3.73.68 $\int \frac{1}{4}{e}^{2x}(155-32x-19x^2-2x^3-2x^4+e^5(-5-10x+3x^2+2x^3))dx$

Optimal. Leaf size=31 $$e^{2x}(4-x-\frac{1}{4}(-3+e^5-x)x)(5-x^2)$$

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Rubi [B]  time = 0.29, antiderivative size = 80, normalized size of antiderivative = 2.58, number of steps used = 30, number of rules used = 4, integrand size = 48, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.083, Rules used = {12, 2196, 2194, 2176} $$\begin{array}{c}-\frac{1}{4}{e}^{2x}{x}^4+\frac{1}{4}{e}^{2x}{x}^3+\frac{1}{4}{e}^{2x+5}{x}^3-\frac{11}{4}{e}^{2x}{x}^2-\frac{5}{4}{e}^{2x}x-\frac{5}{4}{e}^{2x+5}x+20e^{2x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]

[Out]

20*E^(2*x) - (5*E^(2*x)*x)/4 - (5*E^(5 + 2*x)*x)/4 - (11*E^(2*x)*x^2)/4 + (E^(2*x)*x^3)/4 + (E^(5 + 2*x)*x^3)/
4 - (E^(2*x)*x^4)/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int e^{2x}(155-32x-19x^2-2x^3-2x^4+e^5(-5-10x+3x^2+2x^3))dx\\ & =\frac{1}{4}\int (155e^{2x}-32e^{2x}x-19e^{2x}{x}^2-2e^{2x}{x}^3-2e^{2x}{x}^4+e^{5+2x}(-5-10x+3x^2+2x^3))dx\\ & =\frac{1}{4}\int e^{5+2x}(-5-10x+3x^2+2x^3)dx-\frac{1}{2}\int e^{2x}{x}^{3}dx-\frac{1}{2}\int e^{2x}{x}^{4}dx-\frac{19}{4}\int e^{2x}{x}^{2}dx-8\int e^{2x}xdx+\frac{155}{4}\int e^{2x}dx\\ & =\frac{155e^{2x}}{8}-4e^{2x}x-\frac{19}{8}{e}^{2x}{x}^2-\frac{1}{4}{e}^{2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4+\frac{1}{4}\int (-5e^{5+2x}-10e^{5+2x}x+3e^{5+2x}{x}^2+2e^{5+2x}{x}^3)dx+\frac{3}{4}\int e^{2x}{x}^{2}dx+4\int e^{2x}dx+\frac{19}{4}\int e^{2x}xdx+\int e^{2x}{x}^{3}dx\\ & =\frac{171e^{2x}}{8}-\frac{13}{8}{e}^{2x}x-2e^{2x}{x}^2+\frac{1}{4}{e}^{2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4+\frac{1}{2}\int e^{5+2x}{x}^{3}dx-\frac{3}{4}\int e^{2x}xdx+\frac{3}{4}\int e^{5+2x}{x}^{2}dx-\frac{5}{4}\int e^{5+2x}dx-\frac{3}{2}\int e^{2x}{x}^{2}dx-\frac{19}{8}\int e^{2x}dx-\frac{5}{2}\int e^{5+2x}xdx\\ & =\frac{323e^{2x}}{16}-\frac{5}{8}{e}^{5+2x}-2e^{2x}x-\frac{5}{4}{e}^{5+2x}x-\frac{11}{4}{e}^{2x}{x}^2+\frac{3}{8}{e}^{5+2x}{x}^2+\frac{1}{4}{e}^{2x}{x}^3+\frac{1}{4}{e}^{5+2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4+\frac{3}{8}\int e^{2x}dx-\frac{3}{4}\int e^{5+2x}xdx-\frac{3}{4}\int e^{5+2x}{x}^{2}dx+\frac{5}{4}\int e^{5+2x}dx+\frac{3}{2}\int e^{2x}xdx\\ & =\frac{163e^{2x}}{8}-\frac{5}{4}{e}^{2x}x-\frac{13}{8}{e}^{5+2x}x-\frac{11}{4}{e}^{2x}{x}^2+\frac{1}{4}{e}^{2x}{x}^3+\frac{1}{4}{e}^{5+2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4+\frac{3}{8}\int e^{5+2x}dx-\frac{3}{4}\int e^{2x}dx+\frac{3}{4}\int e^{5+2x}xdx\\ & =20e^{2x}+\frac{3}{16}{e}^{5+2x}-\frac{5}{4}{e}^{2x}x-\frac{5}{4}{e}^{5+2x}x-\frac{11}{4}{e}^{2x}{x}^2+\frac{1}{4}{e}^{2x}{x}^3+\frac{1}{4}{e}^{5+2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4-\frac{3}{8}\int e^{5+2x}dx\\ & =20e^{2x}-\frac{5}{4}{e}^{2x}x-\frac{5}{4}{e}^{5+2x}x-\frac{11}{4}{e}^{2x}{x}^2+\frac{1}{4}{e}^{2x}{x}^3+\frac{1}{4}{e}^{5+2x}{x}^3-\frac{1}{4}{e}^{2x}{x}^4\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 38, normalized size = 1.23 $$\begin{array}{c}\frac{1}{4}{e}^{2x}(80-5(1+e^5)x-11x^2+(1+e^5)x^3-x^4)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(155 - 32*x - 19*x^2 - 2*x^3 - 2*x^4 + E^5*(-5 - 10*x + 3*x^2 + 2*x^3)))/4,x]

