3.91.74 \(\int \frac {-50-350 x-1600 x^3-2000 x^4+2400 x^5+1600 x^6+e^{2 x} (-18+18 x)+e^x (-60-180 x+210 x^2-240 x^3-120 x^4)}{9 x^3} \, dx\)

Optimal. Leaf size=29 \[ \left (\frac {e^x}{x}-\frac {5 (-1+x) \left (-3+(2+2 x)^2\right )}{3 x}\right )^2 \]

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Rubi [B]  time = 0.20, antiderivative size = 85, normalized size of antiderivative = 2.93, number of steps used = 17, number of rules used = 8, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {12, 14, 2197, 2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {400 x^4}{9}+\frac {800 x^3}{9}-\frac {1000 x^2}{9}+\frac {10 e^x}{3 x^2}+\frac {e^{2 x}}{x^2}+\frac {25}{9 x^2}-\frac {40 e^x x}{3}-\frac {1600 x}{9}-\frac {40 e^x}{3}+\frac {70 e^x}{3 x}+\frac {350}{9 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 - 350*x - 1600*x^3 - 2000*x^4 + 2400*x^5 + 1600*x^6 + E^(2*x)*(-18 + 18*x) + E^x*(-60 - 180*x + 210*x
^2 - 240*x^3 - 120*x^4))/(9*x^3),x]

[Out]

(-40*E^x)/3 + 25/(9*x^2) + (10*E^x)/(3*x^2) + E^(2*x)/x^2 + 350/(9*x) + (70*E^x)/(3*x) - (1600*x)/9 - (40*E^x*
x)/3 - (1000*x^2)/9 + (800*x^3)/9 + (400*x^4)/9

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 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
]

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 {1}{9} \int \frac {-50-350 x-1600 x^3-2000 x^4+2400 x^5+1600 x^6+e^{2 x} (-18+18 x)+e^x \left (-60-180 x+210 x^2-240 x^3-120 x^4\right )}{x^3} \, dx\\ &=\frac {1}{9} \int \left (\frac {18 e^{2 x} (-1+x)}{x^3}-\frac {30 e^x \left (2+6 x-7 x^2+8 x^3+4 x^4\right )}{x^3}+\frac {50 \left (-1-7 x-32 x^3-40 x^4+48 x^5+32 x^6\right )}{x^3}\right ) \, dx\\ &=2 \int \frac {e^{2 x} (-1+x)}{x^3} \, dx-\frac {10}{3} \int \frac {e^x \left (2+6 x-7 x^2+8 x^3+4 x^4\right )}{x^3} \, dx+\frac {50}{9} \int \frac {-1-7 x-32 x^3-40 x^4+48 x^5+32 x^6}{x^3} \, dx\\ &=\frac {e^{2 x}}{x^2}-\frac {10}{3} \int \left (8 e^x+\frac {2 e^x}{x^3}+\frac {6 e^x}{x^2}-\frac {7 e^x}{x}+4 e^x x\right ) \, dx+\frac {50}{9} \int \left (-32-\frac {1}{x^3}-\frac {7}{x^2}-40 x+48 x^2+32 x^3\right ) \, dx\\ &=\frac {25}{9 x^2}+\frac {e^{2 x}}{x^2}+\frac {350}{9 x}-\frac {1600 x}{9}-\frac {1000 x^2}{9}+\frac {800 x^3}{9}+\frac {400 x^4}{9}-\frac {20}{3} \int \frac {e^x}{x^3} \, dx-\frac {40}{3} \int e^x x \, dx-20 \int \frac {e^x}{x^2} \, dx+\frac {70}{3} \int \frac {e^x}{x} \, dx-\frac {80 \int e^x \, dx}{3}\\ &=-\frac {80 e^x}{3}+\frac {25}{9 x^2}+\frac {10 e^x}{3 x^2}+\frac {e^{2 x}}{x^2}+\frac {350}{9 x}+\frac {20 e^x}{x}-\frac {1600 x}{9}-\frac {40 e^x x}{3}-\frac {1000 x^2}{9}+\frac {800 x^3}{9}+\frac {400 x^4}{9}+\frac {70 \text {Ei}(x)}{3}-\frac {10}{3} \int \frac {e^x}{x^2} \, dx+\frac {40 \int e^x \, dx}{3}-20 \int \frac {e^x}{x} \, dx\\ &=-\frac {40 e^x}{3}+\frac {25}{9 x^2}+\frac {10 e^x}{3 x^2}+\frac {e^{2 x}}{x^2}+\frac {350}{9 x}+\frac {70 e^x}{3 x}-\frac {1600 x}{9}-\frac {40 e^x x}{3}-\frac {1000 x^2}{9}+\frac {800 x^3}{9}+\frac {400 x^4}{9}+\frac {10 \text {Ei}(x)}{3}-\frac {10}{3} \int \frac {e^x}{x} \, dx\\ &=-\frac {40 e^x}{3}+\frac {25}{9 x^2}+\frac {10 e^x}{3 x^2}+\frac {e^{2 x}}{x^2}+\frac {350}{9 x}+\frac {70 e^x}{3 x}-\frac {1600 x}{9}-\frac {40 e^x x}{3}-\frac {1000 x^2}{9}+\frac {800 x^3}{9}+\frac {400 x^4}{9}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 62, normalized size = 2.14 \begin {gather*} \frac {1}{9} \left (e^x \left (-120+\frac {30}{x^2}+\frac {210}{x}-120 x\right )+\frac {25}{x^2}+\frac {9 e^{2 x}}{x^2}+\frac {350}{x}-1600 x-1000 x^2+800 x^3+400 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 - 350*x - 1600*x^3 - 2000*x^4 + 2400*x^5 + 1600*x^6 + E^(2*x)*(-18 + 18*x) + E^x*(-60 - 180*x +
210*x^2 - 240*x^3 - 120*x^4))/(9*x^3),x]

