3.34.91 \(\int \frac {-8 e^{\frac {3}{5}+x}+e^{e^3} (-42+2 e^x-2 x)+e^{3/5} (168+8 x)+(4 e^{3/5} x-4 e^{\frac {3}{5}+x} x+e^{e^3} (-x+e^x x)) \log (x^2)}{x} \, dx\)

Optimal. Leaf size=26 \[ \left (-4 e^{3/5}+e^{e^3}\right ) \left (-21+e^x-x\right ) \log \left (x^2\right ) \]

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Rubi [B]  time = 0.09, antiderivative size = 65, normalized size of antiderivative = 2.50, number of steps used = 8, number of rules used = 4, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {14, 2288, 43, 2295} \begin {gather*} \left (4 e^{3/5}-e^{e^3}\right ) x \log \left (x^2\right )-\left (4 e^{3/5}-e^{e^3}\right ) e^x \log \left (x^2\right )+42 \left (4 e^{3/5}-e^{e^3}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*E^(3/5 + x) + E^E^3*(-42 + 2*E^x - 2*x) + E^(3/5)*(168 + 8*x) + (4*E^(3/5)*x - 4*E^(3/5 + x)*x + E^E^3
*(-x + E^x*x))*Log[x^2])/x,x]

[Out]

42*(4*E^(3/5) - E^E^3)*Log[x] - E^x*(4*E^(3/5) - E^E^3)*Log[x^2] + (4*E^(3/5) - E^E^3)*x*Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^x \left (-4 e^{3/5}+e^{e^3}\right ) \left (2+x \log \left (x^2\right )\right )}{x}-\frac {\left (-4 e^{3/5}+e^{e^3}\right ) \left (42+2 x+x \log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=\left (4 e^{3/5}-e^{e^3}\right ) \int \frac {42+2 x+x \log \left (x^2\right )}{x} \, dx+\left (-4 e^{3/5}+e^{e^3}\right ) \int \frac {e^x \left (2+x \log \left (x^2\right )\right )}{x} \, dx\\ &=-e^x \left (4 e^{3/5}-e^{e^3}\right ) \log \left (x^2\right )+\left (4 e^{3/5}-e^{e^3}\right ) \int \left (\frac {2 (21+x)}{x}+\log \left (x^2\right )\right ) \, dx\\ &=-e^x \left (4 e^{3/5}-e^{e^3}\right ) \log \left (x^2\right )+\left (4 e^{3/5}-e^{e^3}\right ) \int \log \left (x^2\right ) \, dx+\left (2 \left (4 e^{3/5}-e^{e^3}\right )\right ) \int \frac {21+x}{x} \, dx\\ &=-2 \left (4 e^{3/5}-e^{e^3}\right ) x-e^x \left (4 e^{3/5}-e^{e^3}\right ) \log \left (x^2\right )+\left (4 e^{3/5}-e^{e^3}\right ) x \log \left (x^2\right )+\left (2 \left (4 e^{3/5}-e^{e^3}\right )\right ) \int \left (1+\frac {21}{x}\right ) \, dx\\ &=42 \left (4 e^{3/5}-e^{e^3}\right ) \log (x)-e^x \left (4 e^{3/5}-e^{e^3}\right ) \log \left (x^2\right )+\left (4 e^{3/5}-e^{e^3}\right ) x \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 31, normalized size = 1.19 \begin {gather*} \left (-4 e^{3/5}+e^{e^3}\right ) \left (-42 \log (x)+\left (e^x-x\right ) \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E^(3/5 + x) + E^E^3*(-42 + 2*E^x - 2*x) + E^(3/5)*(168 + 8*x) + (4*E^(3/5)*x - 4*E^(3/5 + x)*x +
E^E^3*(-x + E^x*x))*Log[x^2])/x,x]

[Out]

(-4*E^(3/5) + E^E^3)*(-42*Log[x] + (E^x - x)*Log[x^2])

