3.49.48 \(\int \frac {-e^{\frac {1+30 x}{2 x^2}} x^2+e^{\frac {1+30 x}{2 x^2}} (-1-15 x+x^2) \log (x)+(-x^2+e^{\frac {1+30 x}{2 x^2}} (-1-15 x+x^2)) \log ^2(x)}{x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ -x+e^{\frac {5 \left (3+\frac {1}{10 x}\right )}{x}} \left (x+\frac {x}{\log (x)}\right ) \]

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Rubi [A]  time = 1.07, antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 3, number of rules used = 2, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6742, 2288} \begin {gather*} \frac {e^{\frac {1}{2 x^2}+\frac {15}{x}} \left (15 x \log ^2(x)+\log ^2(x)+15 x \log (x)+\log (x)\right )}{\left (\frac {1}{x^3}+\frac {15}{x^2}\right ) x^2 \log ^2(x)}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^((1 + 30*x)/(2*x^2))*x^2) + E^((1 + 30*x)/(2*x^2))*(-1 - 15*x + x^2)*Log[x] + (-x^2 + E^((1 + 30*x)/(
2*x^2))*(-1 - 15*x + x^2))*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

-x + (E^(1/(2*x^2) + 15/x)*(Log[x] + 15*x*Log[x] + Log[x]^2 + 15*x*Log[x]^2))/((x^(-3) + 15/x^2)*x^2*Log[x]^2)

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 39, normalized size = 1.34 \begin {gather*} \left (-1+e^{\frac {1+30 x}{2 x^2}}\right ) x+\frac {e^{\frac {1+30 x}{2 x^2}} x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^((1 + 30*x)/(2*x^2))*x^2) + E^((1 + 30*x)/(2*x^2))*(-1 - 15*x + x^2)*Log[x] + (-x^2 + E^((1 + 3
0*x)/(2*x^2))*(-1 - 15*x + x^2))*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

(-1 + E^((1 + 30*x)/(2*x^2)))*x + (E^((1 + 30*x)/(2*x^2))*x)/Log[x]

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fricas [A]  time = 0.62, size = 39, normalized size = 1.34 \begin {gather*} \frac {x e^{\left (\frac {30 \, x + 1}{2 \, x^{2}}\right )} + {\left (x e^{\left (\frac {30 \, x + 1}{2 \, x^{2}}\right )} - x\right )} \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)-x^2)*log(x)^2+(x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)*log(x)-x^2*exp
(1/2*(30*x+1)/x^2))/x^2/log(x)^2,x, algorithm="fricas")

[Out]

(x*e^(1/2*(30*x + 1)/x^2) + (x*e^(1/2*(30*x + 1)/x^2) - x)*log(x))/log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} e^{\left (\frac {30 \, x + 1}{2 \, x^{2}}\right )} - {\left (x^{2} - 15 \, x - 1\right )} e^{\left (\frac {30 \, x + 1}{2 \, x^{2}}\right )} \log \relax (x) + {\left (x^{2} - {\left (x^{2} - 15 \, x - 1\right )} e^{\left (\frac {30 \, x + 1}{2 \, x^{2}}\right )}\right )} \log \relax (x)^{2}}{x^{2} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)-x^2)*log(x)^2+(x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)*log(x)-x^2*exp
(1/2*(30*x+1)/x^2))/x^2/log(x)^2,x, algorithm="giac")

[Out]

integrate(-(x^2*e^(1/2*(30*x + 1)/x^2) - (x^2 - 15*x - 1)*e^(1/2*(30*x + 1)/x^2)*log(x) + (x^2 - (x^2 - 15*x -
1)*e^(1/2*(30*x + 1)/x^2))*log(x)^2)/(x^2*log(x)^2), x)

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maple [A]  time = 0.05, size = 35, normalized size = 1.21

method	result	size
risch	\(x \,{\mathrm e}^{\frac {30 x +1}{2 x^{2}}}-x +\frac {x \,{\mathrm e}^{\frac {30 x +1}{2 x^{2}}}}{\ln \relax (x )}\)	\(35\)
default	\(-x +\frac {x^{2} {\mathrm e}^{\frac {30 x +1}{2 x^{2}}}+\ln \relax (x ) {\mathrm e}^{\frac {30 x +1}{2 x^{2}}} x^{2}}{x \ln \relax (x )}\)	\(46\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {30 x +1}{2 x^{2}}}+\ln \relax (x ) {\mathrm e}^{\frac {30 x +1}{2 x^{2}}} x^{2}-x^{2} \ln \relax (x )}{x \ln \relax (x )}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)-x^2)*ln(x)^2+(x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)*ln(x)-x^2*exp(1/2*(30
*x+1)/x^2))/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(1/2*(30*x+1)/x^2)-x+x*exp(1/2*(30*x+1)/x^2)/ln(x)

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maxima [A]  time = 0.41, size = 27, normalized size = 0.93 \begin {gather*} -x + \frac {{\left (x \log \relax (x) + x\right )} e^{\left (\frac {15}{x} + \frac {1}{2 \, x^{2}}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)-x^2)*log(x)^2+(x^2-15*x-1)*exp(1/2*(30*x+1)/x^2)*log(x)-x^2*exp
(1/2*(30*x+1)/x^2))/x^2/log(x)^2,x, algorithm="maxima")

[Out]

-x + (x*log(x) + x)*e^(15/x + 1/2/x^2)/log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {x^2\,{\mathrm {e}}^{\frac {15\,x+\frac {1}{2}}{x^2}}+{\ln \relax (x)}^2\,\left ({\mathrm {e}}^{\frac {15\,x+\frac {1}{2}}{x^2}}\,\left (-x^2+15\,x+1\right )+x^2\right )+{\mathrm {e}}^{\frac {15\,x+\frac {1}{2}}{x^2}}\,\ln \relax (x)\,\left (-x^2+15\,x+1\right )}{x^2\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*exp((15*x + 1/2)/x^2) + log(x)^2*(exp((15*x + 1/2)/x^2)*(15*x - x^2 + 1) + x^2) + exp((15*x + 1/2)/x
^2)*log(x)*(15*x - x^2 + 1))/(x^2*log(x)^2),x)

[Out]

int(-(x^2*exp((15*x + 1/2)/x^2) + log(x)^2*(exp((15*x + 1/2)/x^2)*(15*x - x^2 + 1) + x^2) + exp((15*x + 1/2)/x
^2)*log(x)*(15*x - x^2 + 1))/(x^2*log(x)^2), x)

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sympy [A]  time = 0.32, size = 22, normalized size = 0.76 \begin {gather*} - x + \frac {\left (x \log {\relax (x )} + x\right ) e^{\frac {15 x + \frac {1}{2}}{x^{2}}}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-15*x-1)*exp(1/2*(30*x+1)/x**2)-x**2)*ln(x)**2+(x**2-15*x-1)*exp(1/2*(30*x+1)/x**2)*ln(x)-x**
2*exp(1/2*(30*x+1)/x**2))/x**2/ln(x)**2,x)

[Out]

-x + (x*log(x) + x)*exp((15*x + 1/2)/x**2)/log(x)

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