3.90.42 \(\int \frac {9 e^5+x^2+e^x (225-225 x-9 x^2)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ x+\frac {9 \left (-e^5+x-e^x (25+x)\right )}{x} \]

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Rubi [A]  time = 0.08, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {14, 2199, 2194, 2177, 2178} \begin {gather*} x-9 e^x-\frac {225 e^x}{x}-\frac {9 e^5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*E^5 + x^2 + E^x*(225 - 225*x - 9*x^2))/x^2,x]

[Out]

-9*E^x - (9*E^5)/x - (225*E^x)/x + 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 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 {9 e^5+x^2}{x^2}-\frac {9 e^x \left (-25+25 x+x^2\right )}{x^2}\right ) \, dx\\ &=-\left (9 \int \frac {e^x \left (-25+25 x+x^2\right )}{x^2} \, dx\right )+\int \frac {9 e^5+x^2}{x^2} \, dx\\ &=-\left (9 \int \left (e^x-\frac {25 e^x}{x^2}+\frac {25 e^x}{x}\right ) \, dx\right )+\int \left (1+\frac {9 e^5}{x^2}\right ) \, dx\\ &=-\frac {9 e^5}{x}+x-9 \int e^x \, dx+225 \int \frac {e^x}{x^2} \, dx-225 \int \frac {e^x}{x} \, dx\\ &=-9 e^x-\frac {9 e^5}{x}-\frac {225 e^x}{x}+x-225 \text {Ei}(x)+225 \int \frac {e^x}{x} \, dx\\ &=-9 e^x-\frac {9 e^5}{x}-\frac {225 e^x}{x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 1.05 \begin {gather*} -9 e^x-\frac {9 e^5}{x}-\frac {225 e^x}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^5 + x^2 + E^x*(225 - 225*x - 9*x^2))/x^2,x]

[Out]

-9*E^x - (9*E^5)/x - (225*E^x)/x + x

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fricas [A]  time = 0.56, size = 19, normalized size = 0.86 \begin {gather*} \frac {x^{2} - 9 \, {\left (x + 25\right )} e^{x} - 9 \, e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-225*x+225)*exp(x)+9*exp(5)+x^2)/x^2,x, algorithm="fricas")

[Out]

(x^2 - 9*(x + 25)*e^x - 9*e^5)/x

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giac [A]  time = 1.16, size = 21, normalized size = 0.95 \begin {gather*} \frac {x^{2} - 9 \, x e^{x} - 9 \, e^{5} - 225 \, e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-225*x+225)*exp(x)+9*exp(5)+x^2)/x^2,x, algorithm="giac")

[Out]

(x^2 - 9*x*e^x - 9*e^5 - 225*e^x)/x

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maple [A]  time = 0.08, size = 20, normalized size = 0.91

method	result	size
risch	\(x -\frac {9 \,{\mathrm e}^{5}}{x}-\frac {9 \left (x +25\right ) {\mathrm e}^{x}}{x}\)	\(20\)
default	\(x -\frac {9 \,{\mathrm e}^{5}}{x}-\frac {225 \,{\mathrm e}^{x}}{x}-9 \,{\mathrm e}^{x}\)	\(21\)
norman	\(\frac {x^{2}-9 \,{\mathrm e}^{x} x -9 \,{\mathrm e}^{5}-225 \,{\mathrm e}^{x}}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x^2-225*x+225)*exp(x)+9*exp(5)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x-9*exp(5)/x-9*(x+25)/x*exp(x)

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maxima [C]  time = 0.37, size = 24, normalized size = 1.09 \begin {gather*} x - \frac {9 \, e^{5}}{x} - 225 \, {\rm Ei}\relax (x) - 9 \, e^{x} + 225 \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-225*x+225)*exp(x)+9*exp(5)+x^2)/x^2,x, algorithm="maxima")

[Out]

x - 9*e^5/x - 225*Ei(x) - 9*e^x + 225*gamma(-1, -x)

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mupad [B]  time = 6.93, size = 20, normalized size = 0.91 \begin {gather*} x-9\,{\mathrm {e}}^x-\frac {9\,{\mathrm {e}}^5+225\,{\mathrm {e}}^x}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(5) - exp(x)*(225*x + 9*x^2 - 225) + x^2)/x^2,x)

[Out]

x - 9*exp(x) - (9*exp(5) + 225*exp(x))/x

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sympy [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} x + \frac {\left (- 9 x - 225\right ) e^{x}}{x} - \frac {9 e^{5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x**2-225*x+225)*exp(x)+9*exp(5)+x**2)/x**2,x)

[Out]

x + (-9*x - 225)*exp(x)/x - 9*exp(5)/x

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