3.18.76 \(\int \frac {-8 e^{3+x} x^2+2 e^3 x^2 \log (3)+e^{25+2 x} (e^{3+x} (2-6 x)+e^3 x^2 \log (3))}{x^2} \, dx\)

Optimal. Leaf size=30 \[ e^3 \left (\frac {1}{2} e^{25+2 x}+2 x\right ) \left (-\frac {4 e^x}{x}+\log (3)\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 40, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {14, 2194, 2197} \begin {gather*} -8 e^{x+3}-\frac {2 e^{3 x+28}}{x}+e^3 x \log (9)+\frac {1}{2} e^{2 x+28} \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*E^(3 + x)*x^2 + 2*E^3*x^2*Log[3] + E^(25 + 2*x)*(E^(3 + x)*(2 - 6*x) + E^3*x^2*Log[3]))/x^2,x]

[Out]

-8*E^(3 + x) - (2*E^(28 + 3*x))/x + (E^(28 + 2*x)*Log[3])/2 + E^3*x*Log[9]

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 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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-8 e^{3+x}-\frac {2 e^{28+3 x} (-1+3 x)}{x^2}+e^{28+2 x} \log (3)+e^3 \log (9)\right ) \, dx\\ &=e^3 x \log (9)-2 \int \frac {e^{28+3 x} (-1+3 x)}{x^2} \, dx-8 \int e^{3+x} \, dx+\log (3) \int e^{28+2 x} \, dx\\ &=-8 e^{3+x}-\frac {2 e^{28+3 x}}{x}+\frac {1}{2} e^{28+2 x} \log (3)+e^3 x \log (9)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 39, normalized size = 1.30 \begin {gather*} e^3 \left (-8 e^x-\frac {2 e^{25+3 x}}{x}+\frac {1}{2} e^{25+2 x} \log (3)+x \log (9)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^(3 + x)*x^2 + 2*E^3*x^2*Log[3] + E^(25 + 2*x)*(E^(3 + x)*(2 - 6*x) + E^3*x^2*Log[3]))/x^2,x]

[Out]

E^3*(-8*E^x - (2*E^(25 + 3*x))/x + (E^(25 + 2*x)*Log[3])/2 + x*Log[9])

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fricas [A]  time = 0.95, size = 40, normalized size = 1.33 \begin {gather*} \frac {4 \, x^{2} e^{3} \log \relax (3) + x e^{\left (2 \, x + 28\right )} \log \relax (3) - 16 \, x e^{\left (x + 3\right )} - 4 \, e^{\left (3 \, x + 28\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+2)*exp(3)*exp(x)+x^2*exp(3)*log(3))*exp(2*x+25)-8*x^2*exp(3)*exp(x)+2*x^2*exp(3)*log(3))/x^2
,x, algorithm="fricas")

[Out]

1/2*(4*x^2*e^3*log(3) + x*e^(2*x + 28)*log(3) - 16*x*e^(x + 3) - 4*e^(3*x + 28))/x

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giac [A]  time = 0.15, size = 40, normalized size = 1.33 \begin {gather*} \frac {4 \, x^{2} e^{3} \log \relax (3) + x e^{\left (2 \, x + 28\right )} \log \relax (3) - 16 \, x e^{\left (x + 3\right )} - 4 \, e^{\left (3 \, x + 28\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+2)*exp(3)*exp(x)+x^2*exp(3)*log(3))*exp(2*x+25)-8*x^2*exp(3)*exp(x)+2*x^2*exp(3)*log(3))/x^2
,x, algorithm="giac")

[Out]

1/2*(4*x^2*e^3*log(3) + x*e^(2*x + 28)*log(3) - 16*x*e^(x + 3) - 4*e^(3*x + 28))/x

