3.99.93 \(\int \frac {e^{4 x^2+8 x \log (4)+4 \log ^2(4)} (-1+8 x^2+8 x \log (4))}{2 x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{2} \left (9+\frac {e^{4 (x+\log (4))^2}}{x}\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 39, normalized size of antiderivative = 1.95, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {12, 2288} \begin {gather*} \frac {2^{16 x-1} e^{4 x^2+4 \log ^2(4)} \left (x^2+x \log (4)\right )}{x^2 (x+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4*x^2 + 8*x*Log[4] + 4*Log[4]^2)*(-1 + 8*x^2 + 8*x*Log[4]))/(2*x^2),x]

[Out]

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

Rule 12

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

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} &=\frac {1}{2} \int \frac {e^{4 x^2+8 x \log (4)+4 \log ^2(4)} \left (-1+8 x^2+8 x \log (4)\right )}{x^2} \, dx\\ &=\frac {2^{-1+16 x} e^{4 x^2+4 \log ^2(4)} \left (x^2+x \log (4)\right )}{x^2 (x+\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 23, normalized size = 1.15 \begin {gather*} \frac {2^{-1+16 x} e^{4 \left (x^2+\log ^2(4)\right )}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x^2 + 8*x*Log[4] + 4*Log[4]^2)*(-1 + 8*x^2 + 8*x*Log[4]))/(2*x^2),x]

[Out]

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

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fricas [A]  time = 0.72, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{\left (4 \, x^{2} + 16 \, x \log \relax (2) + 16 \, \log \relax (2)^{2}\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(2)+8*x^2-1)*exp(16*log(2)^2+16*x*log(2)+4*x^2)/x^2,x, algorithm="fricas")

[Out]

1/2*e^(4*x^2 + 16*x*log(2) + 16*log(2)^2)/x

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giac [A]  time = 1.60, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{\left (4 \, x^{2} + 16 \, x \log \relax (2) + 16 \, \log \relax (2)^{2}\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(2)+8*x^2-1)*exp(16*log(2)^2+16*x*log(2)+4*x^2)/x^2,x, algorithm="giac")

[Out]

1/2*e^(4*x^2 + 16*x*log(2) + 16*log(2)^2)/x

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maple [A]  time = 0.06, size = 22, normalized size = 1.10

method	result	size
risch	\(\frac {65536^{x} {\mathrm e}^{16 \ln \relax (2)^{2}+4 x^{2}}}{2 x}\)	\(22\)
gosper	\(\frac {{\mathrm e}^{16 \ln \relax (2)^{2}+16 x \ln \relax (2)+4 x^{2}}}{2 x}\)	\(24\)
norman	\(\frac {{\mathrm e}^{16 \ln \relax (2)^{2}+16 x \ln \relax (2)+4 x^{2}}}{2 x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(16*x*ln(2)+8*x^2-1)*exp(16*ln(2)^2+16*x*ln(2)+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/2/x*65536^x*exp(16*ln(2)^2+4*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -i \, \sqrt {\pi } \operatorname {erf}\left (2 i \, x + 4 i \, \log \relax (2)\right ) + \frac {1}{2} \, \int \frac {{\left (16 \, x e^{\left (16 \, \log \relax (2)^{2}\right )} \log \relax (2) - e^{\left (16 \, \log \relax (2)^{2}\right )}\right )} e^{\left (4 \, x^{2} + 16 \, x \log \relax (2)\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*log(2)+8*x^2-1)*exp(16*log(2)^2+16*x*log(2)+4*x^2)/x^2,x, algorithm="maxima")

[Out]

-I*sqrt(pi)*erf(2*I*x + 4*I*log(2)) + 1/2*integrate((16*x*e^(16*log(2)^2)*log(2) - e^(16*log(2)^2))*e^(4*x^2 +
16*x*log(2))/x^2, x)

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mupad [B]  time = 0.09, size = 23, normalized size = 1.15 \begin {gather*} \frac {2^{16\,x}\,{\mathrm {e}}^{16\,{\ln \relax (2)}^2}\,{\mathrm {e}}^{4\,x^2}}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(16*x*log(2) + 16*log(2)^2 + 4*x^2)*(16*x*log(2) + 8*x^2 - 1))/(2*x^2),x)

[Out]

(2^(16*x)*exp(16*log(2)^2)*exp(4*x^2))/(2*x)

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sympy [A]  time = 0.14, size = 22, normalized size = 1.10 \begin {gather*} \frac {e^{4 x^{2} + 16 x \log {\relax (2 )} + 16 \log {\relax (2 )}^{2}}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(16*x*ln(2)+8*x**2-1)*exp(16*ln(2)**2+16*x*ln(2)+4*x**2)/x**2,x)

[Out]

exp(4*x**2 + 16*x*log(2) + 16*log(2)**2)/(2*x)

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