3.74.82 \(\int \frac {-80 x^3-40 x^4+e^x (-72+33 x+3 x^2)}{20 x^3} \, dx\)

Optimal. Leaf size=21 \[ 1-x (4+x)+\frac {3 e^x (12+x)}{20 x^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 11, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 14, 2199, 2177, 2178} \begin {gather*} \frac {9 e^x}{5 x^2}-(x+2)^2+\frac {3 e^x}{20 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-80*x^3 - 40*x^4 + E^x*(-72 + 33*x + 3*x^2))/(20*x^3),x]

[Out]

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

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 2177

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2199

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{20} \int \frac {-80 x^3-40 x^4+e^x \left (-72+33 x+3 x^2\right )}{x^3} \, dx\\ &=\frac {1}{20} \int \left (-40 (2+x)+\frac {3 e^x \left (-24+11 x+x^2\right )}{x^3}\right ) \, dx\\ &=-(2+x)^2+\frac {3}{20} \int \frac {e^x \left (-24+11 x+x^2\right )}{x^3} \, dx\\ &=-(2+x)^2+\frac {3}{20} \int \left (-\frac {24 e^x}{x^3}+\frac {11 e^x}{x^2}+\frac {e^x}{x}\right ) \, dx\\ &=-(2+x)^2+\frac {3}{20} \int \frac {e^x}{x} \, dx+\frac {33}{20} \int \frac {e^x}{x^2} \, dx-\frac {18}{5} \int \frac {e^x}{x^3} \, dx\\ &=\frac {9 e^x}{5 x^2}-\frac {33 e^x}{20 x}-(2+x)^2+\frac {3 \text {Ei}(x)}{20}+\frac {33}{20} \int \frac {e^x}{x} \, dx-\frac {9}{5} \int \frac {e^x}{x^2} \, dx\\ &=\frac {9 e^x}{5 x^2}+\frac {3 e^x}{20 x}-(2+x)^2+\frac {9 \text {Ei}(x)}{5}-\frac {9}{5} \int \frac {e^x}{x} \, dx\\ &=\frac {9 e^x}{5 x^2}+\frac {3 e^x}{20 x}-(2+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 29, normalized size = 1.38 \begin {gather*} \frac {9 e^x}{5 x^2}+\frac {3 e^x}{20 x}-4 x-x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*x^3 - 40*x^4 + E^x*(-72 + 33*x + 3*x^2))/(20*x^3),x]

[Out]

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

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fricas [A]  time = 1.05, size = 23, normalized size = 1.10 \begin {gather*} -\frac {20 \, x^{4} + 80 \, x^{3} - 3 \, {\left (x + 12\right )} e^{x}}{20 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((3*x^2+33*x-72)*exp(x)-40*x^4-80*x^3)/x^3,x, algorithm="fricas")

[Out]

-1/20*(20*x^4 + 80*x^3 - 3*(x + 12)*e^x)/x^2

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giac [A]  time = 0.21, size = 25, normalized size = 1.19 \begin {gather*} -\frac {20 \, x^{4} + 80 \, x^{3} - 3 \, x e^{x} - 36 \, e^{x}}{20 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((3*x^2+33*x-72)*exp(x)-40*x^4-80*x^3)/x^3,x, algorithm="giac")

[Out]

-1/20*(20*x^4 + 80*x^3 - 3*x*e^x - 36*e^x)/x^2

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maple [A]  time = 0.06, size = 20, normalized size = 0.95

method	result	size
risch	\(-x^{2}-4 x +\frac {3 \left (x +12\right ) {\mathrm e}^{x}}{20 x^{2}}\)	\(20\)
default	\(-x^{2}-4 x +\frac {9 \,{\mathrm e}^{x}}{5 x^{2}}+\frac {3 \,{\mathrm e}^{x}}{20 x}\)	\(24\)
norman	\(\frac {-4 x^{3}-x^{4}+\frac {3 \,{\mathrm e}^{x} x}{20}+\frac {9 \,{\mathrm e}^{x}}{5}}{x^{2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*((3*x^2+33*x-72)*exp(x)-40*x^4-80*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

-x^2-4*x+3/20*(x+12)/x^2*exp(x)

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maxima [C]  time = 0.38, size = 27, normalized size = 1.29 \begin {gather*} -x^{2} - 4 \, x + \frac {3}{20} \, {\rm Ei}\relax (x) + \frac {33}{20} \, \Gamma \left (-1, -x\right ) + \frac {18}{5} \, \Gamma \left (-2, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((3*x^2+33*x-72)*exp(x)-40*x^4-80*x^3)/x^3,x, algorithm="maxima")

[Out]

-x^2 - 4*x + 3/20*Ei(x) + 33/20*gamma(-1, -x) + 18/5*gamma(-2, -x)

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mupad [B]  time = 4.56, size = 23, normalized size = 1.10 \begin {gather*} \frac {\frac {9\,{\mathrm {e}}^x}{5}+\frac {3\,x\,{\mathrm {e}}^x}{20}}{x^2}-4\,x-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3 - (exp(x)*(33*x + 3*x^2 - 72))/20 + 2*x^4)/x^3,x)

[Out]

((9*exp(x))/5 + (3*x*exp(x))/20)/x^2 - 4*x - x^2

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sympy [A]  time = 0.10, size = 19, normalized size = 0.90 \begin {gather*} - x^{2} - 4 x + \frac {\left (3 x + 36\right ) e^{x}}{20 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((3*x**2+33*x-72)*exp(x)-40*x**4-80*x**3)/x**3,x)

[Out]

-x**2 - 4*x + (3*x + 36)*exp(x)/(20*x**2)

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