3.53 \(\int \frac {\sinh ^3(x)}{a+a \text {sech}(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\cosh ^3(x)}{3 a}-\frac {\sinh ^2(x)}{2 a} \]

[Out]

1/3*cosh(x)^3/a-1/2*sinh(x)^2/a

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Rubi [A]  time = 0.12, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3872, 2835, 2564, 30, 2565} \[ \frac {\cosh ^3(x)}{3 a}-\frac {\sinh ^2(x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(a + a*Sech[x]),x]

[Out]

Cosh[x]^3/(3*a) - Sinh[x]^2/(2*a)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 23, normalized size = 1.00 \[ \frac {3 \cosh (x)-3 \cosh (2 x)+\cosh (3 x)-7}{12 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(a + a*Sech[x]),x]

[Out]

(-7 + 3*Cosh[x] - 3*Cosh[2*x] + Cosh[3*x])/(12*a)

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fricas [A]  time = 0.38, size = 30, normalized size = 1.30 \[ \frac {\cosh \relax (x)^{3} + 3 \, {\left (\cosh \relax (x) - 1\right )} \sinh \relax (x)^{2} - 3 \, \cosh \relax (x)^{2} + 3 \, \cosh \relax (x)}{12 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+a*sech(x)),x, algorithm="fricas")

[Out]

1/12*(cosh(x)^3 + 3*(cosh(x) - 1)*sinh(x)^2 - 3*cosh(x)^2 + 3*cosh(x))/a

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giac [A]  time = 0.13, size = 37, normalized size = 1.61 \[ \frac {{\left (3 \, e^{\left (2 \, x\right )} - 3 \, e^{x} + 1\right )} e^{\left (-3 \, x\right )} + e^{\left (3 \, x\right )} - 3 \, e^{\left (2 \, x\right )} + 3 \, e^{x}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+a*sech(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.12, size = 67, normalized size = 2.91 \[ \frac {-\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )-1}+\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {8}{8 \tanh \left (\frac {x}{2}\right )+8}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a+a*sech(x)),x)

[Out]

8/a*(-1/24/(tanh(1/2*x)-1)^3-1/8/(tanh(1/2*x)-1)^2-1/8/(tanh(1/2*x)-1)+1/24/(tanh(1/2*x)+1)^3-1/8/(tanh(1/2*x)
+1)^2+1/8/(tanh(1/2*x)+1))

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maxima [B]  time = 0.32, size = 46, normalized size = 2.00 \[ -\frac {{\left (3 \, e^{\left (-x\right )} - 3 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (3 \, x\right )}}{24 \, a} + \frac {3 \, e^{\left (-x\right )} - 3 \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(a+a*sech(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.36, size = 53, normalized size = 2.30 \[ \frac {{\mathrm {e}}^{-x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {{\mathrm {e}}^{2\,x}}{8\,a}+\frac {{\mathrm {e}}^{-3\,x}}{24\,a}+\frac {{\mathrm {e}}^{3\,x}}{24\,a}+\frac {{\mathrm {e}}^x}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(a + a/cosh(x)),x)

[Out]

exp(-x)/(8*a) - exp(-2*x)/(8*a) - exp(2*x)/(8*a) + exp(-3*x)/(24*a) + exp(3*x)/(24*a) + exp(x)/(8*a)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sinh ^{3}{\relax (x )}}{\operatorname {sech}{\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3/(a+a*sech(x)),x)

[Out]

Integral(sinh(x)**3/(sech(x) + 1), x)/a

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