3.16.99 \(\int \frac {-143 x^2+e^{5 x+x^2} (180 x^2+72 x^3)-3 \log (x)}{x^2} \, dx\)

Optimal. Leaf size=27 \[ x+3 \left (12 \left (e^{x (5+x)}-4 x\right )+\frac {1+x+\log (x)}{x}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {14, 2236, 2304} \begin {gather*} 36 e^{x^2+5 x}-143 x+\frac {3}{x}+\frac {3 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-143*x^2 + E^(5*x + x^2)*(180*x^2 + 72*x^3) - 3*Log[x])/x^2,x]

[Out]

36*E^(5*x + x^2) + 3/x - 143*x + (3*Log[x])/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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (36 e^{5 x+x^2} (5+2 x)+\frac {-143 x^2-3 \log (x)}{x^2}\right ) \, dx\\ &=36 \int e^{5 x+x^2} (5+2 x) \, dx+\int \frac {-143 x^2-3 \log (x)}{x^2} \, dx\\ &=36 e^{5 x+x^2}+\int \left (-143-\frac {3 \log (x)}{x^2}\right ) \, dx\\ &=36 e^{5 x+x^2}-143 x-3 \int \frac {\log (x)}{x^2} \, dx\\ &=36 e^{5 x+x^2}+\frac {3}{x}-143 x+\frac {3 \log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.00 \begin {gather*} 36 e^{5 x+x^2}+\frac {3}{x}-143 x+\frac {3 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-143*x^2 + E^(5*x + x^2)*(180*x^2 + 72*x^3) - 3*Log[x])/x^2,x]

[Out]

36*E^(5*x + x^2) + 3/x - 143*x + (3*Log[x])/x

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fricas [A]  time = 0.83, size = 27, normalized size = 1.00 \begin {gather*} -\frac {143 \, x^{2} - 36 \, x e^{\left (x^{2} + 5 \, x\right )} - 3 \, \log \relax (x) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)+(72*x^3+180*x^2)*exp(x^2+5*x)-143*x^2)/x^2,x, algorithm="fricas")

[Out]

-(143*x^2 - 36*x*e^(x^2 + 5*x) - 3*log(x) - 3)/x

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giac [A]  time = 0.24, size = 27, normalized size = 1.00 \begin {gather*} -\frac {143 \, x^{2} - 36 \, x e^{\left (x^{2} + 5 \, x\right )} - 3 \, \log \relax (x) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)+(72*x^3+180*x^2)*exp(x^2+5*x)-143*x^2)/x^2,x, algorithm="giac")

[Out]

-(143*x^2 - 36*x*e^(x^2 + 5*x) - 3*log(x) - 3)/x

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maple [A]  time = 0.02, size = 27, normalized size = 1.00

method	result	size
default	\(-143 x +36 \,{\mathrm e}^{x^{2}+5 x}+\frac {3 \ln \relax (x )}{x}+\frac {3}{x}\)	\(27\)
norman	\(\frac {3-143 x^{2}+36 x \,{\mathrm e}^{x^{2}+5 x}+3 \ln \relax (x )}{x}\)	\(27\)
risch	\(\frac {3 \ln \relax (x )}{x}-\frac {143 x^{2}-36 x \,{\mathrm e}^{\left (5+x \right ) x}-3}{x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*ln(x)+(72*x^3+180*x^2)*exp(x^2+5*x)-143*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-143*x+36*exp(x^2+5*x)+3*ln(x)/x+3/x

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maxima [C]  time = 0.59, size = 83, normalized size = 3.07 \begin {gather*} -90 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {5}{2} i\right ) e^{\left (-\frac {25}{4}\right )} - 18 \, {\left (\frac {5 \, \sqrt {\pi } {\left (2 \, x + 5\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 5\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 5\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 5\right )}^{2}\right )}\right )} e^{\left (-\frac {25}{4}\right )} - 143 \, x + \frac {3 \, \log \relax (x)}{x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x)+(72*x^3+180*x^2)*exp(x^2+5*x)-143*x^2)/x^2,x, algorithm="maxima")

[Out]

-90*I*sqrt(pi)*erf(I*x + 5/2*I)*e^(-25/4) - 18*(5*sqrt(pi)*(2*x + 5)*(erf(1/2*sqrt(-(2*x + 5)^2)) - 1)/sqrt(-(
2*x + 5)^2) - 2*e^(1/4*(2*x + 5)^2))*e^(-25/4) - 143*x + 3*log(x)/x + 3/x

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mupad [B]  time = 1.14, size = 24, normalized size = 0.89 \begin {gather*} 36\,{\mathrm {e}}^{x^2+5\,x}-143\,x+\frac {3\,\ln \relax (x)+3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(x) - exp(5*x + x^2)*(180*x^2 + 72*x^3) + 143*x^2)/x^2,x)

[Out]

36*exp(5*x + x^2) - 143*x + (3*log(x) + 3)/x

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sympy [A]  time = 0.29, size = 22, normalized size = 0.81 \begin {gather*} - 143 x + 36 e^{x^{2} + 5 x} + \frac {3 \log {\relax (x )}}{x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*ln(x)+(72*x**3+180*x**2)*exp(x**2+5*x)-143*x**2)/x**2,x)

[Out]

-143*x + 36*exp(x**2 + 5*x) + 3*log(x)/x + 3/x

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