3.25.37 \(\int \frac {-200+100 x+e^x (850-125 x-300 x^2-100 x^3)}{4 x^3} \, dx\) [2437]

Optimal. Leaf size=24 \[ \frac {25 \left (1-x-e^x \left (2+\left (\frac {3}{2}+x\right )^2\right )\right )}{x^2} \]

[Out]

25*(1-x-((x+3/2)^2+2)*exp(x))/x^2

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Rubi [A]
time = 0.08, antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 13, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {12, 14, 37, 2230, 2225, 2208, 2209} \begin {gather*} \frac {25 (2-x)^2}{4 x^2}-\frac {425 e^x}{4 x^2}-25 e^x-\frac {75 e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-200 + 100*x + E^x*(850 - 125*x - 300*x^2 - 100*x^3))/(4*x^3),x]

[Out]

-25*E^x - (425*E^x)/(4*x^2) + (25*(2 - x)^2)/(4*x^2) - (75*E^x)/x

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2225

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 2230

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] &&  !TrueQ[$UseGamma]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-200+100 x+e^x \left (850-125 x-300 x^2-100 x^3\right )}{x^3} \, dx\\ &=\frac {1}{4} \int \left (\frac {100 (-2+x)}{x^3}-\frac {25 e^x \left (-34+5 x+12 x^2+4 x^3\right )}{x^3}\right ) \, dx\\ &=-\left (\frac {25}{4} \int \frac {e^x \left (-34+5 x+12 x^2+4 x^3\right )}{x^3} \, dx\right )+25 \int \frac {-2+x}{x^3} \, dx\\ &=\frac {25 (2-x)^2}{4 x^2}-\frac {25}{4} \int \left (4 e^x-\frac {34 e^x}{x^3}+\frac {5 e^x}{x^2}+\frac {12 e^x}{x}\right ) \, dx\\ &=\frac {25 (2-x)^2}{4 x^2}-25 \int e^x \, dx-\frac {125}{4} \int \frac {e^x}{x^2} \, dx-75 \int \frac {e^x}{x} \, dx+\frac {425}{2} \int \frac {e^x}{x^3} \, dx\\ &=-25 e^x-\frac {425 e^x}{4 x^2}+\frac {25 (2-x)^2}{4 x^2}+\frac {125 e^x}{4 x}-75 \text {Ei}(x)-\frac {125}{4} \int \frac {e^x}{x} \, dx+\frac {425}{4} \int \frac {e^x}{x^2} \, dx\\ &=-25 e^x-\frac {425 e^x}{4 x^2}+\frac {25 (2-x)^2}{4 x^2}-\frac {75 e^x}{x}-\frac {425 \text {Ei}(x)}{4}+\frac {425}{4} \int \frac {e^x}{x} \, dx\\ &=-25 e^x-\frac {425 e^x}{4 x^2}+\frac {25 (2-x)^2}{4 x^2}-\frac {75 e^x}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.16, size = 27, normalized size = 1.12 \begin {gather*} -\frac {25 \left (4 (-1+x)+e^x \left (17+12 x+4 x^2\right )\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-200 + 100*x + E^x*(850 - 125*x - 300*x^2 - 100*x^3))/(4*x^3),x]

[Out]

(-25*(4*(-1 + x) + E^x*(17 + 12*x + 4*x^2)))/(4*x^2)

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Maple [A]
time = 0.02, size = 30, normalized size = 1.25

method	result	size
norman	\(\frac {25-25 x -75 \,{\mathrm e}^{x} x -25 \,{\mathrm e}^{x} x^{2}-\frac {425 \,{\mathrm e}^{x}}{4}}{x^{2}}\)	\(26\)
risch	\(\frac {-100 x +100}{4 x^{2}}-\frac {25 \left (4 x^{2}+12 x +17\right ) {\mathrm e}^{x}}{4 x^{2}}\)	\(29\)
default	\(\frac {25}{x^{2}}-\frac {25}{x}-\frac {425 \,{\mathrm e}^{x}}{4 x^{2}}-\frac {75 \,{\mathrm e}^{x}}{x}-25 \,{\mathrm e}^{x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-100*x^3-300*x^2-125*x+850)*exp(x)+100*x-200)/x^3,x,method=_RETURNVERBOSE)

[Out]

25/x^2-25/x-425/4*exp(x)/x^2-75*exp(x)/x-25*exp(x)

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.28, size = 33, normalized size = 1.38 \begin {gather*} -\frac {25}{x} + \frac {25}{x^{2}} - 75 \, {\rm Ei}\left (x\right ) - 25 \, e^{x} - \frac {125}{4} \, \Gamma \left (-1, -x\right ) - \frac {425}{2} \, \Gamma \left (-2, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-100*x^3-300*x^2-125*x+850)*exp(x)+100*x-200)/x^3,x, algorithm="maxima")

[Out]

-25/x + 25/x^2 - 75*Ei(x) - 25*e^x - 125/4*gamma(-1, -x) - 425/2*gamma(-2, -x)

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Fricas [A]
time = 0.38, size = 23, normalized size = 0.96 \begin {gather*} -\frac {25 \, {\left ({\left (4 \, x^{2} + 12 \, x + 17\right )} e^{x} + 4 \, x - 4\right )}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-100*x^3-300*x^2-125*x+850)*exp(x)+100*x-200)/x^3,x, algorithm="fricas")

[Out]

-25/4*((4*x^2 + 12*x + 17)*e^x + 4*x - 4)/x^2

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Sympy [A]
time = 0.06, size = 27, normalized size = 1.12 \begin {gather*} \frac {25 - 25 x}{x^{2}} + \frac {\left (- 100 x^{2} - 300 x - 425\right ) e^{x}}{4 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-100*x**3-300*x**2-125*x+850)*exp(x)+100*x-200)/x**3,x)

[Out]

(25 - 25*x)/x**2 + (-100*x**2 - 300*x - 425)*exp(x)/(4*x**2)

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Giac [A]
time = 0.40, size = 26, normalized size = 1.08 \begin {gather*} -\frac {25 \, {\left (4 \, x^{2} e^{x} + 12 \, x e^{x} + 4 \, x + 17 \, e^{x} - 4\right )}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-100*x^3-300*x^2-125*x+850)*exp(x)+100*x-200)/x^3,x, algorithm="giac")

[Out]

-25/4*(4*x^2*e^x + 12*x*e^x + 4*x + 17*e^x - 4)/x^2

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Mupad [B]
time = 0.06, size = 24, normalized size = 1.00 \begin {gather*} -25\,{\mathrm {e}}^x-\frac {\frac {425\,{\mathrm {e}}^x}{4}+x\,\left (75\,{\mathrm {e}}^x+25\right )-25}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(125*x + 300*x^2 + 100*x^3 - 850))/4 - 25*x + 50)/x^3,x)

[Out]

- 25*exp(x) - ((425*exp(x))/4 + x*(75*exp(x) + 25) - 25)/x^2

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