[Out]

(E^x*(-120 + 30/x^2 + 210/x - 120*x) + 25/x^2 + (9*E^(2*x))/x^2 + 350/x - 1600*x - 1000*x^2 + 800*x^3 + 400*x^
4)/9

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fricas [A]  time = 0.60, size = 55, normalized size = 1.90 \begin {gather*} \frac {400 \, x^{6} + 800 \, x^{5} - 1000 \, x^{4} - 1600 \, x^{3} - 30 \, {\left (4 \, x^{3} + 4 \, x^{2} - 7 \, x - 1\right )} e^{x} + 350 \, x + 9 \, e^{\left (2 \, x\right )} + 25}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((18*x-18)*exp(x)^2+(-120*x^4-240*x^3+210*x^2-180*x-60)*exp(x)+1600*x^6+2400*x^5-2000*x^4-1600*x
^3-350*x-50)/x^3,x, algorithm="fricas")

[Out]

1/9*(400*x^6 + 800*x^5 - 1000*x^4 - 1600*x^3 - 30*(4*x^3 + 4*x^2 - 7*x - 1)*e^x + 350*x + 9*e^(2*x) + 25)/x^2

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giac [B]  time = 0.21, size = 59, normalized size = 2.03 \begin {gather*} \frac {400 \, x^{6} + 800 \, x^{5} - 1000 \, x^{4} - 120 \, x^{3} e^{x} - 1600 \, x^{3} - 120 \, x^{2} e^{x} + 210 \, x e^{x} + 350 \, x + 9 \, e^{\left (2 \, x\right )} + 30 \, e^{x} + 25}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((18*x-18)*exp(x)^2+(-120*x^4-240*x^3+210*x^2-180*x-60)*exp(x)+1600*x^6+2400*x^5-2000*x^4-1600*x
^3-350*x-50)/x^3,x, algorithm="giac")

[Out]

1/9*(400*x^6 + 800*x^5 - 1000*x^4 - 120*x^3*e^x - 1600*x^3 - 120*x^2*e^x + 210*x*e^x + 350*x + 9*e^(2*x) + 30*
e^x + 25)/x^2

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maple [B]  time = 0.05, size = 57, normalized size = 1.97

method	result	size
norman	\(\frac {\frac {25}{9}+{\mathrm e}^{2 x}+\frac {350 x}{9}-\frac {1600 x^{3}}{9}-\frac {1000 x^{4}}{9}+\frac {800 x^{5}}{9}+\frac {400 x^{6}}{9}+\frac {70 \,{\mathrm e}^{x} x}{3}-\frac {40 \,{\mathrm e}^{x} x^{2}}{3}-\frac {40 \,{\mathrm e}^{x} x^{3}}{3}+\frac {10 \,{\mathrm e}^{x}}{3}}{x^{2}}\)	\(57\)
risch	\(\frac {400 x^{4}}{9}+\frac {800 x^{3}}{9}-\frac {1000 x^{2}}{9}-\frac {1600 x}{9}+\frac {350 x +25}{9 x^{2}}+\frac {{\mathrm e}^{2 x}}{x^{2}}-\frac {10 \left (4 x^{3}+4 x^{2}-7 x -1\right ) {\mathrm e}^{x}}{3 x^{2}}\)	\(60\)
default	\(-\frac {1000 x^{2}}{9}-\frac {1600 x}{9}+\frac {25}{9 x^{2}}+\frac {350}{9 x}+\frac {800 x^{3}}{9}+\frac {400 x^{4}}{9}+\frac {{\mathrm e}^{2 x}}{x^{2}}+\frac {10 \,{\mathrm e}^{x}}{3 x^{2}}+\frac {70 \,{\mathrm e}^{x}}{3 x}-\frac {40 \,{\mathrm e}^{x} x}{3}-\frac {40 \,{\mathrm e}^{x}}{3}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((18*x-18)*exp(x)^2+(-120*x^4-240*x^3+210*x^2-180*x-60)*exp(x)+1600*x^6+2400*x^5-2000*x^4-1600*x^3-350
*x-50)/x^3,x,method=_RETURNVERBOSE)

