3.19.17 \(\int \frac {-400 x^2-800 x^3+e^{x/400} (-1600+4 x-401 x^2-x^3)}{1200 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {\left (e^{x/400}+x\right ) \left (4-x-x^2\right )}{3 x} \]

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Rubi [A]  time = 0.11, antiderivative size = 49, normalized size of antiderivative = 1.88, number of steps used = 11, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 14, 2199, 2194, 2177, 2178, 2176} \begin {gather*} -\frac {1}{12} (2 x+1)^2-\frac {e^{x/400}}{3}-\frac {1}{3} e^{x/400} x+\frac {4 e^{x/400}}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-400*x^2 - 800*x^3 + E^(x/400)*(-1600 + 4*x - 401*x^2 - x^3))/(1200*x^2),x]

[Out]

-1/3*E^(x/400) + (4*E^(x/400))/(3*x) - (E^(x/400)*x)/3 - (1 + 2*x)^2/12

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {\int \frac {-400 x^2-800 x^3+e^{x/400} \left (-1600+4 x-401 x^2-x^3\right )}{x^2} \, dx}{1200}\\ &=\frac {\int \left (-400 (1+2 x)-\frac {e^{x/400} \left (1600-4 x+401 x^2+x^3\right )}{x^2}\right ) \, dx}{1200}\\ &=-\frac {1}{12} (1+2 x)^2-\frac {\int \frac {e^{x/400} \left (1600-4 x+401 x^2+x^3\right )}{x^2} \, dx}{1200}\\ &=-\frac {1}{12} (1+2 x)^2-\frac {\int \left (401 e^{x/400}+\frac {1600 e^{x/400}}{x^2}-\frac {4 e^{x/400}}{x}+e^{x/400} x\right ) \, dx}{1200}\\ &=-\frac {1}{12} (1+2 x)^2-\frac {\int e^{x/400} x \, dx}{1200}+\frac {1}{300} \int \frac {e^{x/400}}{x} \, dx-\frac {401 \int e^{x/400} \, dx}{1200}-\frac {4}{3} \int \frac {e^{x/400}}{x^2} \, dx\\ &=-\frac {401 e^{x/400}}{3}+\frac {4 e^{x/400}}{3 x}-\frac {1}{3} e^{x/400} x-\frac {1}{12} (1+2 x)^2+\frac {\text {Ei}\left (\frac {x}{400}\right )}{300}-\frac {1}{300} \int \frac {e^{x/400}}{x} \, dx+\frac {1}{3} \int e^{x/400} \, dx\\ &=-\frac {e^{x/400}}{3}+\frac {4 e^{x/400}}{3 x}-\frac {1}{3} e^{x/400} x-\frac {1}{12} (1+2 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 34, normalized size = 1.31 \begin {gather*} -\frac {x}{3}-\frac {x^2}{3}-\frac {e^{x/400} \left (400-\frac {1600}{x}+400 x\right )}{1200} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-400*x^2 - 800*x^3 + E^(x/400)*(-1600 + 4*x - 401*x^2 - x^3))/(1200*x^2),x]

[Out]

-1/3*x - x^2/3 - (E^(x/400)*(400 - 1600/x + 400*x))/1200

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fricas [A]  time = 0.90, size = 23, normalized size = 0.88 \begin {gather*} -\frac {x^{3} + x^{2} + {\left (x^{2} + x - 4\right )} e^{\left (\frac {1}{400} \, x\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/1200*((-x^3-401*x^2+4*x-1600)*exp(1/400*x)-800*x^3-400*x^2)/x^2,x, algorithm="fricas")

[Out]

-1/3*(x^3 + x^2 + (x^2 + x - 4)*e^(1/400*x))/x

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giac [A]  time = 0.23, size = 32, normalized size = 1.23 \begin {gather*} -\frac {x^{3} + x^{2} e^{\left (\frac {1}{400} \, x\right )} + x^{2} + x e^{\left (\frac {1}{400} \, x\right )} - 4 \, e^{\left (\frac {1}{400} \, x\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/1200*((-x^3-401*x^2+4*x-1600)*exp(1/400*x)-800*x^3-400*x^2)/x^2,x, algorithm="giac")

