3.334 \(\int (x+\cosh (x))^3 \, dx\)

Optimal. Leaf size=56 \[ \frac {x^4}{4}+\frac {3 x^2}{4}+3 x^2 \sinh (x)+\frac {\sinh ^3(x)}{3}+7 \sinh (x)-\frac {3 \cosh ^2(x)}{4}-6 x \cosh (x)+\frac {3}{2} x \sinh (x) \cosh (x) \]

[Out]

3/4*x^2+1/4*x^4-6*x*cosh(x)-3/4*cosh(x)^2+7*sinh(x)+3*x^2*sinh(x)+3/2*x*cosh(x)*sinh(x)+1/3*sinh(x)^3

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Rubi [A]  time = 0.07, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6742, 3296, 2637, 3310, 30, 2633} \[ \frac {x^4}{4}+\frac {3 x^2}{4}+3 x^2 \sinh (x)+\frac {\sinh ^3(x)}{3}+7 \sinh (x)-\frac {3 \cosh ^2(x)}{4}-6 x \cosh (x)+\frac {3}{2} x \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[(x + Cosh[x])^3,x]

[Out]

(3*x^2)/4 + x^4/4 - 6*x*Cosh[x] - (3*Cosh[x]^2)/4 + 7*Sinh[x] + 3*x^2*Sinh[x] + (3*x*Cosh[x]*Sinh[x])/2 + Sinh
[x]^3/3

Rule 30

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

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]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3310

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int (x+\cosh (x))^3 \, dx &=\int \left (x^3+3 x^2 \cosh (x)+3 x \cosh ^2(x)+\cosh ^3(x)\right ) \, dx\\ &=\frac {x^4}{4}+3 \int x^2 \cosh (x) \, dx+3 \int x \cosh ^2(x) \, dx+\int \cosh ^3(x) \, dx\\ &=\frac {x^4}{4}-\frac {3 \cosh ^2(x)}{4}+3 x^2 \sinh (x)+\frac {3}{2} x \cosh (x) \sinh (x)+i \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )+\frac {3 \int x \, dx}{2}-6 \int x \sinh (x) \, dx\\ &=\frac {3 x^2}{4}+\frac {x^4}{4}-6 x \cosh (x)-\frac {3 \cosh ^2(x)}{4}+\sinh (x)+3 x^2 \sinh (x)+\frac {3}{2} x \cosh (x) \sinh (x)+\frac {\sinh ^3(x)}{3}+6 \int \cosh (x) \, dx\\ &=\frac {3 x^2}{4}+\frac {x^4}{4}-6 x \cosh (x)-\frac {3 \cosh ^2(x)}{4}+7 \sinh (x)+3 x^2 \sinh (x)+\frac {3}{2} x \cosh (x) \sinh (x)+\frac {\sinh ^3(x)}{3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 51, normalized size = 0.91 \[ \frac {1}{12} \left (3 x^4+9 x^2+9 \left (4 x^2+9\right ) \sinh (x)+9 x \sinh (2 x)+\sinh (3 x)\right )-6 x \cosh (x)-\frac {3}{8} \cosh (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(x + Cosh[x])^3,x]

[Out]

-6*x*Cosh[x] - (3*Cosh[2*x])/8 + (9*x^2 + 3*x^4 + 9*(9 + 4*x^2)*Sinh[x] + 9*x*Sinh[2*x] + Sinh[3*x])/12

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fricas [A]  time = 0.57, size = 54, normalized size = 0.96 \[ \frac {1}{4} \, x^{4} + \frac {1}{12} \, \sinh \relax (x)^{3} + \frac {3}{4} \, x^{2} - 6 \, x \cosh \relax (x) - \frac {3}{8} \, \cosh \relax (x)^{2} + \frac {1}{4} \, {\left (12 \, x^{2} + 6 \, x \cosh \relax (x) + \cosh \relax (x)^{2} + 27\right )} \sinh \relax (x) - \frac {3}{8} \, \sinh \relax (x)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x))^3,x, algorithm="fricas")

[Out]

1/4*x^4 + 1/12*sinh(x)^3 + 3/4*x^2 - 6*x*cosh(x) - 3/8*cosh(x)^2 + 1/4*(12*x^2 + 6*x*cosh(x) + cosh(x)^2 + 27)
*sinh(x) - 3/8*sinh(x)^2

