3.78.70 \(\int \frac {x^2+e^{5+x^2} (-120+240 x^2+(-20+40 x^2) \log (3))}{x^2} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {20 e^{5+x^2} (6+\log (3))}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {14, 2288} \begin {gather*} \frac {20 e^{x^2+5} (6+\log (3))}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + E^(5 + x^2)*(-120 + 240*x^2 + (-20 + 40*x^2)*Log[3]))/x^2,x]

[Out]

x + (20*E^(5 + x^2)*(6 + Log[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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {20 e^{5+x^2} \left (-1+2 x^2\right ) (6+\log (3))}{x^2}\right ) \, dx\\ &=x+(20 (6+\log (3))) \int \frac {e^{5+x^2} \left (-1+2 x^2\right )}{x^2} \, dx\\ &=x+\frac {20 e^{5+x^2} (6+\log (3))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} x+\frac {20 e^{5+x^2} (6+\log (3))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^(5 + x^2)*(-120 + 240*x^2 + (-20 + 40*x^2)*Log[3]))/x^2,x]

[Out]

x + (20*E^(5 + x^2)*(6 + Log[3]))/x

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fricas [A]  time = 0.88, size = 20, normalized size = 1.11 \begin {gather*} \frac {x^{2} + 20 \, {\left (\log \relax (3) + 6\right )} e^{\left (x^{2} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^2-20)*log(3)+240*x^2-120)*exp(x^2+5)+x^2)/x^2,x, algorithm="fricas")

[Out]

(x^2 + 20*(log(3) + 6)*e^(x^2 + 5))/x

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giac [A]  time = 0.17, size = 26, normalized size = 1.44 \begin {gather*} \frac {x^{2} + 20 \, e^{\left (x^{2} + 5\right )} \log \relax (3) + 120 \, e^{\left (x^{2} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^2-20)*log(3)+240*x^2-120)*exp(x^2+5)+x^2)/x^2,x, algorithm="giac")

[Out]

(x^2 + 20*e^(x^2 + 5)*log(3) + 120*e^(x^2 + 5))/x

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maple [A]  time = 0.03, size = 18, normalized size = 1.00

method	result	size
risch	\(x +\frac {20 \left (6+\ln \relax (3)\right ) {\mathrm e}^{x^{2}+5}}{x}\)	\(18\)
norman	\(\frac {x^{2}+\left (20 \ln \relax (3)+120\right ) {\mathrm e}^{x^{2}+5}}{x}\)	\(22\)
default	\(x +120 \,{\mathrm e}^{5} \sqrt {\pi }\, \erfi \relax (x )-120 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \erfi \relax (x )\right )+\frac {20 \,{\mathrm e}^{5} \ln \relax (3) {\mathrm e}^{x^{2}}}{x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((40*x^2-20)*ln(3)+240*x^2-120)*exp(x^2+5)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x+20*(6+ln(3))*exp(x^2+5)/x

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maxima [C]  time = 0.38, size = 70, normalized size = 3.89 \begin {gather*} -20 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{5} \log \relax (3) - 120 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{5} + \frac {10 \, \sqrt {-x^{2}} e^{5} \Gamma \left (-\frac {1}{2}, -x^{2}\right ) \log \relax (3)}{x} + \frac {60 \, \sqrt {-x^{2}} e^{5} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^2-20)*log(3)+240*x^2-120)*exp(x^2+5)+x^2)/x^2,x, algorithm="maxima")

[Out]

-20*I*sqrt(pi)*erf(I*x)*e^5*log(3) - 120*I*sqrt(pi)*erf(I*x)*e^5 + 10*sqrt(-x^2)*e^5*gamma(-1/2, -x^2)*log(3)/
x + 60*sqrt(-x^2)*e^5*gamma(-1/2, -x^2)/x + x

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mupad [B]  time = 0.10, size = 18, normalized size = 1.00 \begin {gather*} x+\frac {{\mathrm {e}}^{x^2+5}\,\left (20\,\ln \relax (3)+120\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + exp(x^2 + 5)*(log(3)*(40*x^2 - 20) + 240*x^2 - 120))/x^2,x)

[Out]

x + (exp(x^2 + 5)*(20*log(3) + 120))/x

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sympy [A]  time = 0.13, size = 15, normalized size = 0.83 \begin {gather*} x + \frac {\left (20 \log {\relax (3 )} + 120\right ) e^{x^{2} + 5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x**2-20)*ln(3)+240*x**2-120)*exp(x**2+5)+x**2)/x**2,x)

[Out]

x + (20*log(3) + 120)*exp(x**2 + 5)/x

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