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fricas [A]  time = 0.60, size = 39, normalized size = 1.50 \begin {gather*} {\left (4 \, {\left (x + 21\right )} e^{\frac {6}{5}} - {\left ({\left (x + 21\right )} e^{\frac {3}{5}} - e^{\left (x + \frac {3}{5}\right )}\right )} e^{\left (e^{3}\right )} - 4 \, e^{\left (x + \frac {6}{5}\right )}\right )} e^{\left (-\frac {3}{5}\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-x)*exp(exp(3))-4*x*exp(3/5)*exp(x)+4*x*exp(3/5))*log(x^2)+(2*exp(x)-2*x-42)*exp(exp(3))-
8*exp(3/5)*exp(x)+(8*x+168)*exp(3/5))/x,x, algorithm="fricas")

[Out]

(4*(x + 21)*e^(6/5) - ((x + 21)*e^(3/5) - e^(x + 3/5))*e^(e^3) - 4*e^(x + 6/5))*e^(-3/5)*log(x^2)

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giac [B]  time = 1.46, size = 53, normalized size = 2.04 \begin {gather*} 4 \, x e^{\frac {3}{5}} \log \left (x^{2}\right ) - x e^{\left (e^{3}\right )} \log \left (x^{2}\right ) + e^{\left (x + e^{3}\right )} \log \left (x^{2}\right ) - 4 \, e^{\left (x + \frac {3}{5}\right )} \log \left (x^{2}\right ) + 168 \, e^{\frac {3}{5}} \log \relax (x) - 42 \, e^{\left (e^{3}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-x)*exp(exp(3))-4*x*exp(3/5)*exp(x)+4*x*exp(3/5))*log(x^2)+(2*exp(x)-2*x-42)*exp(exp(3))-
8*exp(3/5)*exp(x)+(8*x+168)*exp(3/5))/x,x, algorithm="giac")

[Out]

4*x*e^(3/5)*log(x^2) - x*e^(e^3)*log(x^2) + e^(x + e^3)*log(x^2) - 4*e^(x + 3/5)*log(x^2) + 168*e^(3/5)*log(x)
- 42*e^(e^3)*log(x)

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maple [B]  time = 0.10, size = 48, normalized size = 1.85

method	result	size
norman	\(\left (-21 \,{\mathrm e}^{{\mathrm e}^{3}}+84 \,{\mathrm e}^{\frac {3}{5}}\right ) \ln \left (x^{2}\right )+\left (-{\mathrm e}^{{\mathrm e}^{3}}+4 \,{\mathrm e}^{\frac {3}{5}}\right ) x \ln \left (x^{2}\right )+\left ({\mathrm e}^{{\mathrm e}^{3}}-4 \,{\mathrm e}^{\frac {3}{5}}\right ) {\mathrm e}^{x} \ln \left (x^{2}\right )\)	\(48\)
default	\({\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{3}} \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )-4 \,{\mathrm e}^{x} {\mathrm e}^{\frac {3}{5}} \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )+\left (2 \,{\mathrm e}^{{\mathrm e}^{3}}-8 \,{\mathrm e}^{\frac {3}{5}}\right ) {\mathrm e}^{x} \ln \relax (x )-{\mathrm e}^{{\mathrm e}^{3}} \ln \left (x^{2}\right ) x +4 \,{\mathrm e}^{\frac {3}{5}} \ln \left (x^{2}\right ) x -42 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (x )+168 \,{\mathrm e}^{\frac {3}{5}} \ln \relax (x )\)	\(79\)
risch	\(\left (8 x \,{\mathrm e}^{\frac {3}{5}}-8 \,{\mathrm e}^{x +\frac {3}{5}}-2 x \,{\mathrm e}^{{\mathrm e}^{3}}+2 \,{\mathrm e}^{{\mathrm e}^{3}+x}\right ) \ln \relax (x )-2 i \pi \,{\mathrm e}^{\frac {3}{5}} x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 i \pi \,{\mathrm e}^{\frac {3}{5}} x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi \,{\mathrm e}^{\frac {3}{5}} x \mathrm {csgn}\left (i x^{2}\right )^{3}+\frac {i \pi \,{\mathrm e}^{{\mathrm e}^{3}} x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-i \pi \,{\mathrm e}^{{\mathrm e}^{3}} x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi \,{\mathrm e}^{{\mathrm e}^{3}} x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+168 \,{\mathrm e}^{\frac {3}{5}} \ln \relax (x )-42 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \relax (x )+2 i \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2} \pi \,{\mathrm e}^{x +\frac {3}{5}}-4 i \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{x +\frac {3}{5}}+2 i \mathrm {csgn}\left (i x^{2}\right )^{3} \pi \,{\mathrm e}^{x +\frac {3}{5}}-\frac {i \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2} \pi \,{\mathrm e}^{{\mathrm e}^{3}+x}}{2}+i \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{{\mathrm e}^{3}+x}-\frac {i \mathrm {csgn}\left (i x^{2}\right )^{3} \pi \,{\mathrm e}^{{\mathrm e}^{3}+x}}{2}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(x)*x-x)*exp(exp(3))-4*x*exp(3/5)*exp(x)+4*x*exp(3/5))*ln(x^2)+(2*exp(x)-2*x-42)*exp(exp(3))-8*exp(3
/5)*exp(x)+(8*x+168)*exp(3/5))/x,x,method=_RETURNVERBOSE)

