3.48.84 \(\int \frac {11 x^2+e^{5+x^2} (5-10 x^2)}{x^2} \, dx\)

Optimal. Leaf size=21 \[ x+5 \left (3-\frac {e^{5+x^2}}{x}+2 x\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {14, 2288} \begin {gather*} 11 x-\frac {5 e^{x^2+5}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(11*x^2 + E^(5 + x^2)*(5 - 10*x^2))/x^2,x]

[Out]

(-5*E^(5 + x^2))/x + 11*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 (11-\frac {5 e^{5+x^2} \left (-1+2 x^2\right )}{x^2}\right ) \, dx\\ &=11 x-5 \int \frac {e^{5+x^2} \left (-1+2 x^2\right )}{x^2} \, dx\\ &=-\frac {5 e^{5+x^2}}{x}+11 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.76 \begin {gather*} -\frac {5 e^{5+x^2}}{x}+11 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(11*x^2 + E^(5 + x^2)*(5 - 10*x^2))/x^2,x]

[Out]

(-5*E^(5 + x^2))/x + 11*x

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fricas [A]  time = 0.97, size = 18, normalized size = 0.86 \begin {gather*} \frac {11 \, x^{2} - 5 \, e^{\left (x^{2} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+5)*exp(x^2+5)+11*x^2)/x^2,x, algorithm="fricas")

[Out]

(11*x^2 - 5*e^(x^2 + 5))/x

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giac [A]  time = 0.21, size = 18, normalized size = 0.86 \begin {gather*} \frac {11 \, x^{2} - 5 \, e^{\left (x^{2} + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+5)*exp(x^2+5)+11*x^2)/x^2,x, algorithm="giac")

[Out]

(11*x^2 - 5*e^(x^2 + 5))/x

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maple [A]  time = 0.04, size = 16, normalized size = 0.76

method	result	size
risch	\(11 x -\frac {5 \,{\mathrm e}^{x^{2}+5}}{x}\)	\(16\)
norman	\(\frac {11 x^{2}-5 \,{\mathrm e}^{x^{2}+5}}{x}\)	\(19\)
default	\(11 x -5 \,{\mathrm e}^{5} \sqrt {\pi }\, \erfi \relax (x )+5 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x^{2}}}{x}+\sqrt {\pi }\, \erfi \relax (x )\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2+5)*exp(x^2+5)+11*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

11*x-5*exp(x^2+5)/x

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maxima [C]  time = 0.60, size = 36, normalized size = 1.71 \begin {gather*} 5 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) e^{5} - \frac {5 \, \sqrt {-x^{2}} e^{5} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{2 \, x} + 11 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2+5)*exp(x^2+5)+11*x^2)/x^2,x, algorithm="maxima")

[Out]

5*I*sqrt(pi)*erf(I*x)*e^5 - 5/2*sqrt(-x^2)*e^5*gamma(-1/2, -x^2)/x + 11*x

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mupad [B]  time = 0.07, size = 15, normalized size = 0.71 \begin {gather*} 11\,x-\frac {5\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x^2 + 5)*(10*x^2 - 5) - 11*x^2)/x^2,x)

[Out]

11*x - (5*exp(x^2)*exp(5))/x

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sympy [A]  time = 0.09, size = 12, normalized size = 0.57 \begin {gather*} 11 x - \frac {5 e^{x^{2} + 5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2+5)*exp(x**2+5)+11*x**2)/x**2,x)

[Out]

11*x - 5*exp(x**2 + 5)/x

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