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(b*x+a)/b+1/3*cosh(b*x+a)^3/b-sech(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.52, size = 63, normalized size = 1.66 \[ \frac {\cosh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 10\right )} \sinh \left (b x + a\right )^{2} - 20 \, \cosh \left (b x + a\right )^{2} - 45}{24 \, b \cosh \left (b x + a\right )} \]

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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giac [B]  time = 0.15, size = 76, normalized size = 2.00 \[ -\frac {{\left (21 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - {\left (e^{\left (3 \, b x + 24 \, a\right )} - 21 \, e^{\left (b x + 22 \, a\right )}\right )} e^{\left (-21 \, a\right )} + \frac {48 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}}{24 \, b} \]

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

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maple [A]  time = 0.11, size = 52, normalized size = 1.37 \[ \frac {\frac {\sinh ^{4}\left (b x +a \right )}{3 \cosh \left (b x +a \right )}-\frac {4 \left (\sinh ^{2}\left (b x +a \right )\right )}{3 \cosh \left (b x +a \right )}-\frac {8}{3 \cosh \left (b x +a \right )}}{b} \]

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 = 0.43, size = 79, normalized size = 2.08 \[ -\frac {21 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac {20 \, e^{\left (-2 \, b x - 2 \, a\right )} + 69 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1}{24 \, b {\left (e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \]

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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mupad [B]  time = 1.54, size = 78, normalized size = 2.05 \[ \frac {{\mathrm {e}}^{-3\,a-3\,b\,x}}{24\,b}-\frac {7\,{\mathrm {e}}^{-a-b\,x}}{8\,b}-\frac {7\,{\mathrm {e}}^{a+b\,x}}{8\,b}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{24\,b}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(- 3*a - 3*b*x)/(24*b) - (7*exp(- a - b*x))/(8*b) - (7*exp(a + b*x))/(8*b) + exp(3*a + 3*b*x)/(24*b) - (2*e
xp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh ^{3}{\left (a + b x \right )} \tanh ^{2}{\left (a + b x \right )}\, dx \]

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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