\(\int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} (-8 x^2+2 x^3) \log (16) \log (4-x)+(20-5 x) \log (16) \log (2 x^3)}{-20 x^2+5 x^3} \, dx\) [7312]

Optimal result

Integrand size = 78, antiderivative size = 28 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\log (16) \left (e^{2 x/5} \log (4-x)+\frac {\log \left (2 x^3\right )}{x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1607, 6874, 2326, 2340} \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {\log (16) \log \left (2 x^3\right )}{x}+\frac {e^{2 x/5} \log (16) (4 \log (4-x)-x \log (4-x))}{4-x} \]

[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 1607

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 2326

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 2340

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {\log (16) \left (e^{2 x/5} x \log (4-x)+\log \left (2 x^3\right )\right )}{x} \]

[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

Maple [A] (verified)

Time = 8.55 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(\frac {60 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right ) x +60 \ln \left (2\right ) \ln \left (2 x^{3}\right )}{15 x}\)	\(34\)
default	\(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\)	\(51\)
parts	\(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )-\frac {12 \ln \left (2\right )}{x}-4 \ln \left (2\right ) \left (-\frac {\ln \left (x^{3}\right )}{x}-\frac {3}{x}-\frac {\ln \left (2\right )}{x}\right )\)	\(51\)
risch	\(4 \ln \left (2\right ) {\mathrm e}^{\frac {2 x}{5}} \ln \left (-x +4\right )+\frac {2 \ln \left (2\right ) \left (-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}+2 \ln \left (2\right )+6 \ln \left (x \right )\right )}{x}\)	\(155\)

[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]

1/15*(60*ln(2)*exp(1/5*x)^2*ln(-x+4)*x+60*ln(2)*ln(2*x^3))/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4 \, {\left (x e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \log \left (2\right ) \log \left (2 \, x^{3}\right )\right )}}{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 e^{\frac {2 x}{5}} \log {\left (2 \right )} \log {\left (4 - x \right )} + \frac {4 \log {\left (2 \right )} \log {\left (2 x^{3} \right )}}{x} \]

[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

Maxima [F]

\[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\int { \frac {4 \, {\left (5 \, x^{2} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) + 2 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) - 5 \, {\left (x - 4\right )} \log \left (2\right ) \log \left (2 \, x^{3}\right ) + 15 \, {\left (x - 4\right )} \log \left (2\right )\right )}}{5 \, {\left (x^{3} - 4 \, x^{2}\right )}} \,d x } \]

[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

Giac [A] (verification not implemented)

none

Time = 13.68 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=4 \, e^{\left (\frac {2}{5} \, x\right )} \log \left (2\right ) \log \left (-x + 4\right ) + \frac {4 \, \log \left (2\right )^{2}}{x} + \frac {4 \, \log \left (2\right ) \log \left ({\left (x - 4\right )}^{3} + 12 \, {\left (x - 4\right )}^{2} + 48 \, x - 128\right )}{x} \]

[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]

4*e^(2/5*x)*log(2)*log(-x + 4) + 4*log(2)^2/x + 4*log(2)*log((x - 4)^3 + 12*(x - 4)^2 + 48*x - 128)/x

Mupad [B] (verification not implemented)

Time = 13.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {5 e^{2 x/5} x^2 \log (16)+(-60+15 x) \log (16)+e^{2 x/5} \left (-8 x^2+2 x^3\right ) \log (16) \log (4-x)+(20-5 x) \log (16) \log \left (2 x^3\right )}{-20 x^2+5 x^3} \, dx=\frac {4\,\ln \left (2\right )\,\ln \left (2\,x^3\right )}{x}+4\,{\mathrm {e}}^{\frac {2\,x}{5}}\,\ln \left (2\right )\,\ln \left (4-x\right ) \]

[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)