3.45.3 \(\int \frac {-50 x+e^{\frac {9+12 x+4 x^2}{x^2}} (-360-240 x+e^5 (72+48 x))}{x^3} \, dx\)

Optimal. Leaf size=28 \[ 4 e^{\frac {(3+2 x)^2}{x^2}} \left (5-e^5\right )+\frac {50}{x} \]

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Rubi [A]  time = 0.14, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {14, 6706} \begin {gather*} 4 \left (5-e^5\right ) e^{\frac {(2 x+3)^2}{x^2}}+\frac {50}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*x + E^((9 + 12*x + 4*x^2)/x^2)*(-360 - 240*x + E^5*(72 + 48*x)))/x^3,x]

[Out]

4*E^((3 + 2*x)^2/x^2)*(5 - E^5) + 50/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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {50}{x^2}+\frac {24 e^{\frac {(3+2 x)^2}{x^2}} \left (-5+e^5\right ) (3+2 x)}{x^3}\right ) \, dx\\ &=\frac {50}{x}-\left (24 \left (5-e^5\right )\right ) \int \frac {e^{\frac {(3+2 x)^2}{x^2}} (3+2 x)}{x^3} \, dx\\ &=4 e^{\frac {(3+2 x)^2}{x^2}} \left (5-e^5\right )+\frac {50}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 27, normalized size = 0.96 \begin {gather*} -4 e^{4+\frac {9}{x^2}+\frac {12}{x}} \left (-5+e^5\right )+\frac {50}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x + E^((9 + 12*x + 4*x^2)/x^2)*(-360 - 240*x + E^5*(72 + 48*x)))/x^3,x]

[Out]

-4*E^(4 + 9/x^2 + 12/x)*(-5 + E^5) + 50/x

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fricas [A]  time = 0.69, size = 32, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (2 \, {\left (x e^{5} - 5 \, x\right )} e^{\left (\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )} - 25\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x, algorithm="fricas")

[Out]

-2*(2*(x*e^5 - 5*x)*e^((4*x^2 + 12*x + 9)/x^2) - 25)/x

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giac [B]  time = 0.13, size = 92, normalized size = 3.29 \begin {gather*} \frac {2 \, {\left (12 \, x e^{\left (\frac {3 \, {\left (3 \, x^{2} + 4 \, x + 3\right )}}{x^{2}}\right )} + 25 \, e^{\left (\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )} + 18 \, e^{\left (\frac {3 \, {\left (3 \, x^{2} + 4 \, x + 3\right )}}{x^{2}}\right )}\right )} e^{\left (-\frac {4 \, x^{2} + 12 \, x + 9}{x^{2}}\right )}}{x} + 20 \, e^{\left (\frac {12}{x} + \frac {9}{x^{2}} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x, algorithm="giac")

[Out]

2*(12*x*e^(3*(3*x^2 + 4*x + 3)/x^2) + 25*e^((4*x^2 + 12*x + 9)/x^2) + 18*e^(3*(3*x^2 + 4*x + 3)/x^2))*e^(-(4*x
^2 + 12*x + 9)/x^2)/x + 20*e^(12/x + 9/x^2 + 4)

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maple [A]  time = 0.05, size = 34, normalized size = 1.21

method	result	size
norman	\(\frac {\left (-4 \,{\mathrm e}^{5}+20\right ) x^{2} {\mathrm e}^{\frac {4 x^{2}+12 x +9}{x^{2}}}+50 x}{x^{2}}\)	\(34\)
derivativedivides	\(\frac {50}{x}+20 \,{\mathrm e}^{4+\frac {9}{x^{2}}+\frac {12}{x}}-4 \,{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}+9}\)	\(37\)
default	\(\frac {50}{x}+20 \,{\mathrm e}^{4+\frac {9}{x^{2}}+\frac {12}{x}}-4 \,{\mathrm e}^{\frac {9}{x^{2}}+\frac {12}{x}+9}\)	\(37\)
risch	\(\frac {50}{x}-4 \,{\mathrm e}^{\frac {\left (2 x +3\right )^{2}}{x^{2}}} {\mathrm e}^{5}+20 \,{\mathrm e}^{\frac {\left (2 x +3\right )^{2}}{x^{2}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

((-4*exp(5)+20)*x^2*exp((4*x^2+12*x+9)/x^2)+50*x)/x^2

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maxima [A]  time = 0.41, size = 27, normalized size = 0.96 \begin {gather*} -4 \, {\left (e^{9} - 5 \, e^{4}\right )} e^{\left (\frac {12}{x} + \frac {9}{x^{2}}\right )} + \frac {50}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x^2+12*x+9)/x^2)-50*x)/x^3,x, algorithm="maxima")

[Out]

-4*(e^9 - 5*e^4)*e^(12/x + 9/x^2) + 50/x

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mupad [B]  time = 3.18, size = 27, normalized size = 0.96 \begin {gather*} \frac {50}{x}-{\mathrm {e}}^{\frac {12}{x}+\frac {9}{x^2}+4}\,\left (4\,{\mathrm {e}}^5-20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(50*x + exp((12*x + 4*x^2 + 9)/x^2)*(240*x - exp(5)*(48*x + 72) + 360))/x^3,x)

[Out]

50/x - exp(12/x + 9/x^2 + 4)*(4*exp(5) - 20)

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sympy [A]  time = 0.25, size = 24, normalized size = 0.86 \begin {gather*} \left (20 - 4 e^{5}\right ) e^{\frac {4 x^{2} + 12 x + 9}{x^{2}}} + \frac {50}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((48*x+72)*exp(5)-240*x-360)*exp((4*x**2+12*x+9)/x**2)-50*x)/x**3,x)

[Out]

(20 - 4*exp(5))*exp((4*x**2 + 12*x + 9)/x**2) + 50/x

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