3.4.82 \(\int \frac {-x+e^{192 x-48 x^3+3 x^5+(128-32 x^2+2 x^4) \log (x)} (128+192 x-32 x^2-144 x^3+2 x^4+15 x^5+(-64 x^2+8 x^4) \log (x))}{x} \, dx\)

Optimal. Leaf size=23 \[ e^{2 \left (-8+x^2\right )^2 \left (\frac {3 x}{2}+\log (x)\right )}-x \]

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Rubi [B]  time = 0.50, antiderivative size = 76, normalized size of antiderivative = 3.30, number of steps used = 3, number of rules used = 2, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {14, 2288} \begin {gather*} -\frac {e^{3 x \left (8-x^2\right )^2} \left (8-x^2\right ) \left (8 x-5 x^3\right ) x^{2 x^4-32 x^2+127}}{4 x^2 \left (8-x^2\right )-\left (8-x^2\right )^2}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + E^(192*x - 48*x^3 + 3*x^5 + (128 - 32*x^2 + 2*x^4)*Log[x])*(128 + 192*x - 32*x^2 - 144*x^3 + 2*x^4 +
15*x^5 + (-64*x^2 + 8*x^4)*Log[x]))/x,x]

[Out]

-x - (E^(3*x*(8 - x^2)^2)*x^(127 - 32*x^2 + 2*x^4)*(8 - x^2)*(8*x - 5*x^3))/(4*x^2*(8 - x^2) - (8 - x^2)^2)

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+e^{3 x \left (-8+x^2\right )^2} x^{127-32 x^2+2 x^4} \left (-8+x^2\right ) \left (-16-24 x+2 x^2+15 x^3+8 x^2 \log (x)\right )\right ) \, dx\\ &=-x+\int e^{3 x \left (-8+x^2\right )^2} x^{127-32 x^2+2 x^4} \left (-8+x^2\right ) \left (-16-24 x+2 x^2+15 x^3+8 x^2 \log (x)\right ) \, dx\\ &=-x-\frac {e^{3 x \left (8-x^2\right )^2} x^{127-32 x^2+2 x^4} \left (8-x^2\right ) \left (8 x-5 x^3\right )}{4 x^2 \left (8-x^2\right )-\left (-8+x^2\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 28, normalized size = 1.22 \begin {gather*} -x+e^{3 x \left (-8+x^2\right )^2} x^{2 \left (-8+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + E^(192*x - 48*x^3 + 3*x^5 + (128 - 32*x^2 + 2*x^4)*Log[x])*(128 + 192*x - 32*x^2 - 144*x^3 + 2
*x^4 + 15*x^5 + (-64*x^2 + 8*x^4)*Log[x]))/x,x]

[Out]

-x + E^(3*x*(-8 + x^2)^2)*x^(2*(-8 + x^2)^2)

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fricas [A]  time = 0.91, size = 33, normalized size = 1.43 \begin {gather*} -x + e^{\left (3 \, x^{5} - 48 \, x^{3} + 2 \, {\left (x^{4} - 16 \, x^{2} + 64\right )} \log \relax (x) + 192 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-64*x^2)*log(x)+15*x^5+2*x^4-144*x^3-32*x^2+192*x+128)*exp(1/2*(2*x^4-32*x^2+128)*log(x)+3/2
*x^5-24*x^3+96*x)^2-x)/x,x, algorithm="fricas")

[Out]

-x + e^(3*x^5 - 48*x^3 + 2*(x^4 - 16*x^2 + 64)*log(x) + 192*x)

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giac [A]  time = 0.81, size = 37, normalized size = 1.61 \begin {gather*} -x + e^{\left (3 \, x^{5} + 2 \, x^{4} \log \relax (x) - 48 \, x^{3} - 32 \, x^{2} \log \relax (x) + 192 \, x + 128 \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-64*x^2)*log(x)+15*x^5+2*x^4-144*x^3-32*x^2+192*x+128)*exp(1/2*(2*x^4-32*x^2+128)*log(x)+3/2
*x^5-24*x^3+96*x)^2-x)/x,x, algorithm="giac")

[Out]

-x + e^(3*x^5 + 2*x^4*log(x) - 48*x^3 - 32*x^2*log(x) + 192*x + 128*log(x))

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maple [A]  time = 0.08, size = 28, normalized size = 1.22

method	result	size
risch	\(x^{2 \left (x^{2}-8\right )^{2}} {\mathrm e}^{3 x \left (x^{2}-8\right )^{2}}-x\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^4-64*x^2)*ln(x)+15*x^5+2*x^4-144*x^3-32*x^2+192*x+128)*exp(1/2*(2*x^4-32*x^2+128)*ln(x)+3/2*x^5-24*
x^3+96*x)^2-x)/x,x,method=_RETURNVERBOSE)

[Out]

(x^((x^2-8)^2))^2*exp(3*x*(x^2-8)^2)-x

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maxima [A]  time = 0.70, size = 37, normalized size = 1.61 \begin {gather*} x^{128} e^{\left (3 \, x^{5} + 2 \, x^{4} \log \relax (x) - 48 \, x^{3} - 32 \, x^{2} \log \relax (x) + 192 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-64*x^2)*log(x)+15*x^5+2*x^4-144*x^3-32*x^2+192*x+128)*exp(1/2*(2*x^4-32*x^2+128)*log(x)+3/2
*x^5-24*x^3+96*x)^2-x)/x,x, algorithm="maxima")

[Out]

x^128*e^(3*x^5 + 2*x^4*log(x) - 48*x^3 - 32*x^2*log(x) + 192*x) - x

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mupad [B]  time = 0.50, size = 40, normalized size = 1.74 \begin {gather*} \frac {x^{2\,x^4}\,x^{128}\,{\mathrm {e}}^{192\,x}\,{\mathrm {e}}^{3\,x^5}\,{\mathrm {e}}^{-48\,x^3}}{x^{32\,x^2}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - exp(192*x + log(x)*(2*x^4 - 32*x^2 + 128) - 48*x^3 + 3*x^5)*(192*x - log(x)*(64*x^2 - 8*x^4) - 32*x^
2 - 144*x^3 + 2*x^4 + 15*x^5 + 128))/x,x)

[Out]

(x^(2*x^4)*x^128*exp(192*x)*exp(3*x^5)*exp(-48*x^3))/x^(32*x^2) - x

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sympy [A]  time = 0.41, size = 31, normalized size = 1.35 \begin {gather*} - x + e^{3 x^{5} - 48 x^{3} + 192 x + 2 \left (x^{4} - 16 x^{2} + 64\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**4-64*x**2)*ln(x)+15*x**5+2*x**4-144*x**3-32*x**2+192*x+128)*exp(1/2*(2*x**4-32*x**2+128)*ln(
x)+3/2*x**5-24*x**3+96*x)**2-x)/x,x)

[Out]

-x + exp(3*x**5 - 48*x**3 + 192*x + 2*(x**4 - 16*x**2 + 64)*log(x))

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