3.101.87 \(\int \frac {-15 x+e^3 x (-12+138 x+54 x^2)}{5 x} \, dx\)

Optimal. Leaf size=22 \[ -3 x+3 e^3 x (4+x) \left (\frac {1}{5} (-1+x)+x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 14} \begin {gather*} \frac {18 e^3 x^3}{5}+\frac {69 e^3 x^2}{5}-\frac {3}{5} \left (5+4 e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*x + E^3*x*(-12 + 138*x + 54*x^2))/(5*x),x]

[Out]

(-3*(5 + 4*E^3)*x)/5 + (69*E^3*x^2)/5 + (18*E^3*x^3)/5

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}{5} \int \frac {-15 x+e^3 x \left (-12+138 x+54 x^2\right )}{x} \, dx\\ &=\frac {1}{5} \int \left (-3 \left (5+4 e^3\right )+138 e^3 x+54 e^3 x^2\right ) \, dx\\ &=-\frac {3}{5} \left (5+4 e^3\right ) x+\frac {69 e^3 x^2}{5}+\frac {18 e^3 x^3}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.45 \begin {gather*} -3 x-\frac {12 e^3 x}{5}+\frac {69 e^3 x^2}{5}+\frac {18 e^3 x^3}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*x + E^3*x*(-12 + 138*x + 54*x^2))/(5*x),x]

[Out]

-3*x - (12*E^3*x)/5 + (69*E^3*x^2)/5 + (18*E^3*x^3)/5

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fricas [A]  time = 0.66, size = 22, normalized size = 1.00 \begin {gather*} \frac {3}{5} \, {\left (6 \, x^{3} + 23 \, x^{2} - 4 \, x\right )} e^{3} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((54*x^2+138*x-12)*exp(3+log(x))-15*x)/x,x, algorithm="fricas")

[Out]

3/5*(6*x^3 + 23*x^2 - 4*x)*e^3 - 3*x

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giac [A]  time = 0.13, size = 23, normalized size = 1.05 \begin {gather*} \frac {18}{5} \, x^{3} e^{3} + \frac {69}{5} \, x^{2} e^{3} - \frac {12}{5} \, x e^{3} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((54*x^2+138*x-12)*exp(3+log(x))-15*x)/x,x, algorithm="giac")

[Out]

18/5*x^3*e^3 + 69/5*x^2*e^3 - 12/5*x*e^3 - 3*x

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maple [A]  time = 0.02, size = 23, normalized size = 1.05

method	result	size
default	\(-3 x +\frac {6 \,{\mathrm e}^{3} \left (3 x^{3}+\frac {23}{2} x^{2}-2 x \right )}{5}\)	\(23\)
norman	\(\left (-\frac {12 \,{\mathrm e}^{3}}{5}-3\right ) x +\frac {69 x^{2} {\mathrm e}^{3}}{5}+\frac {18 x^{3} {\mathrm e}^{3}}{5}\)	\(24\)
risch	\(\frac {18 x^{3} {\mathrm e}^{3}}{5}+\frac {69 x^{2} {\mathrm e}^{3}}{5}-\frac {12 x \,{\mathrm e}^{3}}{5}-3 x\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((54*x^2+138*x-12)*exp(3+ln(x))-15*x)/x,x,method=_RETURNVERBOSE)

[Out]

-3*x+6/5*exp(3)*(3*x^3+23/2*x^2-2*x)

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maxima [A]  time = 0.35, size = 24, normalized size = 1.09 \begin {gather*} \frac {18}{5} \, x^{3} e^{3} + \frac {69}{5} \, x^{2} e^{3} - \frac {3}{5} \, x {\left (4 \, e^{3} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((54*x^2+138*x-12)*exp(3+log(x))-15*x)/x,x, algorithm="maxima")

[Out]

18/5*x^3*e^3 + 69/5*x^2*e^3 - 3/5*x*(4*e^3 + 5)

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mupad [B]  time = 0.10, size = 24, normalized size = 1.09 \begin {gather*} \frac {18\,{\mathrm {e}}^3\,x^3}{5}+\frac {69\,{\mathrm {e}}^3\,x^2}{5}+\left (-\frac {12\,{\mathrm {e}}^3}{5}-3\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x - (exp(log(x) + 3)*(138*x + 54*x^2 - 12))/5)/x,x)

[Out]

(69*x^2*exp(3))/5 + (18*x^3*exp(3))/5 - x*((12*exp(3))/5 + 3)

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sympy [A]  time = 0.07, size = 31, normalized size = 1.41 \begin {gather*} \frac {18 x^{3} e^{3}}{5} + \frac {69 x^{2} e^{3}}{5} + x \left (- \frac {12 e^{3}}{5} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((54*x**2+138*x-12)*exp(3+ln(x))-15*x)/x,x)

[Out]

18*x**3*exp(3)/5 + 69*x**2*exp(3)/5 + x*(-12*exp(3)/5 - 3)

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