3.100.47 \(\int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log (x^2)}} (-6 x^2-22 x^3-24 x^4-8 x^5+(6 x^2+21 x^3+28 x^4+12 x^5) \log (x^2)+(-8-32 x-32 x^2) \log ^2(x^2))}{(4 x^3+16 x^4+16 x^5) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{\frac {x \left (2 x+x^2+\frac {x}{1+2 x}\right )}{4 \log \left (x^2\right )}}}{x^2} \]

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Rubi [B]  time = 1.27, antiderivative size = 197, normalized size of antiderivative = 5.97, number of steps used = 4, number of rules used = 4, integrand size = 121, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {1594, 27, 12, 2288} \begin {gather*} \frac {\left (8 x^5+24 x^4+22 x^3+6 x^2-\left (12 x^5+28 x^4+21 x^3+6 x^2\right ) \log \left (x^2\right )\right ) \exp \left (\frac {2 x^4+5 x^3+3 x^2}{4 (2 x+1) \log \left (x^2\right )}\right )}{x^3 (2 x+1)^2 \left (-\frac {8 x^3+15 x^2+6 x}{(2 x+1) \log \left (x^2\right )}+\frac {2 \left (2 x^4+5 x^3+3 x^2\right )}{x (2 x+1) \log ^2\left (x^2\right )}+\frac {2 \left (2 x^4+5 x^3+3 x^2\right )}{(2 x+1)^2 \log \left (x^2\right )}\right ) \log ^2\left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3*x^2 + 5*x^3 + 2*x^4)/((4 + 8*x)*Log[x^2]))*(-6*x^2 - 22*x^3 - 24*x^4 - 8*x^5 + (6*x^2 + 21*x^3 + 28
*x^4 + 12*x^5)*Log[x^2] + (-8 - 32*x - 32*x^2)*Log[x^2]^2))/((4*x^3 + 16*x^4 + 16*x^5)*Log[x^2]^2),x]

[Out]

(E^((3*x^2 + 5*x^3 + 2*x^4)/(4*(1 + 2*x)*Log[x^2]))*(6*x^2 + 22*x^3 + 24*x^4 + 8*x^5 - (6*x^2 + 21*x^3 + 28*x^
4 + 12*x^5)*Log[x^2]))/(x^3*(1 + 2*x)^2*((2*(3*x^2 + 5*x^3 + 2*x^4))/(x*(1 + 2*x)*Log[x^2]^2) - (6*x + 15*x^2
+ 8*x^3)/((1 + 2*x)*Log[x^2]) + (2*(3*x^2 + 5*x^3 + 2*x^4))/((1 + 2*x)^2*Log[x^2]))*Log[x^2]^2)

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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 \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{x^3 \left (4+16 x+16 x^2\right ) \log ^2\left (x^2\right )} \, dx\\ &=\int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{4 x^3 (1+2 x)^2 \log ^2\left (x^2\right )} \, dx\\ &=\frac {1}{4} \int \frac {e^{\frac {3 x^2+5 x^3+2 x^4}{(4+8 x) \log \left (x^2\right )}} \left (-6 x^2-22 x^3-24 x^4-8 x^5+\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )+\left (-8-32 x-32 x^2\right ) \log ^2\left (x^2\right )\right )}{x^3 (1+2 x)^2 \log ^2\left (x^2\right )} \, dx\\ &=\frac {\exp \left (\frac {3 x^2+5 x^3+2 x^4}{4 (1+2 x) \log \left (x^2\right )}\right ) \left (6 x^2+22 x^3+24 x^4+8 x^5-\left (6 x^2+21 x^3+28 x^4+12 x^5\right ) \log \left (x^2\right )\right )}{x^3 (1+2 x)^2 \left (\frac {2 \left (3 x^2+5 x^3+2 x^4\right )}{x (1+2 x) \log ^2\left (x^2\right )}-\frac {6 x+15 x^2+8 x^3}{(1+2 x) \log \left (x^2\right )}+\frac {2 \left (3 x^2+5 x^3+2 x^4\right )}{(1+2 x)^2 \log \left (x^2\right )}\right ) \log ^2\left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 36, normalized size = 1.09 \begin {gather*} \frac {e^{\frac {x^2 \left (3+5 x+2 x^2\right )}{4 (1+2 x) \log \left (x^2\right )}}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x^2 + 5*x^3 + 2*x^4)/((4 + 8*x)*Log[x^2]))*(-6*x^2 - 22*x^3 - 24*x^4 - 8*x^5 + (6*x^2 + 21*x^
3 + 28*x^4 + 12*x^5)*Log[x^2] + (-8 - 32*x - 32*x^2)*Log[x^2]^2))/((4*x^3 + 16*x^4 + 16*x^5)*Log[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.61, size = 36, normalized size = 1.09 \begin {gather*} \frac {e^{\left (\frac {2 \, x^{4} + 5 \, x^{3} + 3 \, x^{2}}{4 \, {\left (2 \, x + 1\right )} \log \left (x^{2}\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2
*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(16*x^5+16*x^4+4*x^3)/log(x^2)^2,x, algorithm="fricas")

