3.52 \(\int \frac {\sinh ^4(x)}{a+a \text {sech}(x)} \, dx\)

Optimal. Leaf size=44 \[ -\frac {x}{8 a}-\frac {\sinh ^3(x)}{3 a}+\frac {\sinh (x) \cosh ^3(x)}{4 a}-\frac {\sinh (x) \cosh (x)}{8 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3872, 2839, 2564, 30, 2568, 2635, 8} \[ -\frac {x}{8 a}-\frac {\sinh ^3(x)}{3 a}+\frac {\sinh (x) \cosh ^3(x)}{4 a}-\frac {\sinh (x) \cosh (x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + a*Sech[x]),x]

[Out]

-x/(8*a) - (Cosh[x]*Sinh[x])/(8*a) + (Cosh[x]^3*Sinh[x])/(4*a) - Sinh[x]^3/(3*a)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2568

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2839

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

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 ^4(x)}{a+a \text {sech}(x)} \, dx &=-\int \frac {\cosh (x) \sinh ^4(x)}{-a-a \cosh (x)} \, dx\\ &=-\frac {\int \cosh (x) \sinh ^2(x) \, dx}{a}+\frac {\int \cosh ^2(x) \sinh ^2(x) \, dx}{a}\\ &=\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {i \operatorname {Subst}\left (\int x^2 \, dx,x,i \sinh (x)\right )}{a}-\frac {\int \cosh ^2(x) \, dx}{4 a}\\ &=-\frac {\cosh (x) \sinh (x)}{8 a}+\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {\sinh ^3(x)}{3 a}-\frac {\int 1 \, dx}{8 a}\\ &=-\frac {x}{8 a}-\frac {\cosh (x) \sinh (x)}{8 a}+\frac {\cosh ^3(x) \sinh (x)}{4 a}-\frac {\sinh ^3(x)}{3 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 28, normalized size = 0.64 \[ \frac {24 \sinh (x)-8 \sinh (3 x)+3 (\sinh (4 x)-4 x)}{96 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(a + a*Sech[x]),x]

[Out]

(24*Sinh[x] - 8*Sinh[3*x] + 3*(-4*x + Sinh[4*x]))/(96*a)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 36, normalized size = 0.82 \[ \frac {{\left (3 \, \cosh \relax (x) - 2\right )} \sinh \relax (x)^{3} + 3 \, {\left (\cosh \relax (x)^{3} - 2 \, \cosh \relax (x)^{2} + 2\right )} \sinh \relax (x) - 3 \, x}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+a*sech(x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.11, size = 42, normalized size = 0.95 \[ -\frac {{\left (24 \, e^{\left (3 \, x\right )} - 8 \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} + 24 \, x - 3 \, e^{\left (4 \, x\right )} + 8 \, e^{\left (3 \, x\right )} - 24 \, e^{x}}{192 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+a*sech(x)),x, algorithm="giac")

[Out]

-1/192*((24*e^(3*x) - 8*e^x + 3)*e^(-4*x) + 24*x - 3*e^(4*x) + 8*e^(3*x) - 24*e^x)/a

________________________________________________________________________________________

maple [B]  time = 0.14, size = 130, normalized size = 2.95 \[ \frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {5}{6 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 a}-\frac {1}{4 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {5}{6 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a+a*sech(x)),x)

[Out]

1/4/a/(tanh(1/2*x)-1)^4+5/6/a/(tanh(1/2*x)-1)^3+7/8/a/(tanh(1/2*x)-1)^2+1/8/a/(tanh(1/2*x)-1)+1/8/a*ln(tanh(1/
2*x)-1)-1/4/a/(tanh(1/2*x)+1)^4+5/6/a/(tanh(1/2*x)+1)^3-7/8/a/(tanh(1/2*x)+1)^2+1/8/a/(tanh(1/2*x)+1)-1/8/a*ln
(tanh(1/2*x)+1)

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 54, normalized size = 1.23 \[ -\frac {{\left (8 \, e^{\left (-x\right )} - 24 \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )}}{192 \, a} - \frac {x}{8 \, a} - \frac {24 \, e^{\left (-x\right )} - 8 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )}}{192 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+a*sech(x)),x, algorithm="maxima")

[Out]

-1/192*(8*e^(-x) - 24*e^(-3*x) - 3)*e^(4*x)/a - 1/8*x/a - 1/192*(24*e^(-x) - 8*e^(-3*x) + 3*e^(-4*x))/a

________________________________________________________________________________________

mupad [B]  time = 1.48, size = 59, normalized size = 1.34 \[ \frac {{\mathrm {e}}^{-3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-x}}{8\,a}-\frac {{\mathrm {e}}^{3\,x}}{24\,a}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a}+\frac {{\mathrm {e}}^{4\,x}}{64\,a}-\frac {x}{8\,a}+\frac {{\mathrm {e}}^x}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + a/cosh(x)),x)

[Out]

exp(-3*x)/(24*a) - exp(-x)/(8*a) - exp(3*x)/(24*a) - exp(-4*x)/(64*a) + exp(4*x)/(64*a) - x/(8*a) + exp(x)/(8*
a)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sinh ^{4}{\relax (x )}}{\operatorname {sech}{\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(a+a*sech(x)),x)

[Out]

Integral(sinh(x)**4/(sech(x) + 1), x)/a

________________________________________________________________________________________