\(\int \frac {48-84 x+32 x^2-60 x^3+e^x (9 x-11 x^2+4 x^3+15 x^5)}{x^5} \, dx\) [8841]

Optimal result

Integrand size = 42, antiderivative size = 28 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=5 \left (3-\frac {-4+\frac {3}{x}}{5 x^2}\right ) \left (-4+e^x+\frac {4}{x}\right ) \]

[Out]

(3-1/5*(3/x-4)/x^2)*(20/x+5*exp(x)-20)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 16, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {14, 2230, 2225, 2208, 2209} \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=-\frac {12}{x^4}-\frac {3 e^x}{x^3}+\frac {28}{x^3}+\frac {4 e^x}{x^2}-\frac {16}{x^2}+15 e^x+\frac {60}{x} \]

[In]

Int[(48 - 84*x + 32*x^2 - 60*x^3 + E^x*(9*x - 11*x^2 + 4*x^3 + 15*x^5))/x^5,x]

[Out]

15*E^x - 12/x^4 + 28/x^3 - (3*E^x)/x^3 - 16/x^2 + (4*E^x)/x^2 + 60/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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2225

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 2230

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4 \left (-12+21 x-8 x^2+15 x^3\right )}{x^5}+\frac {e^x \left (9-11 x+4 x^2+15 x^4\right )}{x^4}\right ) \, dx \\ & = -\left (4 \int \frac {-12+21 x-8 x^2+15 x^3}{x^5} \, dx\right )+\int \frac {e^x \left (9-11 x+4 x^2+15 x^4\right )}{x^4} \, dx \\ & = -\left (4 \int \left (-\frac {12}{x^5}+\frac {21}{x^4}-\frac {8}{x^3}+\frac {15}{x^2}\right ) \, dx\right )+\int \left (15 e^x+\frac {9 e^x}{x^4}-\frac {11 e^x}{x^3}+\frac {4 e^x}{x^2}\right ) \, dx \\ & = -\frac {12}{x^4}+\frac {28}{x^3}-\frac {16}{x^2}+\frac {60}{x}+4 \int \frac {e^x}{x^2} \, dx+9 \int \frac {e^x}{x^4} \, dx-11 \int \frac {e^x}{x^3} \, dx+15 \int e^x \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {11 e^x}{2 x^2}+\frac {60}{x}-\frac {4 e^x}{x}+3 \int \frac {e^x}{x^3} \, dx+4 \int \frac {e^x}{x} \, dx-\frac {11}{2} \int \frac {e^x}{x^2} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x}+\frac {3 e^x}{2 x}+4 \operatorname {ExpIntegralEi}(x)+\frac {3}{2} \int \frac {e^x}{x^2} \, dx-\frac {11}{2} \int \frac {e^x}{x} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x}-\frac {3 \operatorname {ExpIntegralEi}(x)}{2}+\frac {3}{2} \int \frac {e^x}{x} \, dx \\ & = 15 e^x-\frac {12}{x^4}+\frac {28}{x^3}-\frac {3 e^x}{x^3}-\frac {16}{x^2}+\frac {4 e^x}{x^2}+\frac {60}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=e^x \left (15-\frac {3}{x^3}+\frac {4}{x^2}\right )-\frac {12}{x^4}+\frac {28}{x^3}-\frac {16}{x^2}+\frac {60}{x} \]

[In]

Integrate[(48 - 84*x + 32*x^2 - 60*x^3 + E^x*(9*x - 11*x^2 + 4*x^3 + 15*x^5))/x^5,x]

[Out]

