3.34.37 \(\int \frac {-1-4 x+e^{-3+x+(4-x+x^2) \log (4)} (x+(-x+2 x^2) \log (4))}{x} \, dx\)

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

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Rubi [A]  time = 0.12, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 43, 2287, 12, 2236} \begin {gather*} \frac {2^{2 x^2+7} \log (16) e^{x (1-\log (4))-3}}{\log (4)}-4 x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x + (2^(7 + 2*x^2)*E^(-3 + x*(1 - Log[4]))*Log[16])/Log[4] - Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1-4 x}{x}-4^{4-x+x^2} e^{-3+x} (-1+\log (4)-x \log (16))\right ) \, dx\\ &=\int \frac {-1-4 x}{x} \, dx-\int 4^{4-x+x^2} e^{-3+x} (-1+\log (4)-x \log (16)) \, dx\\ &=\int \left (-4-\frac {1}{x}\right ) \, dx-\int 256 e^{-3+x (1-\log (4))+x^2 \log (4)} (-1+\log (4)-x \log (16)) \, dx\\ &=-4 x-\log (x)-256 \int e^{-3+x (1-\log (4))+x^2 \log (4)} (-1+\log (4)-x \log (16)) \, dx\\ &=-4 x+\frac {2^{7+2 x^2} e^{-3+x (1-\log (4))} \log (16)}{\log (4)}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.88, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 24, normalized size = 0.80 \begin {gather*} -4 \, x + e^{\left (2 \, {\left (x^{2} - x + 4\right )} \log \relax (2) + x - 3\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x + e^(2*(x^2 - x + 4)*log(2) + x - 3) - log(x)

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giac [A]  time = 0.22, size = 32, normalized size = 1.07 \begin {gather*} -{\left (4 \, x e^{3} + e^{3} \log \relax (x) - 256 \, e^{\left (2 \, x^{2} \log \relax (2) - 2 \, x \log \relax (2) + x\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*x*e^3 + e^3*log(x) - 256*e^(2*x^2*log(2) - 2*x*log(2) + x))*e^(-3)

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maple [A]  time = 0.09, size = 25, normalized size = 0.83

method	result	size
norman	\(-4 x +{\mathrm e}^{2 \left (x^{2}-x +4\right ) \ln \relax (2)+x -3}-\ln \relax (x )\)	\(25\)
risch	\(-4 x +2^{2 x^{2}-2 x +8} {\mathrm e}^{x -3}-\ln \relax (x )\)	\(26\)
default	\({\mathrm e}^{2 x^{2} \ln \relax (2)+\left (1-2 \ln \relax (2)\right ) x +8 \ln \relax (2)-3}+\frac {i \left (1-2 \ln \relax (2)\right ) \sqrt {\pi }\, {\mathrm e}^{8 \ln \relax (2)-3-\frac {\left (1-2 \ln \relax (2)\right )^{2}}{8 \ln \relax (2)}} \sqrt {2}\, \erf \left (i \sqrt {2}\, \sqrt {\ln \relax (2)}\, x +\frac {i \left (1-2 \ln \relax (2)\right ) \sqrt {2}}{4 \sqrt {\ln \relax (2)}}\right )}{4 \sqrt {\ln \relax (2)}}+\frac {i \sqrt {\ln \relax (2)}\, \sqrt {\pi }\, {\mathrm e}^{8 \ln \relax (2)-3-\frac {\left (1-2 \ln \relax (2)\right )^{2}}{8 \ln \relax (2)}} \sqrt {2}\, \erf \left (i \sqrt {2}\, \sqrt {\ln \relax (2)}\, x +\frac {i \left (1-2 \ln \relax (2)\right ) \sqrt {2}}{4 \sqrt {\ln \relax (2)}}\right )}{2}-\frac {i \sqrt {\pi }\, {\mathrm e}^{8 \ln \relax (2)-3-\frac {\left (1-2 \ln \relax (2)\right )^{2}}{8 \ln \relax (2)}} \sqrt {2}\, \erf \left (i \sqrt {2}\, \sqrt {\ln \relax (2)}\, x +\frac {i \left (1-2 \ln \relax (2)\right ) \sqrt {2}}{4 \sqrt {\ln \relax (2)}}\right )}{4 \sqrt {\ln \relax (2)}}-\ln \relax (x )-4 x\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x+exp(2*(x^2-x+4)*ln(2)+x-3)-ln(x)

