3.18.33 \(\int \frac {50 x^3+150 x^4-300 x^5+100 x^6+e^{2 x/5} (-50+10 x)+e^{x/5} (-25 x+5 x^2-30 x^3-10 x^4)}{x^3} \, dx\)

Optimal. Leaf size=26 \[ 25 \left (-\frac {1}{2}-\frac {e^{x/5}}{x}-2 x+x^2\right )^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 62, normalized size of antiderivative = 2.38, number of steps used = 12, number of rules used = 7, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {14, 2197, 2199, 2194, 2177, 2178, 2176} \begin {gather*} 25 x^4-100 x^3+75 x^2+\frac {25 e^{2 x/5}}{x^2}-50 e^{x/5} x+50 x+100 e^{x/5}+\frac {25 e^{x/5}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50*x^3 + 150*x^4 - 300*x^5 + 100*x^6 + E^((2*x)/5)*(-50 + 10*x) + E^(x/5)*(-25*x + 5*x^2 - 30*x^3 - 10*x^
4))/x^3,x]

[Out]

100*E^(x/5) + (25*E^((2*x)/5))/x^2 + (25*E^(x/5))/x + 50*x - 50*E^(x/5)*x + 75*x^2 - 100*x^3 + 25*x^4

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} &=\int \left (\frac {10 e^{2 x/5} (-5+x)}{x^3}+50 \left (1+3 x-6 x^2+2 x^3\right )-\frac {5 e^{x/5} \left (5-x+6 x^2+2 x^3\right )}{x^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{x/5} \left (5-x+6 x^2+2 x^3\right )}{x^2} \, dx\right )+10 \int \frac {e^{2 x/5} (-5+x)}{x^3} \, dx+50 \int \left (1+3 x-6 x^2+2 x^3\right ) \, dx\\ &=\frac {25 e^{2 x/5}}{x^2}+50 x+75 x^2-100 x^3+25 x^4-5 \int \left (6 e^{x/5}+\frac {5 e^{x/5}}{x^2}-\frac {e^{x/5}}{x}+2 e^{x/5} x\right ) \, dx\\ &=\frac {25 e^{2 x/5}}{x^2}+50 x+75 x^2-100 x^3+25 x^4+5 \int \frac {e^{x/5}}{x} \, dx-10 \int e^{x/5} x \, dx-25 \int \frac {e^{x/5}}{x^2} \, dx-30 \int e^{x/5} \, dx\\ &=-150 e^{x/5}+\frac {25 e^{2 x/5}}{x^2}+\frac {25 e^{x/5}}{x}+50 x-50 e^{x/5} x+75 x^2-100 x^3+25 x^4+5 \text {Ei}\left (\frac {x}{5}\right )-5 \int \frac {e^{x/5}}{x} \, dx+50 \int e^{x/5} \, dx\\ &=100 e^{x/5}+\frac {25 e^{2 x/5}}{x^2}+\frac {25 e^{x/5}}{x}+50 x-50 e^{x/5} x+75 x^2-100 x^3+25 x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 50, normalized size = 1.92 \begin {gather*} \frac {25 e^{2 x/5}}{x^2}+50 x+75 x^2-100 x^3+25 x^4-5 e^{x/5} \left (-20-\frac {5}{x}+10 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*x^3 + 150*x^4 - 300*x^5 + 100*x^6 + E^((2*x)/5)*(-50 + 10*x) + E^(x/5)*(-25*x + 5*x^2 - 30*x^3 -
10*x^4))/x^3,x]

[Out]

(25*E^((2*x)/5))/x^2 + 50*x + 75*x^2 - 100*x^3 + 25*x^4 - 5*E^(x/5)*(-20 - 5/x + 10*x)

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fricas [B]  time = 0.77, size = 48, normalized size = 1.85 \begin {gather*} \frac {25 \, {\left (x^{6} - 4 \, x^{5} + 3 \, x^{4} + 2 \, x^{3} - {\left (2 \, x^{3} - 4 \, x^{2} - x\right )} e^{\left (\frac {1}{5} \, x\right )} + e^{\left (\frac {2}{5} \, x\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-50)*exp(1/5*x)^2+(-10*x^4-30*x^3+5*x^2-25*x)*exp(1/5*x)+100*x^6-300*x^5+150*x^4+50*x^3)/x^3,x
, algorithm="fricas")

[Out]

25*(x^6 - 4*x^5 + 3*x^4 + 2*x^3 - (2*x^3 - 4*x^2 - x)*e^(1/5*x) + e^(2/5*x))/x^2

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giac [B]  time = 0.25, size = 52, normalized size = 2.00 \begin {gather*} \frac {25 \, {\left (x^{6} - 4 \, x^{5} + 3 \, x^{4} - 2 \, x^{3} e^{\left (\frac {1}{5} \, x\right )} + 2 \, x^{3} + 4 \, x^{2} e^{\left (\frac {1}{5} \, x\right )} + x e^{\left (\frac {1}{5} \, x\right )} + e^{\left (\frac {2}{5} \, x\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-50)*exp(1/5*x)^2+(-10*x^4-30*x^3+5*x^2-25*x)*exp(1/5*x)+100*x^6-300*x^5+150*x^4+50*x^3)/x^3,x
, algorithm="giac")