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maple [A]  time = 0.08, size = 36, normalized size = 1.20

method	result	size
risch	\(2 x \,{\mathrm e}^{3} \ln \relax (3)-\frac {2 \,{\mathrm e}^{28+3 x}}{x}+\frac {\ln \relax (3) {\mathrm e}^{2 x +28}}{2}-8 \,{\mathrm e}^{3+x}\)	\(36\)
norman	\(\frac {-8 x \,{\mathrm e}^{3} {\mathrm e}^{x}+2 x^{2} {\mathrm e}^{3} \ln \relax (3)-2 \,{\mathrm e}^{3 x} {\mathrm e}^{25} {\mathrm e}^{3}+\frac {{\mathrm e}^{25} {\mathrm e}^{3} \ln \relax (3) x \,{\mathrm e}^{2 x}}{2}}{x}\)	\(45\)
default	\(\frac {{\mathrm e}^{2 x} {\mathrm e}^{25} {\mathrm e}^{3} \ln \relax (3)}{2}+2 x \,{\mathrm e}^{3} \ln \relax (3)-8 \,{\mathrm e}^{x} {\mathrm e}^{3}+2 \,{\mathrm e}^{25} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{3 x}}{x}-3 \expIntegralEi \left (1, -3 x \right )\right )+6 \,{\mathrm e}^{25} {\mathrm e}^{3} \expIntegralEi \left (1, -3 x \right )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x+2)*exp(3)*exp(x)+x^2*exp(3)*ln(3))*exp(2*x+25)-8*x^2*exp(3)*exp(x)+2*x^2*exp(3)*ln(3))/x^2,x,metho
d=_RETURNVERBOSE)

[Out]

2*x*exp(3)*ln(3)-2/x*exp(28+3*x)+1/2*ln(3)*exp(2*x+28)-8*exp(3+x)

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maxima [C]  time = 0.49, size = 41, normalized size = 1.37 \begin {gather*} 2 \, x e^{3} \log \relax (3) - 6 \, {\rm Ei}\left (3 \, x\right ) e^{28} + 6 \, e^{28} \Gamma \left (-1, -3 \, x\right ) + \frac {1}{2} \, e^{\left (2 \, x + 28\right )} \log \relax (3) - 8 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+2)*exp(3)*exp(x)+x^2*exp(3)*log(3))*exp(2*x+25)-8*x^2*exp(3)*exp(x)+2*x^2*exp(3)*log(3))/x^2
,x, algorithm="maxima")

[Out]

2*x*e^3*log(3) - 6*Ei(3*x)*e^28 + 6*e^28*gamma(-1, -3*x) + 1/2*e^(2*x + 28)*log(3) - 8*e^(x + 3)

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mupad [B]  time = 0.10, size = 27, normalized size = 0.90 \begin {gather*} -\frac {{\mathrm {e}}^3\,\left (4\,x+{\mathrm {e}}^{2\,x+25}\right )\,\left (4\,{\mathrm {e}}^x-x\,\ln \relax (3)\right )}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x + 25)*(x^2*exp(3)*log(3) - exp(3)*exp(x)*(6*x - 2)) + 2*x^2*exp(3)*log(3) - 8*x^2*exp(3)*exp(x))/
x^2,x)

[Out]

-(exp(3)*(4*x + exp(2*x + 25))*(4*exp(x) - x*log(3)))/(2*x)

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sympy [A]  time = 0.16, size = 46, normalized size = 1.53 \begin {gather*} 2 x e^{3} \log {\relax (3 )} + \frac {x e^{28} e^{2 x} \log {\relax (3 )} - 16 x e^{3} e^{x} - 4 e^{28} e^{3 x}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x+2)*exp(3)*exp(x)+x**2*exp(3)*ln(3))*exp(2*x+25)-8*x**2*exp(3)*exp(x)+2*x**2*exp(3)*ln(3))/x*
*2,x)

[Out]

2*x*exp(3)*log(3) + (x*exp(28)*exp(2*x)*log(3) - 16*x*exp(3)*exp(x) - 4*exp(28)*exp(3*x))/(2*x)

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