3.96.4 \(\int \frac {96 x^3-192 x^3 \log (x^2)+(8+e^x) \log ^2(x^2)}{\log ^2(x^2)} \, dx\)

Optimal. Leaf size=24 \[ 4+e^x-x+3 x \left (3-\frac {16 x^3}{\log \left (x^2\right )}\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 10, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {6742, 2194, 2306, 2310, 2178} \begin {gather*} -\frac {48 x^4}{\log \left (x^2\right )}+8 x+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(96*x^3 - 192*x^3*Log[x^2] + (8 + E^x)*Log[x^2]^2)/Log[x^2]^2,x]

[Out]

E^x + 8*x - (48*x^4)/Log[x^2]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, 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 (e^x-\frac {8 \left (-12 x^3+24 x^3 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{\log ^2\left (x^2\right )}\right ) \, dx\\ &=-\left (8 \int \frac {-12 x^3+24 x^3 \log \left (x^2\right )-\log ^2\left (x^2\right )}{\log ^2\left (x^2\right )} \, dx\right )+\int e^x \, dx\\ &=e^x-8 \int \left (-1-\frac {12 x^3}{\log ^2\left (x^2\right )}+\frac {24 x^3}{\log \left (x^2\right )}\right ) \, dx\\ &=e^x+8 x+96 \int \frac {x^3}{\log ^2\left (x^2\right )} \, dx-192 \int \frac {x^3}{\log \left (x^2\right )} \, dx\\ &=e^x+8 x-\frac {48 x^4}{\log \left (x^2\right )}-96 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (x^2\right )\right )+192 \int \frac {x^3}{\log \left (x^2\right )} \, dx\\ &=e^x+8 x-96 \text {Ei}\left (2 \log \left (x^2\right )\right )-\frac {48 x^4}{\log \left (x^2\right )}+96 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (x^2\right )\right )\\ &=e^x+8 x-\frac {48 x^4}{\log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 0.75 \begin {gather*} e^x+8 x-\frac {48 x^4}{\log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(96*x^3 - 192*x^3*Log[x^2] + (8 + E^x)*Log[x^2]^2)/Log[x^2]^2,x]

[Out]

E^x + 8*x - (48*x^4)/Log[x^2]

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fricas [A]  time = 0.71, size = 26, normalized size = 1.08 \begin {gather*} -\frac {48 \, x^{4} - {\left (8 \, x + e^{x}\right )} \log \left (x^{2}\right )}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+8)*log(x^2)^2-192*x^3*log(x^2)+96*x^3)/log(x^2)^2,x, algorithm="fricas")

[Out]

-(48*x^4 - (8*x + e^x)*log(x^2))/log(x^2)

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giac [A]  time = 0.28, size = 29, normalized size = 1.21 \begin {gather*} -\frac {48 \, x^{4} - 8 \, x \log \left (x^{2}\right ) - e^{x} \log \left (x^{2}\right )}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+8)*log(x^2)^2-192*x^3*log(x^2)+96*x^3)/log(x^2)^2,x, algorithm="giac")

[Out]

-(48*x^4 - 8*x*log(x^2) - e^x*log(x^2))/log(x^2)

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maple [A]  time = 0.06, size = 18, normalized size = 0.75

method	result	size
default	\(8 x -\frac {48 x^{4}}{\ln \left (x^{2}\right )}+{\mathrm e}^{x}\)	\(18\)
risch	\(8 x +{\mathrm e}^{x}-\frac {96 i x^{4}}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)+8)*ln(x^2)^2-192*x^3*ln(x^2)+96*x^3)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

8*x-48*x^4/ln(x^2)+exp(x)

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maxima [A]  time = 0.39, size = 15, normalized size = 0.62 \begin {gather*} -\frac {24 \, x^{4}}{\log \relax (x)} + 8 \, x + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+8)*log(x^2)^2-192*x^3*log(x^2)+96*x^3)/log(x^2)^2,x, algorithm="maxima")

[Out]

-24*x^4/log(x) + 8*x + e^x

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mupad [B]  time = 7.27, size = 17, normalized size = 0.71 \begin {gather*} 8\,x+{\mathrm {e}}^x-\frac {48\,x^4}{\ln \left (x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)^2*(exp(x) + 8) - 192*x^3*log(x^2) + 96*x^3)/log(x^2)^2,x)

[Out]

8*x + exp(x) - (48*x^4)/log(x^2)

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sympy [A]  time = 0.25, size = 15, normalized size = 0.62 \begin {gather*} - \frac {48 x^{4}}{\log {\left (x^{2} \right )}} + 8 x + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+8)*ln(x**2)**2-192*x**3*ln(x**2)+96*x**3)/ln(x**2)**2,x)

[Out]

-48*x**4/log(x**2) + 8*x + exp(x)

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