[Out]

e^(1/4*(2*x^4 + 5*x^3 + 3*x^2)/((2*x + 1)*log(x^2)))/x^2

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giac [A]  time = 1.47, size = 37, normalized size = 1.12 \begin {gather*} \frac {e^{\left (\frac {2 \, x^{4} + 5 \, x^{3} + 3 \, x^{2}}{4 \, {\left (2 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )\right )}}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2
*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(16*x^5+16*x^4+4*x^3)/log(x^2)^2,x, algorithm="giac")

[Out]

e^(1/4*(2*x^4 + 5*x^3 + 3*x^2)/(2*x*log(x^2) + log(x^2)))/x^2

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maple [A]  time = 0.08, size = 32, normalized size = 0.97

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {x^{2} \left (2 x +3\right ) \left (x +1\right )}{4 \left (2 x +1\right ) \ln \left (x^{2}\right )}}}{x^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^2-32*x-8)*ln(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*ln(x^2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2*x^4+5*x
^3+3*x^2)/(8*x+4)/ln(x^2))/(16*x^5+16*x^4+4*x^3)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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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(((-32*x^2-32*x-8)*log(x^2)^2+(12*x^5+28*x^4+21*x^3+6*x^2)*log(x^2)-8*x^5-24*x^4-22*x^3-6*x^2)*exp((2
*x^4+5*x^3+3*x^2)/(8*x+4)/log(x^2))/(16*x^5+16*x^4+4*x^3)/log(x^2)^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 [B]  time = 7.83, size = 69, normalized size = 2.09 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^4}{2\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}\,{\mathrm {e}}^{\frac {3\,x^2}{4\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}\,{\mathrm {e}}^{\frac {5\,x^3}{4\,\left (\ln \left (x^2\right )+2\,x\,\ln \left (x^2\right )\right )}}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((3*x^2 + 5*x^3 + 2*x^4)/(log(x^2)*(8*x + 4)))*(log(x^2)^2*(32*x + 32*x^2 + 8) + 6*x^2 + 22*x^3 + 24*
x^4 + 8*x^5 - log(x^2)*(6*x^2 + 21*x^3 + 28*x^4 + 12*x^5)))/(log(x^2)^2*(4*x^3 + 16*x^4 + 16*x^5)),x)

[Out]

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

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sympy [A]  time = 0.60, size = 29, normalized size = 0.88 \begin {gather*} \frac {e^{\frac {2 x^{4} + 5 x^{3} + 3 x^{2}}{\left (8 x + 4\right ) \log {\left (x^{2} \right )}}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**2-32*x-8)*ln(x**2)**2+(12*x**5+28*x**4+21*x**3+6*x**2)*ln(x**2)-8*x**5-24*x**4-22*x**3-6*x*
*2)*exp((2*x**4+5*x**3+3*x**2)/(8*x+4)/ln(x**2))/(16*x**5+16*x**4+4*x**3)/ln(x**2)**2,x)

[Out]

exp((2*x**4 + 5*x**3 + 3*x**2)/((8*x + 4)*log(x**2)))/x**2

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