3.33.8 $\int \frac{4x^2+e^{\frac{-4e^4-8x+5x^2}{4x}}(-12e^4-15x^2)}{12x^2}dx$

Optimal. Leaf size=25 $$-e^{-2-\frac{e^4}{x}+\frac{5x}{4}}+\frac{x}{3}$$

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Rubi [A]  time = 0.20, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.062, Rules used = {12, 14, 6706} $$\begin{array}{c}\frac{x}{3}-e^{\frac{5x}{4}-\frac{e^4}{x}-2}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + E^((-4*E^4 - 8*x + 5*x^2)/(4*x))*(-12*E^4 - 15*x^2))/(12*x^2),x]

[Out]

-E^(-2 - E^4/x + (5*x)/4) + x/3

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{12}\int \frac{4x^2+e^{\frac{-4e^4-8x+5x^2}{4x}}(-12e^4-15x^2)}{x^2}dx\\ & =\frac{1}{12}\int (4-\frac{3e^{-2-\frac{e^4}{x}+\frac{5x}{4}}(4e^4+5x^2)}{x^2})dx\\ & =\frac{x}{3}-\frac{1}{4}\int \frac{e^{-2-\frac{e^4}{x}+\frac{5x}{4}}(4e^4+5x^2)}{x^2}dx\\ & =-e^{-2-\frac{e^4}{x}+\frac{5x}{4}}+\frac{x}{3}\end{array}\end{array}$$

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Mathematica [A]  time = 0.20, size = 25, normalized size = 1.00 $$\begin{array}{c}-e^{-2-\frac{e^4}{x}+\frac{5x}{4}}+\frac{x}{3}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + E^((-4*E^4 - 8*x + 5*x^2)/(4*x))*(-12*E^4 - 15*x^2))/(12*x^2),x]

[Out]

-E^(-2 - E^4/x + (5*x)/4) + x/3

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fricas [A]  time = 0.52, size = 25, normalized size = 1.00 $$\begin{array}{c}\frac{1}{3}x-e^{\left(\frac{5x^2-8x-4e^4}{4x}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="fricas")

[Out]

1/3*x - e^(1/4*(5*x^2 - 8*x - 4*e^4)/x)

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giac [A]  time = 0.44, size = 30, normalized size = 1.20 $$\begin{array}{c}\frac{1}{3}(x{e}^4-3e^{\left(\frac{5x^2+8x-4e^4}{4x}\right)})e^{(-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="giac")

[Out]

1/3*(x*e^4 - 3*e^(1/4*(5*x^2 + 8*x - 4*e^4)/x))*e^(-4)

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maple [A]  time = 0.06, size = 26, normalized size = 1.04

method	result	size
risch	$\frac{x}{3}-{\mathrm{e}}^{-\frac{-5x^2+4{\mathrm{e}}^4+8x}{4x}}$	$26$
norman	$\frac{\frac{x^2}{3}-x{\mathrm{e}}^{\frac{-4{\mathrm{e}}^4+5x^2-8x}{4x}}}{x}$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/3*x-exp(-1/4*(-5*x^2+4*exp(4)+8*x)/x)

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maxima [A]  time = 1.26, size = 19, normalized size = 0.76 $$\begin{array}{c}\frac{1}{3}x-e^{(\frac{5}{4}x-\frac{e^4}{x}-2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((-12*exp(4)-15*x^2)*exp(1/4*(-4*exp(4)+5*x^2-8*x)/x)+4*x^2)/x^2,x, algorithm="maxima")

[Out]

1/3*x - e^(5/4*x - e^4/x - 2)

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mupad [B]  time = 1.95, size = 19, normalized size = 0.76 $$\begin{array}{c}\frac{x}{3}-{\mathrm{e}}^{\frac{5x}{4}-\frac{{\mathrm{e}}^4}{x}-2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(-(2*x + exp(4) - (5*x^2)/4)/x)*(12*exp(4) + 15*x^2))/12 - x^2/3)/x^2,x)

[Out]

x/3 - exp((5*x)/4 - exp(4)/x - 2)

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sympy [A]  time = 0.20, size = 19, normalized size = 0.76 $$\begin{array}{c}\frac{x}{3}-e^{\frac{\frac{5x^2}{4}-2x-e^4}{x}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*((-12*exp(4)-15*x**2)*exp(1/4*(-4*exp(4)+5*x**2-8*x)/x)+4*x**2)/x**2,x)

[Out]

x/3 - exp((5*x**2/4 - 2*x - exp(4))/x)

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