∫ x+x2+2x3+e15−3x(−1−3x+x2−3x3)__________________________

3.80.93 $\int \frac {x+x^2+2 x^3+e^{15-3 x} (-1-3 x+x^2-3 x^3)}{x^2} \, dx$

Optimal. Leaf size=24 \[ \log \left (e^{x+\left (e^{3 (5-x)}+x\right ) \left (\frac {1}{x}+x\right )} x\right ) \]

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Rubi [A] time = 0.13, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 6, integrand size = 35, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.171, Rules used = {14, 2199, 2194, 2177, 2178, 2176} \begin {gather*} x^2+e^{15-3 x} x+x+\frac {e^{15-3 x}}{x}+\log (x) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(x + x^2 + 2*x^3 + E^(15 - 3*x)*(-1 - 3*x + x^2 - 3*x^3))/x^2,x]

[Out]

E^(15 - 3*x)/x + x + E^(15 - 3*x)*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 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 (\frac {1+x+2 x^2}{x}-\frac {e^{15-3 x} \left (1+3 x-x^2+3 x^3\right )}{x^2}\right ) \, dx\\ &=\int \frac {1+x+2 x^2}{x} \, dx-\int \frac {e^{15-3 x} \left (1+3 x-x^2+3 x^3\right )}{x^2} \, dx\\ &=\int \left (1+\frac {1}{x}+2 x\right ) \, dx-\int \left (-e^{15-3 x}+\frac {e^{15-3 x}}{x^2}+\frac {3 e^{15-3 x}}{x}+3 e^{15-3 x} x\right ) \, dx\\ &=x+x^2+\log (x)-3 \int \frac {e^{15-3 x}}{x} \, dx-3 \int e^{15-3 x} x \, dx+\int e^{15-3 x} \, dx-\int \frac {e^{15-3 x}}{x^2} \, dx\\ &=-\frac {1}{3} e^{15-3 x}+\frac {e^{15-3 x}}{x}+x+e^{15-3 x} x+x^2-3 e^{15} \text {Ei}(-3 x)+\log (x)+3 \int \frac {e^{15-3 x}}{x} \, dx-\int e^{15-3 x} \, dx\\ &=\frac {e^{15-3 x}}{x}+x+e^{15-3 x} x+x^2+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.05, size = 27, normalized size = 1.12 \begin {gather*} \frac {e^{15-3 x}}{x}+x+e^{15-3 x} x+x^2+\log (x) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x + x^2 + 2*x^3 + E^(15 - 3*x)*(-1 - 3*x + x^2 - 3*x^3))/x^2,x]

[Out]

E^(15 - 3*x)/x + x + E^(15 - 3*x)*x + x^2 + Log[x]

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fricas [A] time = 1.04, size = 27, normalized size = 1.12 \begin {gather*} \frac {x^{3} + x^{2} + {\left (x^{2} + 1\right )} e^{\left (-3 \, x + 15\right )} + x \log \relax (x)}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+x^2-3*x-1)*exp(15-3*x)+2*x^3+x^2+x)/x^2,x, algorithm="fricas")

[Out]

(x^3 + x^2 + (x^2 + 1)*e^(-3*x + 15) + x*log(x))/x

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giac [A] time = 0.12, size = 31, normalized size = 1.29 \begin {gather*} \frac {x^{3} + x^{2} e^{\left (-3 \, x + 15\right )} + x^{2} + x \log \relax (x) + e^{\left (-3 \, x + 15\right )}}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+x^2-3*x-1)*exp(15-3*x)+2*x^3+x^2+x)/x^2,x, algorithm="giac")

[Out]