[Out]

(25/9+exp(x)^2+350/9*x-1600/9*x^3-1000/9*x^4+800/9*x^5+400/9*x^6+70/3*exp(x)*x-40/3*exp(x)*x^2-40/3*exp(x)*x^3
+10/3*exp(x))/x^2

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maxima [C]  time = 0.39, size = 72, normalized size = 2.48 \begin {gather*} \frac {400}{9} \, x^{4} + \frac {800}{9} \, x^{3} - \frac {1000}{9} \, x^{2} - \frac {40}{3} \, {\left (x - 1\right )} e^{x} - \frac {1600}{9} \, x + \frac {350}{9 \, x} + \frac {25}{9 \, x^{2}} + \frac {70}{3} \, {\rm Ei}\relax (x) - \frac {80}{3} \, e^{x} - 20 \, \Gamma \left (-1, -x\right ) + 4 \, \Gamma \left (-1, -2 \, x\right ) + \frac {20}{3} \, \Gamma \left (-2, -x\right ) + 8 \, \Gamma \left (-2, -2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((18*x-18)*exp(x)^2+(-120*x^4-240*x^3+210*x^2-180*x-60)*exp(x)+1600*x^6+2400*x^5-2000*x^4-1600*x
^3-350*x-50)/x^3,x, algorithm="maxima")

[Out]

400/9*x^4 + 800/9*x^3 - 1000/9*x^2 - 40/3*(x - 1)*e^x - 1600/9*x + 350/9/x + 25/9/x^2 + 70/3*Ei(x) - 80/3*e^x
- 20*gamma(-1, -x) + 4*gamma(-1, -2*x) + 20/3*gamma(-2, -x) + 8*gamma(-2, -2*x)

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mupad [B]  time = 6.88, size = 51, normalized size = 1.76 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}+\frac {10\,{\mathrm {e}}^x}{3}+x\,\left (\frac {70\,{\mathrm {e}}^x}{3}+\frac {350}{9}\right )+\frac {25}{9}}{x^2}-x\,\left (\frac {40\,{\mathrm {e}}^x}{3}+\frac {1600}{9}\right )-\frac {40\,{\mathrm {e}}^x}{3}-\frac {1000\,x^2}{9}+\frac {800\,x^3}{9}+\frac {400\,x^4}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((350*x)/9 + (exp(x)*(180*x - 210*x^2 + 240*x^3 + 120*x^4 + 60))/9 - (exp(2*x)*(18*x - 18))/9 + (1600*x^3
)/9 + (2000*x^4)/9 - (800*x^5)/3 - (1600*x^6)/9 + 50/9)/x^3,x)

[Out]

(exp(2*x) + (10*exp(x))/3 + x*((70*exp(x))/3 + 350/9) + 25/9)/x^2 - x*((40*exp(x))/3 + 1600/9) - (40*exp(x))/3
- (1000*x^2)/9 + (800*x^3)/9 + (400*x^4)/9

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sympy [B]  time = 0.18, size = 73, normalized size = 2.52 \begin {gather*} \frac {400 x^{4}}{9} + \frac {800 x^{3}}{9} - \frac {1000 x^{2}}{9} - \frac {1600 x}{9} + \frac {350 x + 25}{9 x^{2}} + \frac {3 x^{2} e^{2 x} + \left (- 40 x^{5} - 40 x^{4} + 70 x^{3} + 10 x^{2}\right ) e^{x}}{3 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((18*x-18)*exp(x)**2+(-120*x**4-240*x**3+210*x**2-180*x-60)*exp(x)+1600*x**6+2400*x**5-2000*x**4
-1600*x**3-350*x-50)/x**3,x)

[Out]

400*x**4/9 + 800*x**3/9 - 1000*x**2/9 - 1600*x/9 + (350*x + 25)/(9*x**2) + (3*x**2*exp(2*x) + (-40*x**5 - 40*x
**4 + 70*x**3 + 10*x**2)*exp(x))/(3*x**4)

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