[Out]

25*(x^6 - 4*x^5 + 3*x^4 - 2*x^3*e^(1/5*x) + 2*x^3 + 4*x^2*e^(1/5*x) + x*e^(1/5*x) + e^(2/5*x))/x^2

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maple [B]  time = 0.12, size = 48, normalized size = 1.85

method	result	size
risch	\(25 x^{4}-100 x^{3}+75 x^{2}+50 x +\frac {25 \,{\mathrm e}^{\frac {2 x}{5}}}{x^{2}}-\frac {25 \left (2 x^{2}-4 x -1\right ) {\mathrm e}^{\frac {x}{5}}}{x}\)	\(48\)
derivativedivides	\(25 x^{4}-100 x^{3}+75 x^{2}+50 x +\frac {25 \,{\mathrm e}^{\frac {2 x}{5}}}{x^{2}}+\frac {25 \,{\mathrm e}^{\frac {x}{5}}}{x}-50 x \,{\mathrm e}^{\frac {x}{5}}+100 \,{\mathrm e}^{\frac {x}{5}}\)	\(53\)
default	\(25 x^{4}-100 x^{3}+75 x^{2}+50 x +\frac {25 \,{\mathrm e}^{\frac {2 x}{5}}}{x^{2}}+\frac {25 \,{\mathrm e}^{\frac {x}{5}}}{x}-50 x \,{\mathrm e}^{\frac {x}{5}}+100 \,{\mathrm e}^{\frac {x}{5}}\)	\(53\)
norman	\(\frac {50 x^{3}+75 x^{4}-100 x^{5}+25 x^{6}+25 \,{\mathrm e}^{\frac {2 x}{5}}+25 x \,{\mathrm e}^{\frac {x}{5}}+100 \,{\mathrm e}^{\frac {x}{5}} x^{2}-50 \,{\mathrm e}^{\frac {x}{5}} x^{3}}{x^{2}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x-50)*exp(1/5*x)^2+(-10*x^4-30*x^3+5*x^2-25*x)*exp(1/5*x)+100*x^6-300*x^5+150*x^4+50*x^3)/x^3,x,metho
d=_RETURNVERBOSE)

[Out]

25*x^4-100*x^3+75*x^2+50*x+25/x^2*exp(2/5*x)-25*(2*x^2-4*x-1)/x*exp(1/5*x)

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maxima [C]  time = 0.50, size = 61, normalized size = 2.35 \begin {gather*} 25 \, x^{4} - 100 \, x^{3} + 75 \, x^{2} - 50 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} + 50 \, x + 5 \, {\rm Ei}\left (\frac {1}{5} \, x\right ) - 150 \, e^{\left (\frac {1}{5} \, x\right )} - 5 \, \Gamma \left (-1, -\frac {1}{5} \, x\right ) + 4 \, \Gamma \left (-1, -\frac {2}{5} \, x\right ) + 8 \, \Gamma \left (-2, -\frac {2}{5} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-50)*exp(1/5*x)^2+(-10*x^4-30*x^3+5*x^2-25*x)*exp(1/5*x)+100*x^6-300*x^5+150*x^4+50*x^3)/x^3,x
, algorithm="maxima")

[Out]

25*x^4 - 100*x^3 + 75*x^2 - 50*(x - 5)*e^(1/5*x) + 50*x + 5*Ei(1/5*x) - 150*e^(1/5*x) - 5*gamma(-1, -1/5*x) +
4*gamma(-1, -2/5*x) + 8*gamma(-2, -2/5*x)

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mupad [B]  time = 1.09, size = 51, normalized size = 1.96 \begin {gather*} 100\,{\mathrm {e}}^{x/5}+\frac {25\,{\mathrm {e}}^{\frac {2\,x}{5}}+25\,x\,{\mathrm {e}}^{x/5}}{x^2}-x\,\left (50\,{\mathrm {e}}^{x/5}-50\right )+75\,x^2-100\,x^3+25\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x)/5)*(10*x - 50) - exp(x/5)*(25*x - 5*x^2 + 30*x^3 + 10*x^4) + 50*x^3 + 150*x^4 - 300*x^5 + 100*x
^6)/x^3,x)

[Out]

100*exp(x/5) + (25*exp((2*x)/5) + 25*x*exp(x/5))/x^2 - x*(50*exp(x/5) - 50) + 75*x^2 - 100*x^3 + 25*x^4

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sympy [B]  time = 0.15, size = 51, normalized size = 1.96 \begin {gather*} 25 x^{4} - 100 x^{3} + 75 x^{2} + 50 x + \frac {25 x e^{\frac {2 x}{5}} + \left (- 50 x^{4} + 100 x^{3} + 25 x^{2}\right ) e^{\frac {x}{5}}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x-50)*exp(1/5*x)**2+(-10*x**4-30*x**3+5*x**2-25*x)*exp(1/5*x)+100*x**6-300*x**5+150*x**4+50*x**
3)/x**3,x)

[Out]

25*x**4 - 100*x**3 + 75*x**2 + 50*x + (25*x*exp(2*x/5) + (-50*x**4 + 100*x**3 + 25*x**2)*exp(x/5))/x**3

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