3.33.43 $\int \frac{x^2-2x^3+6e^{3x}{x}^3+e^{5x}(-2+5x)+e^x(2x^4+x^5)}{x^3}dx$

Optimal. Leaf size=26 $$e^{25}-2x+e^x{(\frac{e^{2x}}{x}+x)}^2+\log (x)$$

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Rubi [A]  time = 0.11, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 6, integrand size = 47, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.128, Rules used = {14, 2194, 43, 2196, 2176, 2197} $$\begin{array}{c}{e}^x{x}^2+\frac{e^{5x}}{x^2}-2x+2e^{3x}+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(x^2 - 2*x^3 + 6*E^(3*x)*x^3 + E^(5*x)*(-2 + 5*x) + E^x*(2*x^4 + x^5))/x^3,x]

[Out]

2*E^(3*x) + E^(5*x)/x^2 - 2*x + E^x*x^2 + Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2196

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

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (6e^{3x}+\frac{1-2x}{x}+e^{x}x(2+x)+\frac{e^{5x}(-2+5x)}{x^3})dx\\ & =6\int e^{3x}dx+\int \frac{1-2x}{x}dx+\int e^{x}x(2+x)dx+\int \frac{e^{5x}(-2+5x)}{x^3}dx\\ & =2e^{3x}+\frac{e^{5x}}{x^2}+\int (-2+\frac{1}{x})dx+\int (2e^{x}x+e^x{x}^2)dx\\ & =2e^{3x}+\frac{e^{5x}}{x^2}-2x+\log (x)+2\int e^{x}xdx+\int e^x{x}^{2}dx\\ & =2e^{3x}+\frac{e^{5x}}{x^2}-2x+2e^{x}x+e^x{x}^2+\log (x)-2\int e^{x}dx-2\int e^{x}xdx\\ & =-2e^x+2e^{3x}+\frac{e^{5x}}{x^2}-2x+e^x{x}^2+\log (x)+2\int e^{x}dx\\ & =2e^{3x}+\frac{e^{5x}}{x^2}-2x+e^x{x}^2+\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.12 $$\begin{array}{c}2e^{3x}+\frac{e^{5x}}{x^2}-2x+e^x{x}^2+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(x^2 - 2*x^3 + 6*E^(3*x)*x^3 + E^(5*x)*(-2 + 5*x) + E^x*(2*x^4 + x^5))/x^3,x]

[Out]

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

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fricas [A]  time = 0.51, size = 35, normalized size = 1.35 $$\begin{array}{c}\frac{x^4{e}^x-2x^3+2x^2{e}^{\left(3x\right)}+x^2\log (x)+e^{\left(5x\right)}}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="fricas")

[Out]

(x^4*e^x - 2*x^3 + 2*x^2*e^(3*x) + x^2*log(x) + e^(5*x))/x^2

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giac [A]  time = 0.27, size = 35, normalized size = 1.35 $$\begin{array}{c}\frac{x^4{e}^x-2x^3+2x^2{e}^{\left(3x\right)}+x^2\log (x)+e^{\left(5x\right)}}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="giac")

[Out]

(x^4*e^x - 2*x^3 + 2*x^2*e^(3*x) + x^2*log(x) + e^(5*x))/x^2

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maple [A]  time = 0.06, size = 27, normalized size = 1.04

method	result	size
default	$\ln (x)-2x+{\mathrm{e}}^x{x}^2+2{\mathrm{e}}^{3x}+\frac{{\mathrm{e}}^{5x}}{x^2}$	$27$
risch	$\ln (x)-2x+{\mathrm{e}}^x{x}^2+2{\mathrm{e}}^{3x}+\frac{{\mathrm{e}}^{5x}}{x^2}$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.47, size = 44, normalized size = 1.69 $$\begin{array}{c}(x^2-2x+2)e^x+2(x-1)e^x-2x+2e^{\left(3x\right)}+25\mathrm{\Gamma }(-1,-5x)+50\mathrm{\Gamma }(-2,-5x)+\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)^5+6*x^3*exp(x)^3+(x^5+2*x^4)*exp(x)-2*x^3+x^2)/x^3,x, algorithm="maxima")

[Out]

(x^2 - 2*x + 2)*e^x + 2*(x - 1)*e^x - 2*x + 2*e^(3*x) + 25*gamma(-1, -5*x) + 50*gamma(-2, -5*x) + log(x)

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mupad [B]  time = 1.88, size = 32, normalized size = 1.23 $$\begin{array}{c}\ln (x)+\frac{{\mathrm{e}}^{5x}+x^4{\mathrm{e}}^x+2x^2{\mathrm{e}}^{3x}-2x^3}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^3*exp(3*x) + exp(x)*(2*x^4 + x^5) + exp(5*x)*(5*x - 2) + x^2 - 2*x^3)/x^3,x)

[Out]

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

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sympy [A]  time = 0.16, size = 31, normalized size = 1.19 $$\begin{array}{c}-2x+\log (x)+\frac{x^4{e}^x+2x^2{e}^{3x}+e^{5x}}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-2)*exp(x)**5+6*x**3*exp(x)**3+(x**5+2*x**4)*exp(x)-2*x**3+x**2)/x**3,x)

[Out]

-2*x + log(x) + (x**4*exp(x) + 2*x**2*exp(3*x) + exp(5*x))/x**2

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