E^x*(15 - 3/x^3 + 4/x^2) - 12/x^4 + 28/x^3 - 16/x^2 + 60/x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {60 x^{3}-16 x^{2}+28 x -12}{x^{4}}+\frac {\left (15 x^{3}+4 x -3\right ) {\mathrm e}^{x}}{x^{3}}\)	\(37\)
norman	\(\frac {-12+28 x -16 x^{2}+60 x^{3}-3 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{x} x^{4}}{x^{4}}\)	\(39\)
parallelrisch	\(\frac {-12+28 x -16 x^{2}+60 x^{3}-3 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{x} x^{4}}{x^{4}}\)	\(39\)
default	\(-\frac {12}{x^{4}}+\frac {28}{x^{3}}-\frac {16}{x^{2}}+\frac {60}{x}-\frac {3 \,{\mathrm e}^{x}}{x^{3}}+\frac {4 \,{\mathrm e}^{x}}{x^{2}}+15 \,{\mathrm e}^{x}\)	\(40\)
parts	\(-\frac {12}{x^{4}}+\frac {28}{x^{3}}-\frac {16}{x^{2}}+\frac {60}{x}-\frac {3 \,{\mathrm e}^{x}}{x^{3}}+\frac {4 \,{\mathrm e}^{x}}{x^{2}}+15 \,{\mathrm e}^{x}\)	\(40\)

[In]

int(((15*x^5+4*x^3-11*x^2+9*x)*exp(x)-60*x^3+32*x^2-84*x+48)/x^5,x,method=_RETURNVERBOSE)

[Out]

(60*x^3-16*x^2+28*x-12)/x^4+(15*x^3+4*x-3)/x^3*exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {60 \, x^{3} - 16 \, x^{2} + {\left (15 \, x^{4} + 4 \, x^{2} - 3 \, x\right )} e^{x} + 28 \, x - 12}{x^{4}} \]

[In]

integrate(((15*x^5+4*x^3-11*x^2+9*x)*exp(x)-60*x^3+32*x^2-84*x+48)/x^5,x, algorithm="fricas")

[Out]

(60*x^3 - 16*x^2 + (15*x^4 + 4*x^2 - 3*x)*e^x + 28*x - 12)/x^4

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {\left (15 x^{3} + 4 x - 3\right ) e^{x}}{x^{3}} - \frac {- 60 x^{3} + 16 x^{2} - 28 x + 12}{x^{4}} \]

[In]

integrate(((15*x**5+4*x**3-11*x**2+9*x)*exp(x)-60*x**3+32*x**2-84*x+48)/x**5,x)

[Out]

(15*x**3 + 4*x - 3)*exp(x)/x**3 - (-60*x**3 + 16*x**2 - 28*x + 12)/x**4

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {60}{x} - \frac {16}{x^{2}} + \frac {28}{x^{3}} - \frac {12}{x^{4}} + 15 \, e^{x} + 4 \, \Gamma \left (-1, -x\right ) + 11 \, \Gamma \left (-2, -x\right ) + 9 \, \Gamma \left (-3, -x\right ) \]

[In]

integrate(((15*x^5+4*x^3-11*x^2+9*x)*exp(x)-60*x^3+32*x^2-84*x+48)/x^5,x, algorithm="maxima")

[Out]

60/x - 16/x^2 + 28/x^3 - 12/x^4 + 15*e^x + 4*gamma(-1, -x) + 11*gamma(-2, -x) + 9*gamma(-3, -x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=\frac {15 \, x^{4} e^{x} + 60 \, x^{3} + 4 \, x^{2} e^{x} - 16 \, x^{2} - 3 \, x e^{x} + 28 \, x - 12}{x^{4}} \]

[In]

integrate(((15*x^5+4*x^3-11*x^2+9*x)*exp(x)-60*x^3+32*x^2-84*x+48)/x^5,x, algorithm="giac")

[Out]

(15*x^4*e^x + 60*x^3 + 4*x^2*e^x - 16*x^2 - 3*x*e^x + 28*x - 12)/x^4

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {48-84 x+32 x^2-60 x^3+e^x \left (9 x-11 x^2+4 x^3+15 x^5\right )}{x^5} \, dx=15\,{\mathrm {e}}^x-\frac {x\,\left (3\,{\mathrm {e}}^x-28\right )-x^2\,\left (4\,{\mathrm {e}}^x-16\right )-60\,x^3+12}{x^4} \]

[In]

int((exp(x)*(9*x - 11*x^2 + 4*x^3 + 15*x^5) - 84*x + 32*x^2 - 60*x^3 + 48)/x^5,x)

[Out]

15*exp(x) - (x*(3*exp(x) - 28) - x^2*(4*exp(x) - 16) - 60*x^3 + 12)/x^4