arctan(cosh(b*x + a) + sinh(b*x + a)) + 15*cosh(b*x + a)^2 + 6*(cosh(b*x
+ a)^5 - 10*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a) - 1)/(b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a)^2*sin
h(b*x + a) + 3*b*cosh(b*x + a)*sinh(b*x + a)^2 + b*sinh(b*x + a)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{3}{\left (a + b x \right )} \tanh{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18248, size = 85, normalized size = 2.24 \begin{align*} \frac{{\left (15 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} +{\left (e^{\left (3 \, b x + 18 \, a\right )} - 15 \, e^{\left (b x + 16 \, a\right )}\right )} e^{\left (-15 \, a\right )} + 48 \, \arctan \left (e^{\left (b x + a\right )}\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((15*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) + (e^(3*b*x + 18*a) - 15*e^(b*x + 16*a))*e^(-15*a) + 48*arctan
(e^(b*x + a)))/b