3.26.12 \(\int \frac {e^{3/x} (-30 e^3-30 x)+e^{6/x} (-6 e^3-6 x)-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx\)

Optimal. Leaf size=31 \[ \log \left (\frac {2 e^{\frac {1}{32} \left (\left (5+e^{3/x}\right )^2-x\right )}}{e^3+x}\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 5, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6, 1593, 6742, 2209, 43} \begin {gather*} -\frac {x}{32}+\frac {5 e^{3/x}}{16}+\frac {e^{6/x}}{32}-\log \left (x+e^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3/x)*(-30*E^3 - 30*x) + E^(6/x)*(-6*E^3 - 6*x) - 32*x^2 - E^3*x^2 - x^3)/(32*E^3*x^2 + 32*x^3),x]

[Out]

(5*E^(3/x))/16 + E^(6/x)/32 - x/32 - Log[E^3 + x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )+\left (-32-e^3\right ) x^2-x^3}{32 e^3 x^2+32 x^3} \, dx\\ &=\int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )+\left (-32-e^3\right ) x^2-x^3}{x^2 \left (32 e^3+32 x\right )} \, dx\\ &=\int \left (-\frac {15 e^{3/x}}{16 x^2}-\frac {3 e^{6/x}}{16 x^2}+\frac {-32-e^3-x}{32 \left (e^3+x\right )}\right ) \, dx\\ &=\frac {1}{32} \int \frac {-32-e^3-x}{e^3+x} \, dx-\frac {3}{16} \int \frac {e^{6/x}}{x^2} \, dx-\frac {15}{16} \int \frac {e^{3/x}}{x^2} \, dx\\ &=\frac {5 e^{3/x}}{16}+\frac {e^{6/x}}{32}+\frac {1}{32} \int \left (-1-\frac {32}{e^3+x}\right ) \, dx\\ &=\frac {5 e^{3/x}}{16}+\frac {e^{6/x}}{32}-\frac {x}{32}-\log \left (e^3+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 32, normalized size = 1.03 \begin {gather*} \frac {1}{32} \left (10 e^{3/x}+e^{6/x}-x-32 \log \left (e^3+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3/x)*(-30*E^3 - 30*x) + E^(6/x)*(-6*E^3 - 6*x) - 32*x^2 - E^3*x^2 - x^3)/(32*E^3*x^2 + 32*x^3),x
]

[Out]

(10*E^(3/x) + E^(6/x) - x - 32*Log[E^3 + x])/32

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fricas [A]  time = 0.76, size = 27, normalized size = 0.87 \begin {gather*} -\frac {1}{32} \, x + \frac {1}{32} \, e^{\frac {6}{x}} + \frac {5}{16} \, e^{\frac {3}{x}} - \log \left (x + e^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3)-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3)
,x, algorithm="fricas")

[Out]

-1/32*x + 1/32*e^(6/x) + 5/16*e^(3/x) - log(x + e^3)

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giac [A]  time = 0.16, size = 47, normalized size = 1.52 \begin {gather*} \frac {1}{32} \, x {\left (\frac {e^{\frac {6}{x}}}{x} + \frac {10 \, e^{\frac {3}{x}}}{x} - \frac {32 \, \log \relax (x)}{x} - \frac {32 \, \log \left (\frac {e^{3}}{x} + 1\right )}{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3)-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3)
,x, algorithm="giac")

[Out]

1/32*x*(e^(6/x)/x + 10*e^(3/x)/x - 32*log(x)/x - 32*log(e^3/x + 1)/x - 1)

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maple [A]  time = 0.40, size = 28, normalized size = 0.90

method	result	size
risch	\(-\frac {x}{32}-\ln \left ({\mathrm e}^{3}+x \right )+\frac {{\mathrm e}^{\frac {6}{x}}}{32}+\frac {5 \,{\mathrm e}^{\frac {3}{x}}}{16}\)	\(28\)
norman	\(\frac {-\frac {x^{2}}{32}+\frac {5 x \,{\mathrm e}^{\frac {3}{x}}}{16}+\frac {{\mathrm e}^{\frac {6}{x}} x}{32}}{x}-\ln \left ({\mathrm e}^{3}+x \right )\)	\(39\)
derivativedivides	\(-\frac {x}{32}-\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )+\ln \left (\frac {3}{x}\right )-\frac {15 \,{\mathrm e}^{-3} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )}{16}-\frac {3 \,{\mathrm e}^{-3} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )}{16}+\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-3} {\mathrm e}^{\frac {3}{x}}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )\right )}{16}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{-3} {\mathrm e}^{\frac {6}{x}}}{2}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )\right )}{16}\)	\(170\)
default	\(-\frac {x}{32}-\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )+\ln \left (\frac {3}{x}\right )-\frac {15 \,{\mathrm e}^{-3} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )}{16}-\frac {3 \,{\mathrm e}^{-3} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )}{16}+\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-3} {\mathrm e}^{\frac {3}{x}}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )\right )}{16}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{-3} {\mathrm e}^{\frac {6}{x}}}{2}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \expIntegralEi \left (1, -\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )\right )}{16}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3)-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3),x,met
hod=_RETURNVERBOSE)

[Out]

-1/32*x-ln(exp(3)+x)+1/32*exp(6/x)+5/16*exp(3/x)

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maxima [A]  time = 0.42, size = 27, normalized size = 0.87 \begin {gather*} -\frac {1}{32} \, x + \frac {1}{32} \, e^{\frac {6}{x}} + \frac {5}{16} \, e^{\frac {3}{x}} - \log \left (x + e^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3)-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3)
,x, algorithm="maxima")

[Out]

-1/32*x + 1/32*e^(6/x) + 5/16*e^(3/x) - log(x + e^3)

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mupad [B]  time = 1.51, size = 27, normalized size = 0.87 \begin {gather*} \frac {5\,{\mathrm {e}}^{3/x}}{16}-\frac {x}{32}+\frac {{\mathrm {e}}^{6/x}}{32}-\ln \left (x+{\mathrm {e}}^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*exp(3) + 32*x^2 + x^3 + exp(6/x)*(6*x + 6*exp(3)) + exp(3/x)*(30*x + 30*exp(3)))/(32*x^2*exp(3) + 32
*x^3),x)

[Out]

(5*exp(3/x))/16 - x/32 + exp(6/x)/32 - log(x + exp(3))

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sympy [A]  time = 0.20, size = 24, normalized size = 0.77 \begin {gather*} - \frac {x}{32} + \frac {e^{\frac {6}{x}}}{32} + \frac {5 e^{\frac {3}{x}}}{16} - \log {\left (x + e^{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*exp(3)-6*x)*exp(3/x)**2+(-30*exp(3)-30*x)*exp(3/x)-x**2*exp(3)-x**3-32*x**2)/(32*x**2*exp(3)+32
*x**3),x)

[Out]

-x/32 + exp(6/x)/32 + 5*exp(3/x)/16 - log(x + exp(3))

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