3.87.17 \(\int \frac {(6 x+12 x^2+4 x^3) \log ^2(3 x)+e^{\frac {2 (-x^2+x^3)}{\log (3 x)}} (6 x^2+2 x^3-6 x^4-2 x^5+(-12 x^2+2 x^3+20 x^4+6 x^5) \log (3 x)+(3+8 x+3 x^2) \log ^2(3 x))}{\log ^2(3 x)} \, dx\)

Optimal. Leaf size=32 \[ \left (e^{\frac {2 x \left (-x+x^2\right )}{\log (3 x)}} x+x^2\right ) \left (-1+(2+x)^2\right ) \]

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Rubi [B]  time = 10.13, antiderivative size = 133, normalized size of antiderivative = 4.16, number of steps used = 5, number of rules used = 3, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6742, 14, 2288} \begin {gather*} x^4+4 x^3+3 x^2+\frac {e^{-\frac {2 (1-x) x^2}{\log (3 x)}} \left (-x^5+3 x^5 \log (3 x)-3 x^4+10 x^4 \log (3 x)+x^3+x^3 \log (3 x)+3 x^2-6 x^2 \log (3 x)\right )}{\left (\frac {x^2}{\log (3 x)}+\frac {(1-x) x}{\log ^2(3 x)}-\frac {2 (1-x) x}{\log (3 x)}\right ) \log ^2(3 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((6*x + 12*x^2 + 4*x^3)*Log[3*x]^2 + E^((2*(-x^2 + x^3))/Log[3*x])*(6*x^2 + 2*x^3 - 6*x^4 - 2*x^5 + (-12*x
^2 + 2*x^3 + 20*x^4 + 6*x^5)*Log[3*x] + (3 + 8*x + 3*x^2)*Log[3*x]^2))/Log[3*x]^2,x]

[Out]

3*x^2 + 4*x^3 + x^4 + (3*x^2 + x^3 - 3*x^4 - x^5 - 6*x^2*Log[3*x] + x^3*Log[3*x] + 10*x^4*Log[3*x] + 3*x^5*Log
[3*x])/(E^((2*(1 - x)*x^2)/Log[3*x])*(((1 - x)*x)/Log[3*x]^2 - (2*(1 - x)*x)/Log[3*x] + x^2/Log[3*x])*Log[3*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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 x \left (3+6 x+2 x^2\right )+\frac {e^{\frac {2 (-1+x) x^2}{\log (3 x)}} \left (6 x^2+2 x^3-6 x^4-2 x^5-12 x^2 \log (3 x)+2 x^3 \log (3 x)+20 x^4 \log (3 x)+6 x^5 \log (3 x)+3 \log ^2(3 x)+8 x \log ^2(3 x)+3 x^2 \log ^2(3 x)\right )}{\log ^2(3 x)}\right ) \, dx\\ &=2 \int x \left (3+6 x+2 x^2\right ) \, dx+\int \frac {e^{\frac {2 (-1+x) x^2}{\log (3 x)}} \left (6 x^2+2 x^3-6 x^4-2 x^5-12 x^2 \log (3 x)+2 x^3 \log (3 x)+20 x^4 \log (3 x)+6 x^5 \log (3 x)+3 \log ^2(3 x)+8 x \log ^2(3 x)+3 x^2 \log ^2(3 x)\right )}{\log ^2(3 x)} \, dx\\ &=\frac {e^{-\frac {2 (1-x) x^2}{\log (3 x)}} \left (3 x^2+x^3-3 x^4-x^5-6 x^2 \log (3 x)+x^3 \log (3 x)+10 x^4 \log (3 x)+3 x^5 \log (3 x)\right )}{\left (\frac {(1-x) x}{\log ^2(3 x)}-\frac {2 (1-x) x}{\log (3 x)}+\frac {x^2}{\log (3 x)}\right ) \log ^2(3 x)}+2 \int \left (3 x+6 x^2+2 x^3\right ) \, dx\\ &=3 x^2+4 x^3+x^4+\frac {e^{-\frac {2 (1-x) x^2}{\log (3 x)}} \left (3 x^2+x^3-3 x^4-x^5-6 x^2 \log (3 x)+x^3 \log (3 x)+10 x^4 \log (3 x)+3 x^5 \log (3 x)\right )}{\left (\frac {(1-x) x}{\log ^2(3 x)}-\frac {2 (1-x) x}{\log (3 x)}+\frac {x^2}{\log (3 x)}\right ) \log ^2(3 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 40, normalized size = 1.25 \begin {gather*} 3 x^2+4 x^3+x^4+e^{\frac {2 (-1+x) x^2}{\log (3 x)}} x \left (3+4 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6*x + 12*x^2 + 4*x^3)*Log[3*x]^2 + E^((2*(-x^2 + x^3))/Log[3*x])*(6*x^2 + 2*x^3 - 6*x^4 - 2*x^5 +
(-12*x^2 + 2*x^3 + 20*x^4 + 6*x^5)*Log[3*x] + (3 + 8*x + 3*x^2)*Log[3*x]^2))/Log[3*x]^2,x]

