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

Optimal. Leaf size=67 \[ \frac{15 x}{8 a}-\frac{4 \sinh ^3(x)}{3 a}-\frac{4 \sinh (x)}{a}+\frac{5 \sinh (x) \cosh ^3(x)}{4 a}+\frac{15 \sinh (x) \cosh (x)}{8 a}-\frac{\sinh (x) \cosh ^3(x)}{a \text{sech}(x)+a} \]

[Out]

(15*x)/(8*a) - (4*Sinh[x])/a + (15*Cosh[x]*Sinh[x])/(8*a) + (5*Cosh[x]^3*Sinh[x])/(4*a) - (Cosh[x]^3*Sinh[x])/
(a + a*Sech[x]) - (4*Sinh[x]^3)/(3*a)

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Rubi [A]  time = 0.0961097, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3819, 3787, 2635, 8, 2633} \[ \frac{15 x}{8 a}-\frac{4 \sinh ^3(x)}{3 a}-\frac{4 \sinh (x)}{a}+\frac{5 \sinh (x) \cosh ^3(x)}{4 a}+\frac{15 \sinh (x) \cosh (x)}{8 a}-\frac{\sinh (x) \cosh ^3(x)}{a \text{sech}(x)+a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + a*Sech[x]),x]

[Out]

(15*x)/(8*a) - (4*Sinh[x])/a + (15*Cosh[x]*Sinh[x])/(8*a) + (5*Cosh[x]^3*Sinh[x])/(4*a) - (Cosh[x]^3*Sinh[x])/
(a + a*Sech[x]) - (4*Sinh[x]^3)/(3*a)

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

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 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \frac{\cosh ^4(x)}{a+a \text{sech}(x)} \, dx &=-\frac{\cosh ^3(x) \sinh (x)}{a+a \text{sech}(x)}-\frac{\int \cosh ^4(x) (-5 a+4 a \text{sech}(x)) \, dx}{a^2}\\ &=-\frac{\cosh ^3(x) \sinh (x)}{a+a \text{sech}(x)}-\frac{4 \int \cosh ^3(x) \, dx}{a}+\frac{5 \int \cosh ^4(x) \, dx}{a}\\ &=\frac{5 \cosh ^3(x) \sinh (x)}{4 a}-\frac{\cosh ^3(x) \sinh (x)}{a+a \text{sech}(x)}-\frac{(4 i) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )}{a}+\frac{15 \int \cosh ^2(x) \, dx}{4 a}\\ &=-\frac{4 \sinh (x)}{a}+\frac{15 \cosh (x) \sinh (x)}{8 a}+\frac{5 \cosh ^3(x) \sinh (x)}{4 a}-\frac{\cosh ^3(x) \sinh (x)}{a+a \text{sech}(x)}-\frac{4 \sinh ^3(x)}{3 a}+\frac{15 \int 1 \, dx}{8 a}\\ &=\frac{15 x}{8 a}-\frac{4 \sinh (x)}{a}+\frac{15 \cosh (x) \sinh (x)}{8 a}+\frac{5 \cosh ^3(x) \sinh (x)}{4 a}-\frac{\cosh ^3(x) \sinh (x)}{a+a \text{sech}(x)}-\frac{4 \sinh ^3(x)}{3 a}\\ \end{align*}

Mathematica [A]  time = 0.0864632, size = 63, normalized size = 0.94 \[ \frac{\text{sech}\left (\frac{x}{2}\right ) \left (-360 \sinh \left (\frac{x}{2}\right )-120 \sinh \left (\frac{3 x}{2}\right )+40 \sinh \left (\frac{5 x}{2}\right )-5 \sinh \left (\frac{7 x}{2}\right )+3 \sinh \left (\frac{9 x}{2}\right )+360 x \cosh \left (\frac{x}{2}\right )\right )}{192 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[x/2]*(360*x*Cosh[x/2] - 360*Sinh[x/2] - 120*Sinh[(3*x)/2] + 40*Sinh[(5*x)/2] - 5*Sinh[(7*x)/2] + 3*Sinh[
(9*x)/2]))/(192*a)

