3.78 \(\int \sinh ^3(a+b x) \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac{\cosh ^3(a+b x)}{3 b}-\frac{2 \cosh (a+b x)}{b}-\frac{\text{sech}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.040545, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2590, 270} \[ \frac{\cosh ^3(a+b x)}{3 b}-\frac{2 \cosh (a+b x)}{b}-\frac{\text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^3*Tanh[a + b*x]^2,x]

[Out]

(-2*Cosh[a + b*x])/b + Cosh[a + b*x]^3/(3*b) - Sech[a + b*x]/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 ^3(a+b x) \tanh ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2+\frac{1}{x^2}+x^2\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac{2 \cosh (a+b x)}{b}+\frac{\cosh ^3(a+b x)}{3 b}-\frac{\text{sech}(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0334827, size = 40, normalized size = 1.05 \[ -\frac{7 \cosh (a+b x)}{4 b}+\frac{\cosh (3 (a+b x))}{12 b}-\frac{\text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^3*Tanh[a + b*x]^2,x]

[Out]

(-7*Cosh[a + b*x])/(4*b) + Cosh[3*(a + b*x)]/(12*b) - Sech[a + b*x]/b

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Maple [A]  time = 0.019, size = 50, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{3\,\cosh \left ( bx+a \right ) }}+{\frac{4\, \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)^3*tanh(b*x+a)^2,x)

[Out]

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

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Maxima [B]  time = 1.06418, size = 107, normalized size = 2.82 \begin{align*} -\frac{21 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac{20 \, e^{\left (-2 \, b x - 2 \, a\right )} + 69 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1}{24 \, b{\left (e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^3*tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/24*(21*e^(-b*x - a) - e^(-3*b*x - 3*a))/b - 1/24*(20*e^(-2*b*x - 2*a) + 69*e^(-4*b*x - 4*a) - 1)/(b*(e^(-3*
b*x - 3*a) + e^(-5*b*x - 5*a)))

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Fricas [A]  time = 1.82863, size = 177, normalized size = 4.66 \begin{align*} \frac{\cosh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{4} + 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} - 10\right )} \sinh \left (b x + a\right )^{2} - 20 \, \cosh \left (b x + a\right )^{2} - 45}{24 \, b \cosh \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^3*tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/24*(cosh(b*x + a)^4 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 10)*sinh(b*x + a)^2 - 20*cosh(b*x + a)^2 - 45
)/(b*cosh(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh ^{3}{\left (a + b x \right )} \tanh ^{2}{\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)**2,x)

[Out]

Integral(sinh(a + b*x)**3*tanh(a + b*x)**2, x)

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Giac [B]  time = 1.2644, size = 103, normalized size = 2.71 \begin{align*} -\frac{{\left (21 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} -{\left (e^{\left (3 \, b x + 24 \, a\right )} - 21 \, e^{\left (b x + 22 \, a\right )}\right )} e^{\left (-21 \, a\right )} + \frac{48 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}}{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)^2,x, algorithm="giac")

[Out]

-1/24*((21*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) - (e^(3*b*x + 24*a) - 21*e^(b*x + 22*a))*e^(-21*a) + 48*e^(b*
x + a)/(e^(2*b*x + 2*a) + 1))/b