[Out]

(E^(2*x)*(80 - 5*(1 + E^5)*x - 11*x^2 + (1 + E^5)*x^3 - x^4))/4

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fricas [A]  time = 0.59, size = 35, normalized size = 1.13 $$\begin{array}{c}-\frac{1}{4}(x^4-x^3+11x^2-(x^3-5x)e^5+5x-80)e^{\left(2x\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="fricas")

[Out]

-1/4*(x^4 - x^3 + 11*x^2 - (x^3 - 5*x)*e^5 + 5*x - 80)*e^(2*x)

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giac [A]  time = 0.23, size = 40, normalized size = 1.29 $$\begin{array}{c}-\frac{1}{4}(x^4-x^3+11x^2+5x-80)e^{\left(2x\right)}+\frac{1}{4}(x^3-5x)e^{(2x+5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="giac")

[Out]

-1/4*(x^4 - x^3 + 11*x^2 + 5*x - 80)*e^(2*x) + 1/4*(x^3 - 5*x)*e^(2*x + 5)

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maple [A]  time = 0.05, size = 36, normalized size = 1.16

method	result	size
gosper	$\frac{{\mathrm{e}}^{2x}(x^3{\mathrm{e}}^5-x^4+x^3-5x{\mathrm{e}}^5-11x^2-5x+80)}{4}$	$36$
risch	$\frac{{\mathrm{e}}^{2x}(x^3{\mathrm{e}}^5-x^4+x^3-5x{\mathrm{e}}^5-11x^2-5x+80)}{4}$	$36$
norman	$(-\frac{5}{4}-\frac{5{\mathrm{e}}^5}{4})x{\mathrm{e}}^{2x}+(\frac{{\mathrm{e}}^5}{4}+\frac{1}{4})x^3{\mathrm{e}}^{2x}+20{\mathrm{e}}^{2x}-\frac{11{\mathrm{e}}^{2x}{x}^2}{4}-\frac{{\mathrm{e}}^{2x}{x}^4}{4}$	$52$
meijerg	$-19+\frac{155{\mathrm{e}}^{2x}}{8}+\frac{(2{\mathrm{e}}^5-2)(6-\frac{(-32x^3+48x^2-48x+24){\mathrm{e}}^{2x}}{4})}{64}-\frac{(3{\mathrm{e}}^5-19)(2-\frac{(12x^2-12x+6){\mathrm{e}}^{2x}}{3})}{32}+\frac{(-10{\mathrm{e}}^5-32)(1-\frac{(-4x+2){\mathrm{e}}^{2x}}{2})}{16}-\frac{(80x^4-160x^3+240x^2-240x+120){\mathrm{e}}^{2x}}{320}+\frac{5{\mathrm{e}}^5(1-{\mathrm{e}}^{2x})}{8}$	$125$
default	$20{\mathrm{e}}^{2x}-\frac{5x{\mathrm{e}}^{2x}}{4}-\frac{5{\mathrm{e}}^5{\mathrm{e}}^{2x}}{8}-\frac{11{\mathrm{e}}^{2x}{x}^2}{4}+\frac{{\mathrm{e}}^{2x}{x}^3}{4}-\frac{{\mathrm{e}}^{2x}{x}^4}{4}-\frac{5{\mathrm{e}}^5(\frac{x{\mathrm{e}}^{2x}}{2}-\frac{{\mathrm{e}}^{2x}}{4})}{2}+\frac{3{\mathrm{e}}^5(\frac{{\mathrm{e}}^{2x}{x}^2}{2}-\frac{x{\mathrm{e}}^{2x}}{2}+\frac{{\mathrm{e}}^{2x}}{4})}{4}+\frac{{\mathrm{e}}^5(\frac{{\mathrm{e}}^{2x}{x}^3}{2}-\frac{3{\mathrm{e}}^{2x}{x}^2}{4}+\frac{3x{\mathrm{e}}^{2x}}{4}-\frac{3{\mathrm{e}}^{2x}}{8})}{2}$	$131$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*exp(x)^2*(x^3*exp(5)-x^4+x^3-5*x*exp(5)-11*x^2-5*x+80)

