3.23.81 \(\int \frac {2 x^3+81 e^{1+25 \log ^2(x)} (-1+50 \log (x))}{x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {81 e^{1+25 \log ^2(x)}}{x}+x^2 \]

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Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14, 2288} \begin {gather*} x^2+\frac {81 e^{25 \log ^2(x)+1}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^3 + 81*E^(1 + 25*Log[x]^2)*(-1 + 50*Log[x]))/x^2,x]

[Out]

(81*E^(1 + 25*Log[x]^2))/x + x^2

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 (2 x+\frac {81 e^{1+25 \log ^2(x)} (-1+50 \log (x))}{x^2}\right ) \, dx\\ &=x^2+81 \int \frac {e^{1+25 \log ^2(x)} (-1+50 \log (x))}{x^2} \, dx\\ &=\frac {81 e^{1+25 \log ^2(x)}}{x}+x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.00 \begin {gather*} \frac {81 e^{1+25 \log ^2(x)}}{x}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^3 + 81*E^(1 + 25*Log[x]^2)*(-1 + 50*Log[x]))/x^2,x]

[Out]

(81*E^(1 + 25*Log[x]^2))/x + x^2

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fricas [A]  time = 0.73, size = 21, normalized size = 1.11 \begin {gather*} \frac {x^{3} + e^{\left (25 \, \log \relax (x)^{2} + 4 \, \log \relax (3) + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)-1)*exp(25*log(x)^2+4*log(3)+1)+2*x^3)/x^2,x, algorithm="fricas")

[Out]

(x^3 + e^(25*log(x)^2 + 4*log(3) + 1))/x

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giac [A]  time = 0.61, size = 19, normalized size = 1.00 \begin {gather*} \frac {x^{3} + 81 \, e^{\left (25 \, \log \relax (x)^{2} + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)-1)*exp(25*log(x)^2+4*log(3)+1)+2*x^3)/x^2,x, algorithm="giac")

[Out]

(x^3 + 81*e^(25*log(x)^2 + 1))/x

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maple [A]  time = 0.04, size = 19, normalized size = 1.00

method	result	size
risch	\(x^{2}+\frac {81 \,{\mathrm e}^{25 \ln \relax (x )^{2}+1}}{x}\)	\(19\)
default	\(x^{2}+\frac {{\mathrm e}^{25 \ln \relax (x )^{2}+4 \ln \relax (3)+1}}{x}\)	\(22\)
norman	\(\frac {x^{3}+{\mathrm e}^{25 \ln \relax (x )^{2}+4 \ln \relax (3)+1}}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*ln(x)-1)*exp(25*ln(x)^2+4*ln(3)+1)+2*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^2+81*exp(25*ln(x)^2+1)/x

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maxima [C]  time = 0.60, size = 75, normalized size = 3.95 \begin {gather*} \frac {81}{10} i \, \sqrt {\pi } \operatorname {erf}\left (5 i \, \log \relax (x) - \frac {1}{10} i\right ) e^{\frac {99}{100}} + x^{2} + \frac {81}{10} \, {\left (\frac {\sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{10} \, \sqrt {-{\left (50 \, \log \relax (x) - 1\right )}^{2}}\right ) - 1\right )} {\left (50 \, \log \relax (x) - 1\right )}}{\sqrt {-{\left (50 \, \log \relax (x) - 1\right )}^{2}}} + 10 \, e^{\left (\frac {1}{100} \, {\left (50 \, \log \relax (x) - 1\right )}^{2}\right )}\right )} e^{\frac {99}{100}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*log(x)-1)*exp(25*log(x)^2+4*log(3)+1)+2*x^3)/x^2,x, algorithm="maxima")

[Out]

81/10*I*sqrt(pi)*erf(5*I*log(x) - 1/10*I)*e^(99/100) + x^2 + 81/10*(sqrt(pi)*(erf(1/10*sqrt(-(50*log(x) - 1)^2
)) - 1)*(50*log(x) - 1)/sqrt(-(50*log(x) - 1)^2) + 10*e^(1/100*(50*log(x) - 1)^2))*e^(99/100)

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mupad [B]  time = 1.31, size = 18, normalized size = 0.95 \begin {gather*} x^2+\frac {81\,{\mathrm {e}}^{25\,{\ln \relax (x)}^2}\,\mathrm {e}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*log(3) + 25*log(x)^2 + 1)*(50*log(x) - 1) + 2*x^3)/x^2,x)

[Out]

x^2 + (81*exp(25*log(x)^2)*exp(1))/x

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sympy [A]  time = 0.24, size = 15, normalized size = 0.79 \begin {gather*} x^{2} + \frac {81 e^{25 \log {\relax (x )}^{2} + 1}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*ln(x)-1)*exp(25*ln(x)**2+4*ln(3)+1)+2*x**3)/x**2,x)

[Out]

x**2 + 81*exp(25*log(x)**2 + 1)/x

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