3.69.4 $\int \frac{-10-25x+5x^2+5x^3+e^x(-x^3-x^4)-20\log (x)}{x^3}dx$

Optimal. Leaf size=30 $$-e^{x}x+\frac{5(5+x^2+\frac{(2+x^2)(1+\log (x))}{x})}{x}$$

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 4, integrand size = 38, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.105, Rules used = {14, 2176, 2194, 2304} $$\begin{array}{c}\frac{10}{x^2}+\frac{10\log (x)}{x^2}+5x+e^x-e^x(x+1)+\frac{25}{x}+5\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-10 - 25*x + 5*x^2 + 5*x^3 + E^x*(-x^3 - x^4) - 20*Log[x])/x^3,x]

[Out]

E^x + 10/x^2 + 25/x + 5*x - E^x*(1 + x) + 5*Log[x] + (10*Log[x])/x^2

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 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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-e^x(1+x)+\frac{5(-2-5x+x^2+x^3-4\log (x))}{x^3})dx\\ & =5\int \frac{-2-5x+x^2+x^3-4\log (x)}{x^3}dx-\int e^x(1+x)dx\\ & =-e^x(1+x)+5\int (\frac{-2-5x+x^2+x^3}{x^3}-\frac{4\log (x)}{x^3})dx+\int e^{x}dx\\ & =e^x-e^x(1+x)+5\int \frac{-2-5x+x^2+x^3}{x^3}dx-20\int \frac{\log (x)}{x^3}dx\\ & =e^x+\frac{5}{x^2}-e^x(1+x)+\frac{10\log (x)}{x^2}+5\int (1-\frac{2}{x^3}-\frac{5}{x^2}+\frac{1}{x})dx\\ & =e^x+\frac{10}{x^2}+\frac{25}{x}+5x-e^x(1+x)+5\log (x)+\frac{10\log (x)}{x^2}\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 28, normalized size = 0.93 $$\begin{array}{c}-\frac{-10-25x+(-5+e^x)x^3-5(2+x^2)\log (x)}{x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 25*x + 5*x^2 + 5*x^3 + E^x*(-x^3 - x^4) - 20*Log[x])/x^3,x]

[Out]

-((-10 - 25*x + (-5 + E^x)*x^3 - 5*(2 + x^2)*Log[x])/x^2)

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fricas [A]  time = 0.61, size = 30, normalized size = 1.00 $$\begin{array}{c}-\frac{x^3{e}^x-5x^3-5(x^2+2)\log (x)-25x-10}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="fricas")

[Out]

-(x^3*e^x - 5*x^3 - 5*(x^2 + 2)*log(x) - 25*x - 10)/x^2

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giac [A]  time = 0.19, size = 32, normalized size = 1.07 $$\begin{array}{c}-\frac{x^3{e}^x-5x^3-5x^2\log (x)-25x-10\log (x)-10}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="giac")

[Out]

-(x^3*e^x - 5*x^3 - 5*x^2*log(x) - 25*x - 10*log(x) - 10)/x^2

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maple [A]  time = 0.04, size = 31, normalized size = 1.03

method	result	size
default	$-{\mathrm{e}}^{x}x+5x+5\ln (x)+\frac{25}{x}+\frac{10}{x^2}+\frac{10\ln (x)}{x^2}$	$31$
norman	$\frac{10+25x+5x^3-{\mathrm{e}}^x{x}^3+10\ln (x)}{x^2}+5\ln (x)$	$31$
risch	$\frac{10\ln (x)}{x^2}+\frac{-{\mathrm{e}}^x{x}^3+5x^2\ln (x)+5x^3+25x+10}{x^2}$	$37$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*ln(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x,method=_RETURNVERBOSE)

[Out]

-exp(x)*x+5*x+5*ln(x)+25/x+10/x^2+10*ln(x)/x^2

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maxima [A]  time = 0.38, size = 36, normalized size = 1.20 $$\begin{array}{c}-(x-1)e^x+5x+\frac{25}{x}+\frac{10\log (x)}{x^2}+\frac{10}{x^2}-e^x+5\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(x)+(-x^4-x^3)*exp(x)+5*x^3+5*x^2-25*x-10)/x^3,x, algorithm="maxima")

[Out]

-(x - 1)*e^x + 5*x + 25/x + 10*log(x)/x^2 + 10/x^2 - e^x + 5*log(x)

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mupad [B]  time = 4.24, size = 25, normalized size = 0.83 $$\begin{array}{c}5\ln (x)-x({\mathrm{e}}^x-5)+\frac{25x+10\ln (x)+10}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x + 20*log(x) + exp(x)*(x^3 + x^4) - 5*x^2 - 5*x^3 + 10)/x^3,x)

[Out]

5*log(x) - x*(exp(x) - 5) + (25*x + 10*log(x) + 10)/x^2

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sympy [A]  time = 0.30, size = 29, normalized size = 0.97 $$\begin{array}{c}-x{e}^x+5x+5\log (x)+\frac{25x+10}{x^2}+\frac{10\log (x)}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*ln(x)+(-x**4-x**3)*exp(x)+5*x**3+5*x**2-25*x-10)/x**3,x)

[Out]

-x*exp(x) + 5*x + 5*log(x) + (25*x + 10)/x**2 + 10*log(x)/x**2

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