3.86.38 \(\int \frac {e^x (-15-23 x+14 x^2+2 x^4+2 x^5)}{x^4} \, dx\)

Optimal. Leaf size=17 \[ e^x \left (2-\frac {-5-14 x}{x^4}\right ) x \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 23, normalized size of antiderivative = 1.35, number of steps used = 14, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {5 e^x}{x^3}+\frac {14 e^x}{x^2}+2 e^x x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-15 - 23*x + 14*x^2 + 2*x^4 + 2*x^5))/x^4,x]

[Out]

(5*E^x)/x^3 + (14*E^x)/x^2 + 2*E^x*x

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 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]

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} &=\int \left (2 e^x-\frac {15 e^x}{x^4}-\frac {23 e^x}{x^3}+\frac {14 e^x}{x^2}+2 e^x x\right ) \, dx\\ &=2 \int e^x \, dx+2 \int e^x x \, dx+14 \int \frac {e^x}{x^2} \, dx-15 \int \frac {e^x}{x^4} \, dx-23 \int \frac {e^x}{x^3} \, dx\\ &=2 e^x+\frac {5 e^x}{x^3}+\frac {23 e^x}{2 x^2}-\frac {14 e^x}{x}+2 e^x x-2 \int e^x \, dx-5 \int \frac {e^x}{x^3} \, dx-\frac {23}{2} \int \frac {e^x}{x^2} \, dx+14 \int \frac {e^x}{x} \, dx\\ &=\frac {5 e^x}{x^3}+\frac {14 e^x}{x^2}-\frac {5 e^x}{2 x}+2 e^x x+14 \text {Ei}(x)-\frac {5}{2} \int \frac {e^x}{x^2} \, dx-\frac {23}{2} \int \frac {e^x}{x} \, dx\\ &=\frac {5 e^x}{x^3}+\frac {14 e^x}{x^2}+2 e^x x+\frac {5 \text {Ei}(x)}{2}-\frac {5}{2} \int \frac {e^x}{x} \, dx\\ &=\frac {5 e^x}{x^3}+\frac {14 e^x}{x^2}+2 e^x x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 18, normalized size = 1.06 \begin {gather*} e^x \left (\frac {5}{x^3}+\frac {14}{x^2}+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-15 - 23*x + 14*x^2 + 2*x^4 + 2*x^5))/x^4,x]

[Out]

E^x*(5/x^3 + 14/x^2 + 2*x)

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 16, normalized size = 0.94 \begin {gather*} \frac {{\left (2 \, x^{4} + 14 \, x + 5\right )} e^{x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^5+2*x^4+14*x^2-23*x-15)*exp(x)/x^4,x, algorithm="fricas")

[Out]

(2*x^4 + 14*x + 5)*e^x/x^3

________________________________________________________________________________________

giac [A]  time = 0.24, size = 21, normalized size = 1.24 \begin {gather*} \frac {2 \, x^{4} e^{x} + 14 \, x e^{x} + 5 \, e^{x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^5+2*x^4+14*x^2-23*x-15)*exp(x)/x^4,x, algorithm="giac")

[Out]

(2*x^4*e^x + 14*x*e^x + 5*e^x)/x^3

________________________________________________________________________________________

maple [A]  time = 0.10, size = 17, normalized size = 1.00

method	result	size
gosper	\(\frac {\left (2 x^{4}+14 x +5\right ) {\mathrm e}^{x}}{x^{3}}\)	\(17\)
risch	\(\frac {\left (2 x^{4}+14 x +5\right ) {\mathrm e}^{x}}{x^{3}}\)	\(17\)
default	\(\frac {5 \,{\mathrm e}^{x}}{x^{3}}+\frac {14 \,{\mathrm e}^{x}}{x^{2}}+2 \,{\mathrm e}^{x} x\)	\(21\)
norman	\(\frac {14 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x} x^{4}+5 \,{\mathrm e}^{x}}{x^{3}}\)	\(22\)
meijerg	\(-\left (-2 x +2\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x}+\frac {33}{2 x}+\frac {47}{6}+\frac {14 x +14}{x}-\frac {14 \,{\mathrm e}^{x}}{x}+\frac {19}{x^{2}}-\frac {23 \left (9 x^{2}+12 x +6\right )}{12 x^{2}}+\frac {23 \left (3 x +3\right ) {\mathrm e}^{x}}{6 x^{2}}+\frac {5}{x^{3}}-\frac {5 \left (22 x^{3}+36 x^{2}+36 x +24\right )}{24 x^{3}}+\frac {5 \left (4 x^{2}+4 x +8\right ) {\mathrm e}^{x}}{8 x^{3}}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^5+2*x^4+14*x^2-23*x-15)*exp(x)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/x^3*(2*x^4+14*x+5)*exp(x)

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 33, normalized size = 1.94 \begin {gather*} 2 \, {\left (x - 1\right )} e^{x} + 2 \, e^{x} + 14 \, \Gamma \left (-1, -x\right ) + 23 \, \Gamma \left (-2, -x\right ) - 15 \, \Gamma \left (-3, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^5+2*x^4+14*x^2-23*x-15)*exp(x)/x^4,x, algorithm="maxima")

[Out]

2*(x - 1)*e^x + 2*e^x + 14*gamma(-1, -x) + 23*gamma(-2, -x) - 15*gamma(-3, -x)

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 16, normalized size = 0.94 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (2\,x^4+14\,x+5\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(14*x^2 - 23*x + 2*x^4 + 2*x^5 - 15))/x^4,x)

[Out]

(exp(x)*(14*x + 2*x^4 + 5))/x^3

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 15, normalized size = 0.88 \begin {gather*} \frac {\left (2 x^{4} + 14 x + 5\right ) e^{x}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**5+2*x**4+14*x**2-23*x-15)*exp(x)/x**4,x)

[Out]

(2*x**4 + 14*x + 5)*exp(x)/x**3

________________________________________________________________________________________