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Maple [B]  time = 0.032, size = 139, normalized size = 2.1 \begin{align*} -{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) }-{\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{15}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{25}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{15}{8\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\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{15}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{25}{8\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{15}{8\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*tanh(1/2*x)-1/4/a/(tanh(1/2*x)+1)^4+5/6/a/(tanh(1/2*x)+1)^3-15/8/a/(tanh(1/2*x)+1)^2+25/8/a/(tanh(1/2*x)+
1)+15/8/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+15/8/a/(tanh(1/2*x)-1)^2+25/8/a/(t
anh(1/2*x)-1)-15/8/a*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.07872, size = 108, normalized size = 1.61 \begin{align*} \frac{15 \, x}{8 \, a} + \frac{168 \, e^{\left (-x\right )} - 48 \, e^{\left (-2 \, x\right )} + 8 \, e^{\left (-3 \, x\right )} - 3 \, e^{\left (-4 \, x\right )}}{192 \, a} - \frac{5 \, e^{\left (-x\right )} - 40 \, e^{\left (-2 \, x\right )} + 120 \, e^{\left (-3 \, x\right )} + 552 \, e^{\left (-4 \, x\right )} - 3}{192 \,{\left (a e^{\left (-4 \, x\right )} + a e^{\left (-5 \, x\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*x/a + 1/192*(168*e^(-x) - 48*e^(-2*x) + 8*e^(-3*x) - 3*e^(-4*x))/a - 1/192*(5*e^(-x) - 40*e^(-2*x) + 120*
e^(-3*x) + 552*e^(-4*x) - 3)/(a*e^(-4*x) + a*e^(-5*x))

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Fricas [B]  time = 2.54816, size = 491, normalized size = 7.33 \begin{align*} \frac{3 \, \cosh \left (x\right )^{5} +{\left (15 \, \cosh \left (x\right ) - 8\right )} \sinh \left (x\right )^{4} + 3 \, \sinh \left (x\right )^{5} - 8 \, \cosh \left (x\right )^{4} +{\left (30 \, \cosh \left (x\right )^{2} - 8 \, \cosh \left (x\right ) + 35\right )} \sinh \left (x\right )^{3} + 45 \, \cosh \left (x\right )^{3} +{\left (30 \, \cosh \left (x\right )^{3} - 48 \, \cosh \left (x\right )^{2} + 135 \, \cosh \left (x\right ) - 160\right )} \sinh \left (x\right )^{2} + 24 \,{\left (15 \, x - 2\right )} \cosh \left (x\right ) - 160 \, \cosh \left (x\right )^{2} +{\left (15 \, \cosh \left (x\right )^{4} - 8 \, \cosh \left (x\right )^{3} + 105 \, \cosh \left (x\right )^{2} + 360 \, x - 160 \, \cosh \left (x\right ) - 288\right )} \sinh \left (x\right ) + 360 \, x + 552}{192 \,{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*cosh(x)^5 + (15*cosh(x) - 8)*sinh(x)^4 + 3*sinh(x)^5 - 8*cosh(x)^4 + (30*cosh(x)^2 - 8*cosh(x) + 35)*
sinh(x)^3 + 45*cosh(x)^3 + (30*cosh(x)^3 - 48*cosh(x)^2 + 135*cosh(x) - 160)*sinh(x)^2 + 24*(15*x - 2)*cosh(x)
- 160*cosh(x)^2 + (15*cosh(x)^4 - 8*cosh(x)^3 + 105*cosh(x)^2 + 360*x - 160*cosh(x) - 288)*sinh(x) + 360*x +
552)/(a*cosh(x) + a*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cosh ^{4}{\left (x \right )}}{\operatorname{sech}{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18051, size = 116, normalized size = 1.73 \begin{align*} \frac{15 \, x}{8 \, a} + \frac{{\left (552 \, e^{\left (4 \, x\right )} + 120 \, e^{\left (3 \, x\right )} - 40 \, e^{\left (2 \, x\right )} + 5 \, e^{x} - 3\right )} e^{\left (-4 \, x\right )}}{192 \, a{\left (e^{x} + 1\right )}} + \frac{3 \, a^{3} e^{\left (4 \, x\right )} - 8 \, a^{3} e^{\left (3 \, x\right )} + 48 \, a^{3} e^{\left (2 \, x\right )} - 168 \, a^{3} e^{x}}{192 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/8*x/a + 1/192*(552*e^(4*x) + 120*e^(3*x) - 40*e^(2*x) + 5*e^x - 3)*e^(-4*x)/(a*(e^x + 1)) + 1/192*(3*a^3*e^
(4*x) - 8*a^3*e^(3*x) + 48*a^3*e^(2*x) - 168*a^3*e^x)/a^4