3.56.36 \(\int \frac {5-2 e^3 x^2+(-1+2 e^3 x^2) \log (x^2)}{2 e^3 x^2} \, dx\)

Optimal. Leaf size=21 \[ -3+\left (\frac {1}{2 e^3 x}+x\right ) \left (-3+\log \left (x^2\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 7, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 14, 2334} \begin {gather*} \frac {\left (2 e^3 x+\frac {1}{x}\right ) \log \left (x^2\right )}{2 e^3}-3 x-\frac {3}{2 e^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(2*E^3*x) - 3*x + ((x^(-1) + 2*E^3*x)*Log[x^2])/(2*E^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 34, normalized size = 1.62 \begin {gather*} -\frac {3}{2 e^3 x}-3 x+\frac {\log \left (x^2\right )}{2 e^3 x}+x \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.72, size = 31, normalized size = 1.48 \begin {gather*} -\frac {{\left (6 \, x^{2} e^{3} - {\left (2 \, x^{2} e^{3} + 1\right )} \log \left (x^{2}\right ) + 3\right )} e^{\left (-3\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*x^2*e^3 - (2*x^2*e^3 + 1)*log(x^2) + 3)*e^(-3)/x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 31, normalized size = 1.48 \begin {gather*} \frac {{\left (2 \, x^{2} e^{3} \log \left (x^{2}\right ) - 6 \, x^{2} e^{3} + \log \left (x^{2}\right ) - 3\right )} e^{\left (-3\right )}}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x^2*e^3*log(x^2) - 6*x^2*e^3 + log(x^2) - 3)*e^(-3)/x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 35, normalized size = 1.67

method	result	size
default	\(\frac {{\mathrm e}^{-3} \left (2 x \,{\mathrm e}^{3} \ln \left (x^{2}\right )-6 x \,{\mathrm e}^{3}+\frac {\ln \left (x^{2}\right )}{x}-\frac {3}{x}\right )}{2}\)	\(35\)
norman	\(\frac {x^{2} \ln \left (x^{2}\right )-3 x^{2}-\frac {3 \,{\mathrm e}^{-3}}{2}+\frac {{\mathrm e}^{-3} \ln \left (x^{2}\right )}{2}}{x}\)	\(35\)
risch	\(\frac {{\mathrm e}^{-3} \left (2 x^{2} {\mathrm e}^{3}+1\right ) \ln \left (x^{2}\right )}{2 x}-\frac {3 \,{\mathrm e}^{-3} \left (2 x^{2} {\mathrm e}^{3}+1\right )}{2 x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/exp(3)*(2*x*exp(3)*ln(x^2)-6*x*exp(3)+ln(x^2)/x-3/x)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 37, normalized size = 1.76 \begin {gather*} \frac {1}{2} \, {\left (2 \, {\left (x \log \left (x^{2}\right ) - 2 \, x\right )} e^{3} - 2 \, x e^{3} + \frac {\log \left (x^{2}\right )}{x} - \frac {3}{x}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.42, size = 22, normalized size = 1.05 \begin {gather*} \frac {{\mathrm {e}}^{-3}\,\left (\ln \left (x^2\right )-3\right )\,\left (2\,{\mathrm {e}}^3\,x^2+1\right )}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(-3)*(log(x^2) - 3)*(2*x^2*exp(3) + 1))/(2*x)

________________________________________________________________________________________

sympy [B]  time = 0.14, size = 37, normalized size = 1.76 \begin {gather*} \frac {- 6 x e^{3} - \frac {3}{x}}{2 e^{3}} + \frac {\left (2 x^{2} e^{3} + 1\right ) \log {\left (x^{2} \right )}}{2 x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-6*x*exp(3) - 3/x)*exp(-3)/2 + (2*x**2*exp(3) + 1)*exp(-3)*log(x**2)/(2*x)

________________________________________________________________________________________