[Out]

3*x^2 + 4*x^3 + x^4 + E^((2*(-1 + x)*x^2)/Log[3*x])*x*(3 + 4*x + x^2)

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fricas [A]  time = 1.18, size = 45, normalized size = 1.41 \begin {gather*} x^{4} + 4 \, x^{3} + 3 \, x^{2} + {\left (x^{3} + 4 \, x^{2} + 3 \, x\right )} e^{\left (\frac {2 \, {\left (x^{3} - x^{2}\right )}}{\log \left (3 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+8*x+3)*log(3*x)^2+(6*x^5+20*x^4+2*x^3-12*x^2)*log(3*x)-2*x^5-6*x^4+2*x^3+6*x^2)*exp((x^3-x^
2)/log(3*x))^2+(4*x^3+12*x^2+6*x)*log(3*x)^2)/log(3*x)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.16, size = 80, normalized size = 2.50 \begin {gather*} x^{4} + x^{3} e^{\left (\frac {2 \, {\left (x^{3} - x^{2}\right )}}{\log \left (3 \, x\right )}\right )} + 4 \, x^{3} + 4 \, x^{2} e^{\left (\frac {2 \, {\left (x^{3} - x^{2}\right )}}{\log \left (3 \, x\right )}\right )} + 3 \, x^{2} + 3 \, x e^{\left (\frac {2 \, {\left (x^{3} - x^{2}\right )}}{\log \left (3 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+8*x+3)*log(3*x)^2+(6*x^5+20*x^4+2*x^3-12*x^2)*log(3*x)-2*x^5-6*x^4+2*x^3+6*x^2)*exp((x^3-x^
2)/log(3*x))^2+(4*x^3+12*x^2+6*x)*log(3*x)^2)/log(3*x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 43, normalized size = 1.34

method	result	size
risch	\(x^{4}+4 x^{3}+3 x^{2}+\left (x^{3}+4 x^{2}+3 x \right ) {\mathrm e}^{\frac {2 x^{2} \left (x -1\right )}{\ln \left (3 x \right )}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2+8*x+3)*ln(3*x)^2+(6*x^5+20*x^4+2*x^3-12*x^2)*ln(3*x)-2*x^5-6*x^4+2*x^3+6*x^2)*exp((x^3-x^2)/ln(3*
x))^2+(4*x^3+12*x^2+6*x)*ln(3*x)^2)/ln(3*x)^2,x,method=_RETURNVERBOSE)

[Out]

x^4+4*x^3+3*x^2+(x^3+4*x^2+3*x)*exp(2*x^2*(x-1)/ln(3*x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+8*x+3)*log(3*x)^2+(6*x^5+20*x^4+2*x^3-12*x^2)*log(3*x)-2*x^5-6*x^4+2*x^3+6*x^2)*exp((x^3-x^
2)/log(3*x))^2+(4*x^3+12*x^2+6*x)*log(3*x)^2)/log(3*x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\ln \left (3\,x\right )}^2\,\left (4\,x^3+12\,x^2+6\,x\right )+{\mathrm {e}}^{-\frac {2\,\left (x^2-x^3\right )}{\ln \left (3\,x\right )}}\,\left ({\ln \left (3\,x\right )}^2\,\left (3\,x^2+8\,x+3\right )+6\,x^2+2\,x^3-6\,x^4-2\,x^5+\ln \left (3\,x\right )\,\left (6\,x^5+20\,x^4+2\,x^3-12\,x^2\right )\right )}{{\ln \left (3\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3*x)^2*(6*x + 12*x^2 + 4*x^3) + exp(-(2*(x^2 - x^3))/log(3*x))*(log(3*x)^2*(8*x + 3*x^2 + 3) + 6*x^2
+ 2*x^3 - 6*x^4 - 2*x^5 + log(3*x)*(2*x^3 - 12*x^2 + 20*x^4 + 6*x^5)))/log(3*x)^2,x)

[Out]

int((log(3*x)^2*(6*x + 12*x^2 + 4*x^3) + exp(-(2*(x^2 - x^3))/log(3*x))*(log(3*x)^2*(8*x + 3*x^2 + 3) + 6*x^2
+ 2*x^3 - 6*x^4 - 2*x^5 + log(3*x)*(2*x^3 - 12*x^2 + 20*x^4 + 6*x^5)))/log(3*x)^2, x)

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sympy [A]  time = 15.85, size = 41, normalized size = 1.28 \begin {gather*} x^{4} + 4 x^{3} + 3 x^{2} + \left (x^{3} + 4 x^{2} + 3 x\right ) e^{\frac {2 \left (x^{3} - x^{2}\right )}{\log {\left (3 x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2+8*x+3)*ln(3*x)**2+(6*x**5+20*x**4+2*x**3-12*x**2)*ln(3*x)-2*x**5-6*x**4+2*x**3+6*x**2)*exp
((x**3-x**2)/ln(3*x))**2+(4*x**3+12*x**2+6*x)*ln(3*x)**2)/ln(3*x)**2,x)

[Out]

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

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