3.89.41 $\int \frac{48-84x+32x^2-60x^3+e^x(9x-11x^2+4x^3+15x^5)}{x^5}dx$

Optimal. Leaf size=28 $$5(3-\frac{-4+\frac{3}{x}}{5x^2})(-4+e^x+\frac{4}{x})$$

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Rubi [A]  time = 0.17, antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 16, number of rules used = 5, integrand size = 42, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.119, Rules used = {14, 2199, 2194, 2177, 2178} $$\begin{array}{c}-\frac{12}{x^4}-\frac{3e^x}{x^3}+\frac{28}{x^3}+\frac{4e^x}{x^2}-\frac{16}{x^2}+15e^x+\frac{60}{x}\end{array}$$

Antiderivative was successfully verified.

[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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{4(-12+21x-8x^2+15x^3)}{x^5}+\frac{e^x(9-11x+4x^2+15x^4)}{x^4})dx\\ & =-(4\int \frac{-12+21x-8x^2+15x^3}{x^5}dx)+\int \frac{e^x(9-11x+4x^2+15x^4)}{x^4}dx\\ & =-(4\int (-\frac{12}{x^5}+\frac{21}{x^4}-\frac{8}{x^3}+\frac{15}{x^2})dx)+\int (15e^x+\frac{9e^x}{x^4}-\frac{11e^x}{x^3}+\frac{4e^x}{x^2})dx\\ & =-\frac{12}{x^4}+\frac{28}{x^3}-\frac{16}{x^2}+\frac{60}{x}+4\int \frac{e^x}{x^2}dx+9\int \frac{e^x}{x^4}dx-11\int \frac{e^x}{x^3}dx+15\int e^{x}dx\\ & =15e^x-\frac{12}{x^4}+\frac{28}{x^3}-\frac{3e^x}{x^3}-\frac{16}{x^2}+\frac{11e^x}{2x^2}+\frac{60}{x}-\frac{4e^x}{x}+3\int \frac{e^x}{x^3}dx+4\int \frac{e^x}{x}dx-\frac{11}{2}\int \frac{e^x}{x^2}dx\\ & =15e^x-\frac{12}{x^4}+\frac{28}{x^3}-\frac{3e^x}{x^3}-\frac{16}{x^2}+\frac{4e^x}{x^2}+\frac{60}{x}+\frac{3e^x}{2x}+4\text{Ei}(x)+\frac{3}{2}\int \frac{e^x}{x^2}dx-\frac{11}{2}\int \frac{e^x}{x}dx\\ & =15e^x-\frac{12}{x^4}+\frac{28}{x^3}-\frac{3e^x}{x^3}-\frac{16}{x^2}+\frac{4e^x}{x^2}+\frac{60}{x}-\frac{3\text{Ei}(x)}{2}+\frac{3}{2}\int \frac{e^x}{x}dx\\ & =15e^x-\frac{12}{x^4}+\frac{28}{x^3}-\frac{3e^x}{x^3}-\frac{16}{x^2}+\frac{4e^x}{x^2}+\frac{60}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.09, size = 38, normalized size = 1.36 $$\begin{array}{c}\frac{-12+(28-3e^x)x+4(-4+e^x)x^2+60x^3+15e^x{x}^4}{x^4}\end{array}$$

Antiderivative was successfully verified.

[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]

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

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fricas [A]  time = 0.51, size = 36, normalized size = 1.29 $$\begin{array}{c}\frac{60x^3-16x^2+(15x^4+4x^2-3x)e^x+28x-12}{x^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 38, normalized size = 1.36 $$\begin{array}{c}\frac{15x^4{e}^x+60x^3+4x^2{e}^x-16x^2-3x{e}^x+28x-12}{x^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 37, normalized size = 1.32

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [C]  time = 0.41, size = 46, normalized size = 1.64 $$\begin{array}{c}\frac{60}{x}-\frac{16}{x^2}+\frac{28}{x^3}-\frac{12}{x^4}+15e^x+4\mathrm{\Gamma }(-1,-x)+11\mathrm{\Gamma }(-2,-x)+9\mathrm{\Gamma }(-3,-x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.09, size = 36, normalized size = 1.29 $$\begin{array}{c}15{\mathrm{e}}^x-\frac{x(3{\mathrm{e}}^x-28)-x^2(4{\mathrm{e}}^x-16)-60x^3+12}{x^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.13, size = 34, normalized size = 1.21 $$\begin{array}{c}\frac{(15x^3+4x-3)e^x}{x^3}-\frac{-60x^3+16x^2-28x+12}{x^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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