[Out]

(-21*exp(exp(3))+84*exp(3/5))*ln(x^2)+(-exp(exp(3))+4*exp(3/5))*x*ln(x^2)+(exp(exp(3))-4*exp(3/5))*exp(x)*ln(x
^2)

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maxima [B]  time = 0.57, size = 74, normalized size = 2.85 \begin {gather*} 4 \, {\left (x \log \left (x^{2}\right ) - 2 \, x\right )} e^{\frac {3}{5}} + 8 \, x e^{\frac {3}{5}} - {\left (x \log \left (x^{2}\right ) - 2 \, x\right )} e^{\left (e^{3}\right )} - 2 \, x e^{\left (e^{3}\right )} + e^{\left (x + e^{3}\right )} \log \left (x^{2}\right ) - 4 \, e^{\left (x + \frac {3}{5}\right )} \log \left (x^{2}\right ) + 168 \, e^{\frac {3}{5}} \log \relax (x) - 42 \, e^{\left (e^{3}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-x)*exp(exp(3))-4*x*exp(3/5)*exp(x)+4*x*exp(3/5))*log(x^2)+(2*exp(x)-2*x-42)*exp(exp(3))-
8*exp(3/5)*exp(x)+(8*x+168)*exp(3/5))/x,x, algorithm="maxima")

[Out]

4*(x*log(x^2) - 2*x)*e^(3/5) + 8*x*e^(3/5) - (x*log(x^2) - 2*x)*e^(e^3) - 2*x*e^(e^3) + e^(x + e^3)*log(x^2) -
4*e^(x + 3/5)*log(x^2) + 168*e^(3/5)*log(x) - 42*e^(e^3)*log(x)

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mupad [B]  time = 2.09, size = 22, normalized size = 0.85 \begin {gather*} \ln \left (x^2\right )\,\left (4\,{\mathrm {e}}^{3/5}-{\mathrm {e}}^{{\mathrm {e}}^3}\right )\,\left (x-{\mathrm {e}}^x+21\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(exp(exp(3))*(x - x*exp(x)) - 4*x*exp(3/5) + 4*x*exp(3/5)*exp(x)) + exp(exp(3))*(2*x - 2*exp(x)
+ 42) + 8*exp(3/5)*exp(x) - exp(3/5)*(8*x + 168))/x,x)

[Out]

log(x^2)*(4*exp(3/5) - exp(exp(3)))*(x - exp(x) + 21)

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sympy [B]  time = 0.52, size = 61, normalized size = 2.35 \begin {gather*} \left (- x e^{e^{3}} + 4 x e^{\frac {3}{5}}\right ) \log {\left (x^{2} \right )} + \left (- 4 e^{\frac {3}{5}} \log {\left (x^{2} \right )} + e^{e^{3}} \log {\left (x^{2} \right )}\right ) e^{x} + \left (- 42 e^{e^{3}} + 168 e^{\frac {3}{5}}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-x)*exp(exp(3))-4*x*exp(3/5)*exp(x)+4*x*exp(3/5))*ln(x**2)+(2*exp(x)-2*x-42)*exp(exp(3))-
8*exp(3/5)*exp(x)+(8*x+168)*exp(3/5))/x,x)

[Out]

(-x*exp(exp(3)) + 4*x*exp(3/5))*log(x**2) + (-4*exp(3/5)*log(x**2) + exp(exp(3))*log(x**2))*exp(x) + (-42*exp(
exp(3)) + 168*exp(3/5))*log(x)

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