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maxima [C]  time = 1.14, size = 269, normalized size = 8.97 \begin {gather*} -\frac {128 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \relax (2)} + \frac {\sqrt {2} {\left (2 \, \log \relax (2) - 1\right )}}{4 \, \sqrt {-\log \relax (2)}}\right ) e^{\left (-\frac {{\left (2 \, \log \relax (2) - 1\right )}^{2}}{8 \, \log \relax (2)} - 3\right )} \log \relax (2)}{\sqrt {-\log \relax (2)}} + 64 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\pi } {\left (4 \, x \log \relax (2) - 2 \, \log \relax (2) + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {{\left (4 \, x \log \relax (2) - 2 \, \log \relax (2) + 1\right )}^{2}}{\log \relax (2)}}\right ) - 1\right )} {\left (2 \, \log \relax (2) - 1\right )}}{\sqrt {-\frac {{\left (4 \, x \log \relax (2) - 2 \, \log \relax (2) + 1\right )}^{2}}{\log \relax (2)}} \log \relax (2)^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} e^{\left (\frac {{\left (4 \, x \log \relax (2) - 2 \, \log \relax (2) + 1\right )}^{2}}{8 \, \log \relax (2)}\right )}}{\sqrt {\log \relax (2)}}\right )} e^{\left (-\frac {{\left (2 \, \log \relax (2) - 1\right )}^{2}}{8 \, \log \relax (2)} - 3\right )} \sqrt {\log \relax (2)} + \frac {64 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \relax (2)} + \frac {\sqrt {2} {\left (2 \, \log \relax (2) - 1\right )}}{4 \, \sqrt {-\log \relax (2)}}\right ) e^{\left (-\frac {{\left (2 \, \log \relax (2) - 1\right )}^{2}}{8 \, \log \relax (2)} - 3\right )}}{\sqrt {-\log \relax (2)}} - 4 \, x - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-128*sqrt(2)*sqrt(pi)*erf(sqrt(2)*x*sqrt(-log(2)) + 1/4*sqrt(2)*(2*log(2) - 1)/sqrt(-log(2)))*e^(-1/8*(2*log(2
) - 1)^2/log(2) - 3)*log(2)/sqrt(-log(2)) + 64*sqrt(2)*(sqrt(2)*sqrt(1/2)*sqrt(pi)*(4*x*log(2) - 2*log(2) + 1)
*(erf(1/2*sqrt(1/2)*sqrt(-(4*x*log(2) - 2*log(2) + 1)^2/log(2))) - 1)*(2*log(2) - 1)/(sqrt(-(4*x*log(2) - 2*lo
g(2) + 1)^2/log(2))*log(2)^(3/2)) + 2*sqrt(2)*e^(1/8*(4*x*log(2) - 2*log(2) + 1)^2/log(2))/sqrt(log(2)))*e^(-1
/8*(2*log(2) - 1)^2/log(2) - 3)*sqrt(log(2)) + 64*sqrt(2)*sqrt(pi)*erf(sqrt(2)*x*sqrt(-log(2)) + 1/4*sqrt(2)*(
2*log(2) - 1)/sqrt(-log(2)))*e^(-1/8*(2*log(2) - 1)^2/log(2) - 3)/sqrt(-log(2)) - 4*x - log(x)

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mupad [B]  time = 2.07, size = 28, normalized size = 0.93 \begin {gather*} \frac {256\,2^{2\,x^2}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{2^{2\,x}}-\ln \relax (x)-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(256*2^(2*x^2)*exp(-3)*exp(x))/2^(2*x) - log(x) - 4*x

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sympy [A]  time = 0.18, size = 24, normalized size = 0.80 \begin {gather*} - 4 x + e^{x + \left (2 x^{2} - 2 x + 8\right ) \log {\relax (2 )} - 3} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x**2-x)*ln(2)+x)*exp(2*(x**2-x+4)*ln(2)+x-3)-4*x-1)/x,x)

[Out]

-4*x + exp(x + (2*x**2 - 2*x + 8)*log(2) - 3) - log(x)

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