3.14.77 \(\int \frac {x^2+e^{\frac {-48+12 x^2-12 x^3+e^5 (36 x^2+12 x^3)}{x}} (48+12 x^2-24 x^3+e^5 (36 x^2+24 x^3))}{x^2} \, dx\)

Optimal. Leaf size=27 \[ -1+e^{12 \left (-\frac {4}{x}+x-x^2+e^5 x (3+x)\right )}+x \]

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Rubi [A]  time = 0.51, antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 2, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {14, 6706} \begin {gather*} \exp \left (-12 \left (1-e^5\right ) x^2+12 \left (1+3 e^5\right ) x-\frac {48}{x}\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + E^((-48 + 12*x^2 - 12*x^3 + E^5*(36*x^2 + 12*x^3))/x)*(48 + 12*x^2 - 24*x^3 + E^5*(36*x^2 + 24*x^3)
))/x^2,x]

[Out]

E^(-48/x + 12*(1 + 3*E^5)*x - 12*(1 - E^5)*x^2) + 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 (1+\frac {12 \exp \left (-\frac {48}{x}+12 \left (1+3 e^5\right ) x+12 \left (-1+e^5\right ) x^2\right ) \left (4+\left (1+3 e^5\right ) x^2-2 \left (1-e^5\right ) x^3\right )}{x^2}\right ) \, dx\\ &=x+12 \int \frac {\exp \left (-\frac {48}{x}+12 \left (1+3 e^5\right ) x+12 \left (-1+e^5\right ) x^2\right ) \left (4+\left (1+3 e^5\right ) x^2-2 \left (1-e^5\right ) x^3\right )}{x^2} \, dx\\ &=e^{-\frac {48}{x}+12 \left (1+3 e^5\right ) x-12 \left (1-e^5\right ) x^2}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 30, normalized size = 1.11 \begin {gather*} e^{-\frac {48}{x}+12 \left (1+3 e^5\right ) x+12 \left (-1+e^5\right ) x^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^((-48 + 12*x^2 - 12*x^3 + E^5*(36*x^2 + 12*x^3))/x)*(48 + 12*x^2 - 24*x^3 + E^5*(36*x^2 + 2
4*x^3)))/x^2,x]

[Out]

E^(-48/x + 12*(1 + 3*E^5)*x + 12*(-1 + E^5)*x^2) + x

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fricas [A]  time = 0.58, size = 31, normalized size = 1.15 \begin {gather*} x + e^{\left (-\frac {12 \, {\left (x^{3} - x^{2} - {\left (x^{3} + 3 \, x^{2}\right )} e^{5} + 4\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^3+36*x^2)*exp(5)-24*x^3+12*x^2+48)*exp(((12*x^3+36*x^2)*exp(5)-12*x^3+12*x^2-48)/x)+x^2)/x^2
,x, algorithm="fricas")

[Out]

x + e^(-12*(x^3 - x^2 - (x^3 + 3*x^2)*e^5 + 4)/x)

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giac [A]  time = 0.43, size = 31, normalized size = 1.15 \begin {gather*} x + e^{\left (\frac {12 \, {\left (x^{3} e^{5} - x^{3} + 3 \, x^{2} e^{5} + x^{2} - 4\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^3+36*x^2)*exp(5)-24*x^3+12*x^2+48)*exp(((12*x^3+36*x^2)*exp(5)-12*x^3+12*x^2-48)/x)+x^2)/x^2
,x, algorithm="giac")

[Out]

x + e^(12*(x^3*e^5 - x^3 + 3*x^2*e^5 + x^2 - 4)/x)

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maple [A]  time = 0.26, size = 32, normalized size = 1.19

method	result	size
risch	\(x +{\mathrm e}^{\frac {12 x^{3} {\mathrm e}^{5}+36 x^{2} {\mathrm e}^{5}-12 x^{3}+12 x^{2}-48}{x}}\)	\(32\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{\frac {\left (12 x^{3}+36 x^{2}\right ) {\mathrm e}^{5}-12 x^{3}+12 x^{2}-48}{x}}}{x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((24*x^3+36*x^2)*exp(5)-24*x^3+12*x^2+48)*exp(((12*x^3+36*x^2)*exp(5)-12*x^3+12*x^2-48)/x)+x^2)/x^2,x,met
hod=_RETURNVERBOSE)

[Out]

x+exp(12*(x^3*exp(5)+3*x^2*exp(5)-x^3+x^2-4)/x)

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maxima [A]  time = 0.74, size = 29, normalized size = 1.07 \begin {gather*} x + e^{\left (12 \, x^{2} e^{5} - 12 \, x^{2} + 36 \, x e^{5} + 12 \, x - \frac {48}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^3+36*x^2)*exp(5)-24*x^3+12*x^2+48)*exp(((12*x^3+36*x^2)*exp(5)-12*x^3+12*x^2-48)/x)+x^2)/x^2
,x, algorithm="maxima")

[Out]

x + e^(12*x^2*e^5 - 12*x^2 + 36*x*e^5 + 12*x - 48/x)

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mupad [B]  time = 1.19, size = 33, normalized size = 1.22 \begin {gather*} x+{\mathrm {e}}^{12\,x^2\,{\mathrm {e}}^5}\,{\mathrm {e}}^{12\,x}\,{\mathrm {e}}^{-12\,x^2}\,{\mathrm {e}}^{-\frac {48}{x}}\,{\mathrm {e}}^{36\,x\,{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(5)*(36*x^2 + 12*x^3) + 12*x^2 - 12*x^3 - 48)/x)*(exp(5)*(36*x^2 + 24*x^3) + 12*x^2 - 24*x^3 + 48
) + x^2)/x^2,x)

[Out]

x + exp(12*x^2*exp(5))*exp(12*x)*exp(-12*x^2)*exp(-48/x)*exp(36*x*exp(5))

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sympy [A]  time = 0.22, size = 29, normalized size = 1.07 \begin {gather*} x + e^{\frac {- 12 x^{3} + 12 x^{2} + \left (12 x^{3} + 36 x^{2}\right ) e^{5} - 48}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x**3+36*x**2)*exp(5)-24*x**3+12*x**2+48)*exp(((12*x**3+36*x**2)*exp(5)-12*x**3+12*x**2-48)/x)+
x**2)/x**2,x)

[Out]

x + exp((-12*x**3 + 12*x**2 + (12*x**3 + 36*x**2)*exp(5) - 48)/x)

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