3.1.93 \(\int \frac {-5+2 x^2-5 e^{1+5 x} x^2+2 x^3}{x^2} \, dx\)

Optimal. Leaf size=28 \[ 2-e^{1+5 x}+\frac {5-x}{x}-2 x+(2+x)^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 0.75, number of steps used = 5, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14, 2194} \begin {gather*} x^2+2 x-e^{5 x+1}+\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 2*x^2 - 5*E^(1 + 5*x)*x^2 + 2*x^3)/x^2,x]

[Out]

-E^(1 + 5*x) + 5/x + 2*x + x^2

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5 e^{1+5 x}+\frac {-5+2 x^2+2 x^3}{x^2}\right ) \, dx\\ &=-\left (5 \int e^{1+5 x} \, dx\right )+\int \frac {-5+2 x^2+2 x^3}{x^2} \, dx\\ &=-e^{1+5 x}+\int \left (2-\frac {5}{x^2}+2 x\right ) \, dx\\ &=-e^{1+5 x}+\frac {5}{x}+2 x+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.75 \begin {gather*} -e^{1+5 x}+\frac {5}{x}+2 x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 2*x^2 - 5*E^(1 + 5*x)*x^2 + 2*x^3)/x^2,x]

[Out]

-E^(1 + 5*x) + 5/x + 2*x + x^2

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fricas [A]  time = 0.66, size = 23, normalized size = 0.82 \begin {gather*} \frac {x^{3} + 2 \, x^{2} - x e^{\left (5 \, x + 1\right )} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3 + 2*x^2 - x*e^(5*x + 1) + 5)/x

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giac [A]  time = 0.30, size = 23, normalized size = 0.82 \begin {gather*} \frac {x^{3} + 2 \, x^{2} - x e^{\left (5 \, x + 1\right )} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3 + 2*x^2 - x*e^(5*x + 1) + 5)/x

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maple [A]  time = 0.04, size = 21, normalized size = 0.75

method	result	size
default	\(x^{2}+2 x +\frac {5}{x}-{\mathrm e}^{5 x} {\mathrm e}\)	\(21\)
risch	\(x^{2}+2 x +\frac {5}{x}-{\mathrm e}^{1+5 x}\)	\(21\)
norman	\(\frac {5+x^{3}+2 x^{2}-x \,{\mathrm e} \,{\mathrm e}^{5 x}}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 20, normalized size = 0.71 \begin {gather*} x^{2} + 2 \, x + \frac {5}{x} - e^{\left (5 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 2*x + 5/x - e^(5*x + 1)

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mupad [B]  time = 0.25, size = 20, normalized size = 0.71 \begin {gather*} 2\,x-{\mathrm {e}}^{5\,x+1}+\frac {5}{x}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - exp(5*x + 1) + 5/x + x^2

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sympy [A]  time = 0.14, size = 17, normalized size = 0.61 \begin {gather*} x^{2} + 2 x - e e^{5 x} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x**2*exp(x)*exp(4*x+1)+2*x**3+2*x**2-5)/x**2,x)

[Out]

x**2 + 2*x - E*exp(5*x) + 5/x

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