3.25.38 \(\int \frac {-1-x+e^{9+6 x^2+x^4} (x+2 x^2+12 x^3+12 x^4+4 x^5+4 x^6) \log (4)}{x} \, dx\)

Optimal. Leaf size=28 \[ -1-x+e^{\left (3+x^2\right )^2} \left (x+x^2\right ) \log (4)-\log (5 x) \]

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Rubi [A]  time = 0.14, antiderivative size = 49, normalized size of antiderivative = 1.75, number of steps used = 5, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14, 43, 2288} \begin {gather*} \frac {e^{x^4+6 x^2+9} \left (x^5+x^4+3 x^3+3 x^2\right ) \log (4)}{x^3+3 x}-x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - x + E^(9 + 6*x^2 + x^4)*(x + 2*x^2 + 12*x^3 + 12*x^4 + 4*x^5 + 4*x^6)*Log[4])/x,x]

[Out]

-x + (E^(9 + 6*x^2 + x^4)*(3*x^2 + 3*x^3 + x^4 + x^5)*Log[4])/(3*x + x^3) - Log[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 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 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 (\frac {-1-x}{x}+e^{9+6 x^2+x^4} \left (1+2 x+12 x^2+12 x^3+4 x^4+4 x^5\right ) \log (4)\right ) \, dx\\ &=\log (4) \int e^{9+6 x^2+x^4} \left (1+2 x+12 x^2+12 x^3+4 x^4+4 x^5\right ) \, dx+\int \frac {-1-x}{x} \, dx\\ &=\frac {e^{9+6 x^2+x^4} \left (3 x^2+3 x^3+x^4+x^5\right ) \log (4)}{3 x+x^3}+\int \left (-1-\frac {1}{x}\right ) \, dx\\ &=-x+\frac {e^{9+6 x^2+x^4} \left (3 x^2+3 x^3+x^4+x^5\right ) \log (4)}{3 x+x^3}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 24, normalized size = 0.86 \begin {gather*} x \left (-1+e^{\left (3+x^2\right )^2} (1+x) \log (4)\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - x + E^(9 + 6*x^2 + x^4)*(x + 2*x^2 + 12*x^3 + 12*x^4 + 4*x^5 + 4*x^6)*Log[4])/x,x]

[Out]

x*(-1 + E^(3 + x^2)^2*(1 + x)*Log[4]) - Log[x]

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fricas [A]  time = 0.63, size = 28, normalized size = 1.00 \begin {gather*} 2 \, {\left (x^{2} + x\right )} e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \relax (2) - x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*x^6+4*x^5+12*x^4+12*x^3+2*x^2+x)*log(2)*exp(x^4+6*x^2+9)-x-1)/x,x, algorithm="fricas")

[Out]

2*(x^2 + x)*e^(x^4 + 6*x^2 + 9)*log(2) - x - log(x)

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giac [A]  time = 0.22, size = 42, normalized size = 1.50 \begin {gather*} 2 \, x^{2} e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \relax (2) + 2 \, x e^{\left (x^{4} + 6 \, x^{2} + 9\right )} \log \relax (2) - x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*x^6+4*x^5+12*x^4+12*x^3+2*x^2+x)*log(2)*exp(x^4+6*x^2+9)-x-1)/x,x, algorithm="giac")

[Out]

2*x^2*e^(x^4 + 6*x^2 + 9)*log(2) + 2*x*e^(x^4 + 6*x^2 + 9)*log(2) - x - log(x)

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maple [A]  time = 0.08, size = 31, normalized size = 1.11

method	result	size
risch	\(-x -\ln \relax (x )+\left (2 x^{2} \ln \relax (2)+2 x \ln \relax (2)\right ) {\mathrm e}^{\left (x^{2}+3\right )^{2}}\)	\(31\)
norman	\(-x +2 \ln \relax (2) {\mathrm e}^{x^{4}+6 x^{2}+9} x +2 \ln \relax (2) {\mathrm e}^{x^{4}+6 x^{2}+9} x^{2}-\ln \relax (x )\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(4*x^6+4*x^5+12*x^4+12*x^3+2*x^2+x)*ln(2)*exp(x^4+6*x^2+9)-x-1)/x,x,method=_RETURNVERBOSE)

[Out]

-x-ln(x)+(2*x^2*ln(2)+2*x*ln(2))*exp((x^2+3)^2)

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maxima [A]  time = 0.65, size = 35, normalized size = 1.25 \begin {gather*} 2 \, {\left (x^{2} e^{9} \log \relax (2) + x e^{9} \log \relax (2)\right )} e^{\left (x^{4} + 6 \, x^{2}\right )} - x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*x^6+4*x^5+12*x^4+12*x^3+2*x^2+x)*log(2)*exp(x^4+6*x^2+9)-x-1)/x,x, algorithm="maxima")

[Out]

2*(x^2*e^9*log(2) + x*e^9*log(2))*e^(x^4 + 6*x^2) - x - log(x)

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mupad [B]  time = 0.10, size = 42, normalized size = 1.50 \begin {gather*} 2\,x\,{\mathrm {e}}^{x^4+6\,x^2+9}\,\ln \relax (2)-\ln \relax (x)-x+2\,x^2\,{\mathrm {e}}^{x^4+6\,x^2+9}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 2*exp(6*x^2 + x^4 + 9)*log(2)*(x + 2*x^2 + 12*x^3 + 12*x^4 + 4*x^5 + 4*x^6) + 1)/x,x)

[Out]

2*x*exp(6*x^2 + x^4 + 9)*log(2) - log(x) - x + 2*x^2*exp(6*x^2 + x^4 + 9)*log(2)

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sympy [A]  time = 0.17, size = 31, normalized size = 1.11 \begin {gather*} - x + \left (2 x^{2} \log {\relax (2 )} + 2 x \log {\relax (2 )}\right ) e^{x^{4} + 6 x^{2} + 9} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(4*x**6+4*x**5+12*x**4+12*x**3+2*x**2+x)*ln(2)*exp(x**4+6*x**2+9)-x-1)/x,x)

[Out]

-x + (2*x**2*log(2) + 2*x*log(2))*exp(x**4 + 6*x**2 + 9) - log(x)

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