3.49.56 \(\int \frac {10 e^{36 x^4}+x^2 \log (4 x^2)+e^{36 x^4} (-5+720 x^4) \log (4 x^2) \log (\log (4 x^2))}{x^2 \log (4 x^2)} \, dx\)

Optimal. Leaf size=21 \[ x+\frac {5 e^{36 x^4} \log \left (\log \left (4 x^2\right )\right )}{x} \]

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Rubi [A]  time = 0.55, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6742, 2288} \begin {gather*} \frac {5 e^{36 x^4} \log \left (\log \left (4 x^2\right )\right )}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E^(36*x^4) + x^2*Log[4*x^2] + E^(36*x^4)*(-5 + 720*x^4)*Log[4*x^2]*Log[Log[4*x^2]])/(x^2*Log[4*x^2]),x
]

[Out]

x + (5*E^(36*x^4)*Log[Log[4*x^2]])/x

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 (1+\frac {5 e^{36 x^4} \left (2-\log \left (4 x^2\right ) \log \left (\log \left (4 x^2\right )\right )+144 x^4 \log \left (4 x^2\right ) \log \left (\log \left (4 x^2\right )\right )\right )}{x^2 \log \left (4 x^2\right )}\right ) \, dx\\ &=x+5 \int \frac {e^{36 x^4} \left (2-\log \left (4 x^2\right ) \log \left (\log \left (4 x^2\right )\right )+144 x^4 \log \left (4 x^2\right ) \log \left (\log \left (4 x^2\right )\right )\right )}{x^2 \log \left (4 x^2\right )} \, dx\\ &=x+\frac {5 e^{36 x^4} \log \left (\log \left (4 x^2\right )\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.00 \begin {gather*} x+\frac {5 e^{36 x^4} \log \left (\log \left (4 x^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^(36*x^4) + x^2*Log[4*x^2] + E^(36*x^4)*(-5 + 720*x^4)*Log[4*x^2]*Log[Log[4*x^2]])/(x^2*Log[4*x
^2]),x]

[Out]

x + (5*E^(36*x^4)*Log[Log[4*x^2]])/x

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fricas [A]  time = 0.69, size = 23, normalized size = 1.10 \begin {gather*} \frac {x^{2} + 5 \, e^{\left (36 \, x^{4}\right )} \log \left (\log \left (4 \, x^{2}\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((720*x^4-5)*exp(36*x^4)*log(4*x^2)*log(log(4*x^2))+x^2*log(4*x^2)+10*exp(36*x^4))/x^2/log(4*x^2),x,
algorithm="fricas")

[Out]

(x^2 + 5*e^(36*x^4)*log(log(4*x^2)))/x

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giac [A]  time = 0.23, size = 23, normalized size = 1.10 \begin {gather*} \frac {x^{2} + 5 \, e^{\left (36 \, x^{4}\right )} \log \left (\log \left (4 \, x^{2}\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((720*x^4-5)*exp(36*x^4)*log(4*x^2)*log(log(4*x^2))+x^2*log(4*x^2)+10*exp(36*x^4))/x^2/log(4*x^2),x,
algorithm="giac")

[Out]

(x^2 + 5*e^(36*x^4)*log(log(4*x^2)))/x

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maple [C]  time = 0.45, size = 52, normalized size = 2.48

method	result	size
risch	\(\frac {5 \,{\mathrm e}^{36 x^{4}} \ln \left (2 \ln \relax (2)+2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}{x}+x\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((720*x^4-5)*exp(36*x^4)*ln(4*x^2)*ln(ln(4*x^2))+x^2*ln(4*x^2)+10*exp(36*x^4))/x^2/ln(4*x^2),x,method=_RET
URNVERBOSE)

[Out]

5/x*exp(36*x^4)*ln(2*ln(2)+2*ln(x)-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2)+x

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maxima [A]  time = 0.50, size = 30, normalized size = 1.43 \begin {gather*} x + \frac {5 \, {\left (e^{\left (36 \, x^{4}\right )} \log \relax (2) + e^{\left (36 \, x^{4}\right )} \log \left (\log \relax (2) + \log \relax (x)\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((720*x^4-5)*exp(36*x^4)*log(4*x^2)*log(log(4*x^2))+x^2*log(4*x^2)+10*exp(36*x^4))/x^2/log(4*x^2),x,
algorithm="maxima")

[Out]

x + 5*(e^(36*x^4)*log(2) + e^(36*x^4)*log(log(2) + log(x)))/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {10\,{\mathrm {e}}^{36\,x^4}+x^2\,\ln \left (4\,x^2\right )+\ln \left (\ln \left (4\,x^2\right )\right )\,{\mathrm {e}}^{36\,x^4}\,\ln \left (4\,x^2\right )\,\left (720\,x^4-5\right )}{x^2\,\ln \left (4\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(36*x^4) + x^2*log(4*x^2) + log(log(4*x^2))*exp(36*x^4)*log(4*x^2)*(720*x^4 - 5))/(x^2*log(4*x^2)),
x)

[Out]

int((10*exp(36*x^4) + x^2*log(4*x^2) + log(log(4*x^2))*exp(36*x^4)*log(4*x^2)*(720*x^4 - 5))/(x^2*log(4*x^2)),
x)

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sympy [A]  time = 0.40, size = 19, normalized size = 0.90 \begin {gather*} x + \frac {5 e^{36 x^{4}} \log {\left (\log {\left (4 x^{2} \right )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((720*x**4-5)*exp(36*x**4)*ln(4*x**2)*ln(ln(4*x**2))+x**2*ln(4*x**2)+10*exp(36*x**4))/x**2/ln(4*x**2
),x)

[Out]

x + 5*exp(36*x**4)*log(log(4*x**2))/x

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