3.85.19 $\int \frac{e^{x^2}x+5x^2+e^{\frac{1-x^4}{x^2}}{x}^2+(5x^2+2e^{x^2}{x}^3+e^{\frac{1-x^4}{x^2}}(-2+x^2-2x^4))\log (x)}{x^2}dx$

Optimal. Leaf size=24 $$(e^{x^2}+(5+e^{\frac{1}{x^2}-x^2})x)\log (x)$$

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Rubi [A]  time = 0.17, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 6, number of rules used = 3, integrand size = 77, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.039, Rules used = {14, 2295, 2288} $$\begin{array}{c}{e}^{x^2}\log (x)+\frac{e^{\frac{1}{x^2}-x^2}(x^4\log (x)+\log (x))}{x^2(\frac{1}{x^3}+x)}+5x\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^x^2*x + 5*x^2 + E^((1 - x^4)/x^2)*x^2 + (5*x^2 + 2*E^x^2*x^3 + E^((1 - x^4)/x^2)*(-2 + x^2 - 2*x^4))*Lo
g[x])/x^2,x]

[Out]

E^x^2*Log[x] + 5*x*Log[x] + (E^(x^(-2) - x^2)*(Log[x] + x^4*Log[x]))/(x^2*(x^(-3) + 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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (5(1+\log (x))+\frac{e^{x^2}(1+2x^2\log (x))}{x}-\frac{e^{\frac{1}{x^2}-x^2}(-x^2+2\log (x)-x^2\log (x)+2x^4\log (x))}{x^2})dx\\ & =5\int (1+\log (x))dx+\int \frac{e^{x^2}(1+2x^2\log (x))}{x}dx-\int \frac{e^{\frac{1}{x^2}-x^2}(-x^2+2\log (x)-x^2\log (x)+2x^4\log (x))}{x^2}dx\\ & =5x+e^{x^2}\log (x)+\frac{e^{\frac{1}{x^2}-x^2}(\log (x)+x^4\log (x))}{x^2(\frac{1}{x^3}+x)}+5\int \log (x)dx\\ & =e^{x^2}\log (x)+5x\log (x)+\frac{e^{\frac{1}{x^2}-x^2}(\log (x)+x^4\log (x))}{x^2(\frac{1}{x^3}+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.23, size = 25, normalized size = 1.04 $$\begin{array}{c}(e^{x^2}+5x+e^{\frac{1}{x^2}-x^2}x)\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x^2 + 5*x + E^(x^(-2) - x^2)*x)*Log[x]

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fricas [A]  time = 0.83, size = 24, normalized size = 1.00 $$\begin{array}{c}(x{e}^{(-\frac{x^4-1}{x^2})}+5x+e^{\left(x^2\right)})\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^(-(x^4 - 1)/x^2) + 5*x + e^(x^2))*log(x)

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giac [A]  time = 0.20, size = 28, normalized size = 1.17 $$\begin{array}{c}x{e}^{(-\frac{x^4-1}{x^2})}\log (x)+5x\log (x)+e^{\left(x^2\right)}\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-(x^4 - 1)/x^2)*log(x) + 5*x*log(x) + e^(x^2)*log(x)

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maple [A]  time = 0.08, size = 30, normalized size = 1.25

method	result	size
default	$x{\mathrm{e}}^{\frac{-x^4+1}{x^2}}\ln (x)+{\mathrm{e}}^{x^2}\ln (x)+5x\ln (x)$	$30$
risch	$(x{\mathrm{e}}^{-\frac{(x-1)(x+1)(x^2+1)}{x^2}}+5x+{\mathrm{e}}^{x^2})\ln (x)$	$31$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp((-x^4+1)/x^2)*ln(x)+exp(x^2)*ln(x)+5*x*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}-\frac{1}{4}\sqrt{\pi }(\mathrm{erfc}(x+\frac{i}{x})e^{\left(4i\right)}-\mathrm{erfc}(-x+\frac{i}{x}))e^{(-2i)}+5x\log (x)+e^{\left(x^2\right)}\log (x)+\frac{1}{2}\mathrm{E}\mathrm{i}\left(x^2\right)-\int \frac{(2x^4-x^2+2)e^{(-x^2+\frac{1}{x^2})}\log (x)}{x^2}dx-\int \frac{e^{\left(x^2\right)}}{x}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(pi)*(erfc(x + I/x)*e^(4*I) - erfc(-x + I/x))*e^(-2*I) + 5*x*log(x) + e^(x^2)*log(x) + 1/2*Ei(x^2) -
integrate((2*x^4 - x^2 + 2)*e^(-x^2 + 1/x^2)*log(x)/x^2, x) - integrate(e^(x^2)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{x^2{\mathrm{e}}^{-\frac{x^4-1}{x^2}}+x{\mathrm{e}}^{x^2}+\ln (x)(2x^3{\mathrm{e}}^{x^2}+5x^2-{\mathrm{e}}^{-\frac{x^4-1}{x^2}}(2x^4-x^2+2))+5x^2}{x^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.58, size = 29, normalized size = 1.21 $$\begin{array}{c}x{e}^{\frac{1-x^4}{x^2}}\log (x)+5x\log (x)+e^{x^2}\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp((1 - x**4)/x**2)*log(x) + 5*x*log(x) + exp(x**2)*log(x)

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