3.82.21 \(\int \frac {e^4 (-45+10 x^3)}{18 x^2} \, dx\)

Optimal. Leaf size=18 \[ \frac {5}{2} e^4 \left (\frac {1}{x}+\frac {x^2}{9}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 14} \begin {gather*} \frac {5 e^4 x^2}{18}+\frac {5 e^4}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(-45 + 10*x^3))/(18*x^2),x]

[Out]

(5*E^4)/(2*x) + (5*E^4*x^2)/18

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{18} e^4 \int \frac {-45+10 x^3}{x^2} \, dx\\ &=\frac {1}{18} e^4 \int \left (-\frac {45}{x^2}+10 x\right ) \, dx\\ &=\frac {5 e^4}{2 x}+\frac {5 e^4 x^2}{18}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.89 \begin {gather*} \frac {5}{18} e^4 \left (\frac {9}{x}+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(-45 + 10*x^3))/(18*x^2),x]

[Out]

(5*E^4*(9/x + x^2))/18

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fricas [A]  time = 0.79, size = 12, normalized size = 0.67 \begin {gather*} \frac {5 \, {\left (x^{3} + 9\right )} e^{4}}{18 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(10*x^3-45)*exp(4)/x^2,x, algorithm="fricas")

[Out]

5/18*(x^3 + 9)*e^4/x

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giac [A]  time = 0.13, size = 13, normalized size = 0.72 \begin {gather*} \frac {5}{18} \, {\left (x^{2} + \frac {9}{x}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(10*x^3-45)*exp(4)/x^2,x, algorithm="giac")

[Out]

5/18*(x^2 + 9/x)*e^4

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maple [A]  time = 0.02, size = 13, normalized size = 0.72

method	result	size
gosper	\(\frac {5 \,{\mathrm e}^{4} \left (x^{3}+9\right )}{18 x}\)	\(13\)
default	\(\frac {5 \,{\mathrm e}^{4} \left (x^{2}+\frac {9}{x}\right )}{18}\)	\(14\)
risch	\(\frac {5 x^{2} {\mathrm e}^{4}}{18}+\frac {5 \,{\mathrm e}^{4}}{2 x}\)	\(16\)
norman	\(\frac {\frac {5 x^{3} {\mathrm e}^{4}}{18}+\frac {5 \,{\mathrm e}^{4}}{2}}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*(10*x^3-45)*exp(4)/x^2,x,method=_RETURNVERBOSE)

[Out]

5/18*exp(4)*(x^3+9)/x

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maxima [A]  time = 0.38, size = 13, normalized size = 0.72 \begin {gather*} \frac {5}{18} \, {\left (x^{2} + \frac {9}{x}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(10*x^3-45)*exp(4)/x^2,x, algorithm="maxima")

[Out]

5/18*(x^2 + 9/x)*e^4

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mupad [B]  time = 5.23, size = 12, normalized size = 0.67 \begin {gather*} \frac {5\,{\mathrm {e}}^4\,\left (x^3+9\right )}{18\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*(10*x^3 - 45))/(18*x^2),x)

[Out]

(5*exp(4)*(x^3 + 9))/(18*x)

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sympy [A]  time = 0.08, size = 17, normalized size = 0.94 \begin {gather*} \frac {5 x^{2} e^{4}}{18} + \frac {5 e^{4}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(10*x**3-45)*exp(4)/x**2,x)

[Out]

5*x**2*exp(4)/18 + 5*exp(4)/(2*x)

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