(x^3 + x^2*e^(-3*x + 15) + x^2 + x*log(x) + e^(-3*x + 15))/x

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maple [A] time = 0.10, size = 23, normalized size = 0.96


method	result	size

risch	$\ln \relax (x )+x^{2}+x +\frac {\left (x^{2}+1\right ) {\mathrm e}^{15-3 x}}{x}$	$23$
norman	$\frac {x^{2}+x^{3}+{\mathrm e}^{15-3 x} x^{2}+{\mathrm e}^{15-3 x}}{x}+\ln \relax (x )$	$31$
derivativedivides	$\ln \left (-3 x \right )-55+11 x +\frac {\left (15-3 x \right )^{2}}{9}+\frac {{\mathrm e}^{15-3 x}}{x}+5 \,{\mathrm e}^{15-3 x}-\frac {{\mathrm e}^{15-3 x} \left (15-3 x \right )}{3}$	$50$
default	$\ln \left (-3 x \right )-55+11 x +\frac {\left (15-3 x \right )^{2}}{9}+\frac {{\mathrm e}^{15-3 x}}{x}+5 \,{\mathrm e}^{15-3 x}-\frac {{\mathrm e}^{15-3 x} \left (15-3 x \right )}{3}$	$50$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^3+x^2-3*x-1)*exp(15-3*x)+2*x^3+x^2+x)/x^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)+x^2+x+(x^2+1)/x*exp(15-3*x)

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maxima [C] time = 0.39, size = 46, normalized size = 1.92 \begin {gather*} x^{2} - 3 \, {\rm Ei}\left (-3 \, x\right ) e^{15} + \frac {1}{3} \, {\left (3 \, x e^{15} + e^{15}\right )} e^{\left (-3 \, x\right )} + 3 \, e^{15} \Gamma \left (-1, 3 \, x\right ) + x - \frac {1}{3} \, e^{\left (-3 \, x + 15\right )} + \log \relax (x) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^3+x^2-3*x-1)*exp(15-3*x)+2*x^3+x^2+x)/x^2,x, algorithm="maxima")

[Out]

x^2 - 3*Ei(-3*x)*e^15 + 1/3*(3*x*e^15 + e^15)*e^(-3*x) + 3*e^15*gamma(-1, 3*x) + x - 1/3*e^(-3*x + 15) + log(x

)

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mupad [B] time = 5.27, size = 30, normalized size = 1.25 \begin {gather*} \ln \relax (x)+\frac {{\mathrm {e}}^{15-3\,x}+x^2\,{\mathrm {e}}^{15-3\,x}+x^2+x^3}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((x - exp(15 - 3*x)*(3*x - x^2 + 3*x^3 + 1) + x^2 + 2*x^3)/x^2,x)

[Out]

log(x) + (exp(15 - 3*x) + x^2*exp(15 - 3*x) + x^2 + x^3)/x

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sympy [A] time = 0.11, size = 20, normalized size = 0.83 \begin {gather*} x^{2} + x + \log {\relax (x )} + \frac {\left (x^{2} + 1\right ) e^{15 - 3 x}}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**3+x**2-3*x-1)*exp(15-3*x)+2*x**3+x**2+x)/x**2,x)

[Out]

x**2 + x + log(x) + (x**2 + 1)*exp(15 - 3*x)/x

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method	result	size

risch	\(\ln \relax (x )+x^{2}+x +\frac {\left (x^{2}+1\right ) {\mathrm e}^{15-3 x}}{x}\)	\(23\)
norman	\(\frac {x^{2}+x^{3}+{\mathrm e}^{15-3 x} x^{2}+{\mathrm e}^{15-3 x}}{x}+\ln \relax (x )\)	\(31\)
derivativedivides	\(\ln \left (-3 x \right )-55+11 x +\frac {\left (15-3 x \right )^{2}}{9}+\frac {{\mathrm e}^{15-3 x}}{x}+5 \,{\mathrm e}^{15-3 x}-\frac {{\mathrm e}^{15-3 x} \left (15-3 x \right )}{3}\)	\(50\)
default	\(\ln \left (-3 x \right )-55+11 x +\frac {\left (15-3 x \right )^{2}}{9}+\frac {{\mathrm e}^{15-3 x}}{x}+5 \,{\mathrm e}^{15-3 x}-\frac {{\mathrm e}^{15-3 x} \left (15-3 x \right )}{3}\)	\(50\)

3.80.93 \(\int \frac {x+x^2+2 x^3+e^{15-3 x} (-1-3 x+x^2-3 x^3)}{x^2} \, dx\)