[Out]

-1/3*(x^3 + x^2*e^(1/400*x) + x^2 + x*e^(1/400*x) - 4*e^(1/400*x))/x

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maple [A]  time = 0.08, size = 25, normalized size = 0.96

method	result	size
risch	\(-\frac {x^{2}}{3}-\frac {x}{3}-\frac {\left (x^{2}+x -4\right ) {\mathrm e}^{\frac {x}{400}}}{3 x}\)	\(25\)
derivativedivides	\(-\frac {x^{2}}{3}-\frac {x}{3}+\frac {4 \,{\mathrm e}^{\frac {x}{400}}}{3 x}-\frac {{\mathrm e}^{\frac {x}{400}} x}{3}-\frac {{\mathrm e}^{\frac {x}{400}}}{3}\)	\(32\)
default	\(-\frac {x^{2}}{3}-\frac {x}{3}+\frac {4 \,{\mathrm e}^{\frac {x}{400}}}{3 x}-\frac {{\mathrm e}^{\frac {x}{400}} x}{3}-\frac {{\mathrm e}^{\frac {x}{400}}}{3}\)	\(32\)
norman	\(\frac {-\frac {x^{2}}{3}-\frac {x^{3}}{3}-\frac {{\mathrm e}^{\frac {x}{400}} x}{3}-\frac {{\mathrm e}^{\frac {x}{400}} x^{2}}{3}+\frac {4 \,{\mathrm e}^{\frac {x}{400}}}{3}}{x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/1200*((-x^3-401*x^2+4*x-1600)*exp(1/400*x)-800*x^3-400*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*x^2-1/3*x-1/3*(x^2+x-4)/x*exp(1/400*x)

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maxima [C]  time = 0.39, size = 37, normalized size = 1.42 \begin {gather*} -\frac {1}{3} \, x^{2} - \frac {1}{3} \, {\left (x - 400\right )} e^{\left (\frac {1}{400} \, x\right )} - \frac {1}{3} \, x + \frac {1}{300} \, {\rm Ei}\left (\frac {1}{400} \, x\right ) - \frac {401}{3} \, e^{\left (\frac {1}{400} \, x\right )} - \frac {1}{300} \, \Gamma \left (-1, -\frac {1}{400} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/1200*((-x^3-401*x^2+4*x-1600)*exp(1/400*x)-800*x^3-400*x^2)/x^2,x, algorithm="maxima")

[Out]

-1/3*x^2 - 1/3*(x - 400)*e^(1/400*x) - 1/3*x + 1/300*Ei(1/400*x) - 401/3*e^(1/400*x) - 1/300*gamma(-1, -1/400*
x)

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mupad [B]  time = 1.14, size = 32, normalized size = 1.23 \begin {gather*} \frac {4\,{\mathrm {e}}^{x/400}}{3\,x}-\frac {{\mathrm {e}}^{x/400}}{3}-x\,\left (\frac {{\mathrm {e}}^{x/400}}{3}+\frac {1}{3}\right )-\frac {x^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2/3 + (2*x^3)/3 + (exp(x/400)*(401*x^2 - 4*x + x^3 + 1600))/1200)/x^2,x)

[Out]

(4*exp(x/400))/(3*x) - exp(x/400)/3 - x*(exp(x/400)/3 + 1/3) - x^2/3

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sympy [A]  time = 0.11, size = 22, normalized size = 0.85 \begin {gather*} - \frac {x^{2}}{3} - \frac {x}{3} + \frac {\left (- x^{2} - x + 4\right ) e^{\frac {x}{400}}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/1200*((-x**3-401*x**2+4*x-1600)*exp(1/400*x)-800*x**3-400*x**2)/x**2,x)

[Out]

-x**2/3 - x/3 + (-x**2 - x + 4)*exp(x/400)/(3*x)

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