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maxima [B]  time = 0.35, size = 156, normalized size = 5.03 $$\begin{array}{c}-\frac{1}{8}(2x^4-4x^3+6x^2-6x+3)e^{\left(2x\right)}+\frac{1}{16}(4x^3{e}^5-6x^2{e}^5+6x{e}^5-3e^5)e^{\left(2x\right)}-\frac{1}{16}(4x^3-6x^2+6x-3)e^{\left(2x\right)}+\frac{3}{16}(2x^2{e}^5-2x{e}^5+e^5)e^{\left(2x\right)}-\frac{19}{16}(2x^2-2x+1)e^{\left(2x\right)}-\frac{5}{8}(2x{e}^5-e^5)e^{\left(2x\right)}-2(2x-1)e^{\left(2x\right)}+\frac{155}{8}{e}^{\left(2x\right)}-\frac{5}{8}{e}^{(2x+5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x^3+3*x^2-10*x-5)*exp(5)-2*x^4-2*x^3-19*x^2-32*x+155)*exp(x)^2,x, algorithm="maxima")

[Out]

-1/8*(2*x^4 - 4*x^3 + 6*x^2 - 6*x + 3)*e^(2*x) + 1/16*(4*x^3*e^5 - 6*x^2*e^5 + 6*x*e^5 - 3*e^5)*e^(2*x) - 1/16
*(4*x^3 - 6*x^2 + 6*x - 3)*e^(2*x) + 3/16*(2*x^2*e^5 - 2*x*e^5 + e^5)*e^(2*x) - 19/16*(2*x^2 - 2*x + 1)*e^(2*x
) - 5/8*(2*x*e^5 - e^5)*e^(2*x) - 2*(2*x - 1)*e^(2*x) + 155/8*e^(2*x) - 5/8*e^(2*x + 5)

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mupad [B]  time = 4.26, size = 23, normalized size = 0.74 $$\begin{array}{c}\frac{{\mathrm{e}}^{2x}(x^2-5)(x+x{\mathrm{e}}^5-x^2-16)}{4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x)*(32*x + exp(5)*(10*x - 3*x^2 - 2*x^3 + 5) + 19*x^2 + 2*x^3 + 2*x^4 - 155))/4,x)

[Out]

(exp(2*x)*(x^2 - 5)*(x + x*exp(5) - x^2 - 16))/4

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sympy [A]  time = 0.18, size = 36, normalized size = 1.16 $$\begin{array}{c}\frac{(-x^4+x^3+x^3{e}^5-11x^2-5x{e}^5-5x+80)e^{2x}}{4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2*x**3+3*x**2-10*x-5)*exp(5)-2*x**4-2*x**3-19*x**2-32*x+155)*exp(x)**2,x)

[Out]

(-x**4 + x**3 + x**3*exp(5) - 11*x**2 - 5*x*exp(5) - 5*x + 80)*exp(2*x)/4

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