3.22.91 \(\int \frac {-59 x+e^{\frac {4 x^2+4 x \log (5)+\log ^2(5)}{x}} (3 x+12 x^2-3 \log ^2(5))}{x} \, dx\)

Optimal. Leaf size=24 \[ -4+x+3 \left (-20+e^{\frac {(2 x+\log (5))^2}{x}}\right ) x+\log (9) \]

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Rubi [A]  time = 0.13, antiderivative size = 48, normalized size of antiderivative = 2.00, number of steps used = 3, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {14, 2288} \begin {gather*} \frac {1875 e^{4 x+\frac {\log ^2(5)}{x}} \left (4 x^2-\log ^2(5)\right )}{x \left (4-\frac {\log ^2(5)}{x^2}\right )}-59 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-59*x + E^((4*x^2 + 4*x*Log[5] + Log[5]^2)/x)*(3*x + 12*x^2 - 3*Log[5]^2))/x,x]

[Out]

-59*x + (1875*E^(4*x + Log[5]^2/x)*(4*x^2 - Log[5]^2))/(x*(4 - Log[5]^2/x^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 (-59+\frac {1875 e^{4 x+\frac {\log ^2(5)}{x}} \left (x+4 x^2-\log ^2(5)\right )}{x}\right ) \, dx\\ &=-59 x+1875 \int \frac {e^{4 x+\frac {\log ^2(5)}{x}} \left (x+4 x^2-\log ^2(5)\right )}{x} \, dx\\ &=-59 x+\frac {1875 e^{4 x+\frac {\log ^2(5)}{x}} \left (4 x^2-\log ^2(5)\right )}{x \left (4-\frac {\log ^2(5)}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 0.83 \begin {gather*} \left (-59+1875 e^{4 x+\frac {\log ^2(5)}{x}}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-59*x + E^((4*x^2 + 4*x*Log[5] + Log[5]^2)/x)*(3*x + 12*x^2 - 3*Log[5]^2))/x,x]

[Out]

(-59 + 1875*E^(4*x + Log[5]^2/x))*x

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fricas [A]  time = 0.82, size = 27, normalized size = 1.12 \begin {gather*} 3 \, x e^{\left (\frac {4 \, x^{2} + 4 \, x \log \relax (5) + \log \relax (5)^{2}}{x}\right )} - 59 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(5)^2+12*x^2+3*x)*exp((log(5)^2+4*x*log(5)+4*x^2)/x)-59*x)/x,x, algorithm="fricas")

[Out]

3*x*e^((4*x^2 + 4*x*log(5) + log(5)^2)/x) - 59*x

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giac [A]  time = 0.22, size = 22, normalized size = 0.92 \begin {gather*} 1875 \, x e^{\left (\frac {4 \, x^{2} + \log \relax (5)^{2}}{x}\right )} - 59 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(5)^2+12*x^2+3*x)*exp((log(5)^2+4*x*log(5)+4*x^2)/x)-59*x)/x,x, algorithm="giac")

[Out]

1875*x*e^((4*x^2 + log(5)^2)/x) - 59*x

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maple [A]  time = 0.10, size = 21, normalized size = 0.88

method	result	size
risch	\(-59 x +3 \,{\mathrm e}^{\frac {\left (\ln \relax (5)+2 x \right )^{2}}{x}} x\)	\(21\)
norman	\(-59 x +3 \,{\mathrm e}^{\frac {\ln \relax (5)^{2}+4 x \ln \relax (5)+4 x^{2}}{x}} x\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*ln(5)^2+12*x^2+3*x)*exp((ln(5)^2+4*x*ln(5)+4*x^2)/x)-59*x)/x,x,method=_RETURNVERBOSE)

[Out]

-59*x+3*exp((ln(5)+2*x)^2/x)*x

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maxima [A]  time = 0.61, size = 20, normalized size = 0.83 \begin {gather*} 1875 \, x e^{\left (4 \, x + \frac {\log \relax (5)^{2}}{x}\right )} - 59 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*log(5)^2+12*x^2+3*x)*exp((log(5)^2+4*x*log(5)+4*x^2)/x)-59*x)/x,x, algorithm="maxima")

[Out]

1875*x*e^(4*x + log(5)^2/x) - 59*x

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mupad [B]  time = 1.26, size = 20, normalized size = 0.83 \begin {gather*} 1875\,x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{\frac {{\ln \relax (5)}^2}{x}}-59\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(59*x - exp((4*x*log(5) + log(5)^2 + 4*x^2)/x)*(3*x - 3*log(5)^2 + 12*x^2))/x,x)

[Out]

1875*x*exp(4*x)*exp(log(5)^2/x) - 59*x

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sympy [A]  time = 0.17, size = 26, normalized size = 1.08 \begin {gather*} 3 x e^{\frac {4 x^{2} + 4 x \log {\relax (5 )} + \log {\relax (5 )}^{2}}{x}} - 59 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*ln(5)**2+12*x**2+3*x)*exp((ln(5)**2+4*x*ln(5)+4*x**2)/x)-59*x)/x,x)

[Out]

3*x*exp((4*x**2 + 4*x*log(5) + log(5)**2)/x) - 59*x

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