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giac [A]  time = 0.12, size = 75, normalized size = 1.34 \[ \frac {1}{4} \, x^{4} + \frac {3}{4} \, x^{2} + \frac {3}{16} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {3}{8} \, {\left (4 \, x^{2} + 8 \, x + 9\right )} e^{\left (-x\right )} - \frac {3}{16} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {3}{8} \, {\left (4 \, x^{2} - 8 \, x + 9\right )} e^{x} + \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {1}{24} \, e^{\left (-3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x))^3,x, algorithm="giac")

[Out]

1/4*x^4 + 3/4*x^2 + 3/16*(2*x - 1)*e^(2*x) - 3/8*(4*x^2 + 8*x + 9)*e^(-x) - 3/16*(2*x + 1)*e^(-2*x) + 3/8*(4*x
^2 - 8*x + 9)*e^x + 1/24*e^(3*x) - 1/24*e^(-3*x)

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maple [A]  time = 0.22, size = 52, normalized size = 0.93 \[ \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\relax (x )\right )}{3}\right ) \sinh \relax (x )+\frac {3 x \cosh \relax (x ) \sinh \relax (x )}{2}+\frac {3 x^{2}}{4}-\frac {3 \left (\cosh ^{2}\relax (x )\right )}{4}+3 x^{2} \sinh \relax (x )-6 x \cosh \relax (x )+6 \sinh \relax (x )+\frac {x^{4}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+cosh(x))^3,x)

[Out]

(2/3+1/3*cosh(x)^2)*sinh(x)+3/2*x*cosh(x)*sinh(x)+3/4*x^2-3/4*cosh(x)^2+3*x^2*sinh(x)-6*x*cosh(x)+6*sinh(x)+1/
4*x^4

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maxima [A]  time = 0.32, size = 81, normalized size = 1.45 \[ \frac {1}{4} \, x^{4} + \frac {3}{4} \, x^{2} + \frac {3}{16} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - \frac {3}{2} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} - \frac {3}{16} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {3}{2} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {3}{8} \, e^{\left (-x\right )} - \frac {1}{24} \, e^{\left (-3 \, x\right )} + \frac {3}{8} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x))^3,x, algorithm="maxima")

[Out]

1/4*x^4 + 3/4*x^2 + 3/16*(2*x - 1)*e^(2*x) - 3/2*(x^2 + 2*x + 2)*e^(-x) - 3/16*(2*x + 1)*e^(-2*x) + 3/2*(x^2 -
2*x + 2)*e^x + 1/24*e^(3*x) - 3/8*e^(-x) - 1/24*e^(-3*x) + 3/8*e^x

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mupad [B]  time = 0.07, size = 48, normalized size = 0.86 \[ \frac {20\,\mathrm {sinh}\relax (x)}{3}+3\,x^2\,\mathrm {sinh}\relax (x)-\frac {3\,{\mathrm {cosh}\relax (x)}^2}{4}+\frac {{\mathrm {cosh}\relax (x)}^2\,\mathrm {sinh}\relax (x)}{3}-6\,x\,\mathrm {cosh}\relax (x)+\frac {3\,x^2}{4}+\frac {x^4}{4}+\frac {3\,x\,\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + cosh(x))^3,x)

[Out]

(20*sinh(x))/3 + 3*x^2*sinh(x) - (3*cosh(x)^2)/4 + (cosh(x)^2*sinh(x))/3 - 6*x*cosh(x) + (3*x^2)/4 + x^4/4 + (
3*x*cosh(x)*sinh(x))/2

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sympy [A]  time = 0.31, size = 85, normalized size = 1.52 \[ \frac {x^{4}}{4} - \frac {3 x^{2} \sinh ^{2}{\relax (x )}}{4} + 3 x^{2} \sinh {\relax (x )} + \frac {3 x^{2} \cosh ^{2}{\relax (x )}}{4} + \frac {3 x \sinh {\relax (x )} \cosh {\relax (x )}}{2} - 6 x \cosh {\relax (x )} - \frac {2 \sinh ^{3}{\relax (x )}}{3} + \sinh {\relax (x )} \cosh ^{2}{\relax (x )} + 6 \sinh {\relax (x )} - \frac {3 \cosh ^{2}{\relax (x )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+cosh(x))**3,x)

[Out]

x**4/4 - 3*x**2*sinh(x)**2/4 + 3*x**2*sinh(x) + 3*x**2*cosh(x)**2/4 + 3*x*sinh(x)*cosh(x)/2 - 6*x*cosh(x) - 2*
sinh(x)**3/3 + sinh(x)*cosh(x)**2 + 6*sinh(x) - 3*cosh(x)**2/4

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