3.74.12 \(\int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} (-8 x^2+2 x^3) \log (16) \log (4-x)+(20-5 x) \log (16) \log (2 x^3)}{-20 x^2+5 x^3} \, dx\)

Optimal. Leaf size=28 \[ \log (16) \left (e^{2 x/5} \log (4-x)+\frac {\log \left (2 x^3\right )}{x}\right ) \]

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Rubi [A]  time = 0.67, antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 5, number of rules used = 4, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1593, 6742, 2288, 2303} \begin {gather*} \frac {\log (16) \log \left (2 x^3\right )}{x}+\frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^((2*x)/5)*x^2*Log[16] + (-60 + 15*x)*Log[16] + E^((2*x)/5)*(-8*x^2 + 2*x^3)*Log[16]*Log[4 - x] + (20
- 5*x)*Log[16]*Log[2*x^3])/(-20*x^2 + 5*x^3),x]

[Out]

(E^((2*x)/5)*Log[16]*(4*Log[4 - x] - x*Log[4 - x]))/(4 - x) + (Log[16]*Log[2*x^3])/x

Rule 1593

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

Rule 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

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 \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{x^2 (-20+5 x)} \, dx\\ &=\int \left (\frac {e^{2 x/5} \log (16) (5-8 \log (4-x)+2 x \log (4-x))}{5 (-4+x)}-\frac {\log (16) \left (-3+\log \left (2 x^3\right )\right )}{x^2}\right ) \, dx\\ &=\frac {1}{5} \log (16) \int \frac {e^{2 x/5} (5-8 \log (4-x)+2 x \log (4-x))}{-4+x} \, dx-\log (16) \int \frac {-3+\log \left (2 x^3\right )}{x^2} \, dx\\ &=\frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x}+\frac {\log (16) \log \left (2 x^3\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 28, normalized size = 1.00 \begin {gather*} \frac {\log (16) \left (e^{2 x/5} x \log (4-x)+\log \left (2 x^3\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^((2*x)/5)*x^2*Log[16] + (-60 + 15*x)*Log[16] + E^((2*x)/5)*(-8*x^2 + 2*x^3)*Log[16]*Log[4 - x]
+ (20 - 5*x)*Log[16]*Log[2*x^3])/(-20*x^2 + 5*x^3),x]

[Out]

(Log[16]*(E^((2*x)/5)*x*Log[4 - x] + Log[2*x^3]))/x

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fricas [A]  time = 0.55, size = 29, normalized size = 1.04 \begin {gather*} \frac {4 \, {\left (x e^{\left (\frac {2}{5} \, x\right )} \log \relax (2) \log \left (-x + 4\right ) + \log \relax (2) \log \left (2 \, x^{3}\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x)^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x
)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x)^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x
)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.23, size = 38, normalized size = 1.36

method	result	size
default	\(4 \ln \relax (2) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )+\frac {4 \ln \relax (2) \ln \left (x^{3}\right )}{x}+\frac {4 \ln \relax (2)^{2}}{x}\)	\(38\)
risch	\(4 \ln \relax (2) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )+\frac {2 \ln \relax (2) \left (-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+2 \ln \relax (2)+6 \ln \relax (x )\right )}{x}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(-5*x+20)*ln(2)*ln(2*x^3)+4*(2*x^3-8*x^2)*ln(2)*exp(1/5*x)^2*ln(-x+4)+20*x^2*ln(2)*exp(1/5*x)^2+4*(15*x
-60)*ln(2))/(5*x^3-20*x^2),x,method=_RETURNVERBOSE)

[Out]

4*ln(2)*exp(1/5*x)^2*ln(-x+4)+4*ln(2)/x*ln(x^3)+4*ln(2)^2/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -4 \, e^{\frac {8}{5}} E_{1}\left (-\frac {2}{5} \, x + \frac {8}{5}\right ) \log \relax (2) - 3 \, {\left (\frac {4}{x} + \log \left (x - 4\right ) - \log \relax (x)\right )} \log \relax (2) + 3 \, {\left (\log \left (x - 4\right ) - \log \relax (x)\right )} \log \relax (2) - 4 \, \int \frac {e^{\left (\frac {2}{5} \, x\right )}}{x - 4}\,{d x} \log \relax (2) + \frac {4 \, {\left (x e^{\left (\frac {2}{5} \, x\right )} \log \relax (2) \log \left (-x + 4\right ) + \log \relax (2)^{2} + 3 \, \log \relax (2) \log \relax (x) + 3 \, \log \relax (2)\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(-5*x+20)*log(2)*log(2*x^3)+4*(2*x^3-8*x^2)*log(2)*exp(1/5*x)^2*log(-x+4)+20*x^2*log(2)*exp(1/5*x
)^2+4*(15*x-60)*log(2))/(5*x^3-20*x^2),x, algorithm="maxima")

[Out]

-4*e^(8/5)*exp_integral_e(1, -2/5*x + 8/5)*log(2) - 3*(4/x + log(x - 4) - log(x))*log(2) + 3*(log(x - 4) - log
(x))*log(2) - 4*integrate(e^(2/5*x)/(x - 4), x)*log(2) + 4*(x*e^(2/5*x)*log(2)*log(-x + 4) + log(2)^2 + 3*log(
2)*log(x) + 3*log(2))/x

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mupad [B]  time = 4.87, size = 28, normalized size = 1.00 \begin {gather*} \frac {4\,\ln \relax (2)\,\ln \left (2\,x^3\right )}{x}+4\,{\mathrm {e}}^{\frac {2\,x}{5}}\,\ln \relax (2)\,\ln \left (4-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(2)*(15*x - 60) - 4*log(2)*log(2*x^3)*(5*x - 20) + 20*x^2*exp((2*x)/5)*log(2) - 4*exp((2*x)/5)*log(
2)*log(4 - x)*(8*x^2 - 2*x^3))/(20*x^2 - 5*x^3),x)

[Out]

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

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sympy [A]  time = 0.97, size = 29, normalized size = 1.04 \begin {gather*} 4 e^{\frac {2 x}{5}} \log {\relax (2 )} \log {\left (4 - x \right )} + \frac {4 \log {\relax (2 )} \log {\left (2 x^{3} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(-5*x+20)*ln(2)*ln(2*x**3)+4*(2*x**3-8*x**2)*ln(2)*exp(1/5*x)**2*ln(-x+4)+20*x**2*ln(2)*exp(1/5*x
)**2+4*(15*x-60)*ln(2))/(5*x**3-20*x**2),x)

[Out]

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

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