3.79.59 \(\int \frac {160 x-200 x^2+e^{x/5} (320 x-168 x^2-40 x^3)+(320 x-200 x^2) \log (x)}{16-40 x+25 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {8 x^2 \left (e^{x/5}+\log (x)\right )}{\frac {4}{5}-x} \]

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Rubi [B]  time = 0.45, antiderivative size = 57, normalized size of antiderivative = 2.38, number of steps used = 20, number of rules used = 12, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {27, 6742, 2199, 2194, 2176, 2177, 2178, 43, 2357, 2295, 2314, 31} \begin {gather*} -8 e^{x/5} x-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}+\frac {32 x \log (x)}{4-5 x}-8 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(160*x - 200*x^2 + E^(x/5)*(320*x - 168*x^2 - 40*x^3) + (320*x - 200*x^2)*Log[x])/(16 - 40*x + 25*x^2),x]

[Out]

(-32*E^(x/5))/5 + (128*E^(x/5))/(5*(4 - 5*x)) - 8*E^(x/5)*x - 8*x*Log[x] + (32*x*Log[x])/(4 - 5*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 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 {160 x-200 x^2+e^{x/5} \left (320 x-168 x^2-40 x^3\right )+\left (320 x-200 x^2\right ) \log (x)}{(-4+5 x)^2} \, dx\\ &=\int \left (-\frac {8 e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2}-\frac {40 x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{x/5} x \left (-40+21 x+5 x^2\right )}{(-4+5 x)^2} \, dx\right )-40 \int \frac {x (-4+5 x-8 \log (x)+5 x \log (x))}{(-4+5 x)^2} \, dx\\ &=-\left (8 \int \left (\frac {29 e^{x/5}}{25}+\frac {1}{5} e^{x/5} x-\frac {16 e^{x/5}}{(-4+5 x)^2}+\frac {16 e^{x/5}}{25 (-4+5 x)}\right ) \, dx\right )-40 \int \left (\frac {x}{-4+5 x}+\frac {x (-8+5 x) \log (x)}{(-4+5 x)^2}\right ) \, dx\\ &=-\left (\frac {8}{5} \int e^{x/5} x \, dx\right )-\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx-\frac {232}{25} \int e^{x/5} \, dx-40 \int \frac {x}{-4+5 x} \, dx-40 \int \frac {x (-8+5 x) \log (x)}{(-4+5 x)^2} \, dx+128 \int \frac {e^{x/5}}{(-4+5 x)^2} \, dx\\ &=-\frac {232 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {128}{125} e^{4/25} \text {Ei}\left (\frac {1}{25} (-4+5 x)\right )+\frac {128}{25} \int \frac {e^{x/5}}{-4+5 x} \, dx+8 \int e^{x/5} \, dx-40 \int \left (\frac {1}{5}+\frac {4}{5 (-4+5 x)}\right ) \, dx-40 \int \left (\frac {\log (x)}{5}-\frac {16 \log (x)}{5 (-4+5 x)^2}\right ) \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 x-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 \int \log (x) \, dx+128 \int \frac {\log (x)}{(-4+5 x)^2} \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-\frac {32}{5} \log (4-5 x)-8 x \log (x)+\frac {32 x \log (x)}{4-5 x}+32 \int \frac {1}{-4+5 x} \, dx\\ &=-\frac {32 e^{x/5}}{5}+\frac {128 e^{x/5}}{5 (4-5 x)}-8 e^{x/5} x-8 x \log (x)+\frac {32 x \log (x)}{4-5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 22, normalized size = 0.92 \begin {gather*} -\frac {40 x^2 \left (e^{x/5}+\log (x)\right )}{-4+5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(160*x - 200*x^2 + E^(x/5)*(320*x - 168*x^2 - 40*x^3) + (320*x - 200*x^2)*Log[x])/(16 - 40*x + 25*x^
2),x]

[Out]

(-40*x^2*(E^(x/5) + Log[x]))/(-4 + 5*x)

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fricas [A]  time = 1.02, size = 24, normalized size = 1.00 \begin {gather*} -\frac {40 \, {\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} + x^{2} \log \relax (x)\right )}}{5 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x^2+320*x)*log(x)+(-40*x^3-168*x^2+320*x)*exp(1/5*x)-200*x^2+160*x)/(25*x^2-40*x+16),x, algor
ithm="fricas")

[Out]

-40*(x^2*e^(1/5*x) + x^2*log(x))/(5*x - 4)

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giac [A]  time = 0.19, size = 24, normalized size = 1.00 \begin {gather*} -\frac {40 \, {\left (x^{2} e^{\left (\frac {1}{5} \, x\right )} + x^{2} \log \relax (x)\right )}}{5 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x^2+320*x)*log(x)+(-40*x^3-168*x^2+320*x)*exp(1/5*x)-200*x^2+160*x)/(25*x^2-40*x+16),x, algor
ithm="giac")

[Out]

-40*(x^2*e^(1/5*x) + x^2*log(x))/(5*x - 4)

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maple [A]  time = 0.29, size = 26, normalized size = 1.08

method	result	size
norman	\(\frac {-40 x^{2} \ln \relax (x )-40 \,{\mathrm e}^{\frac {x}{5}} x^{2}}{5 x -4}\)	\(26\)
default	\(-8 x \ln \relax (x )-\frac {32 \ln \relax (x ) x}{5 x -4}-\frac {128 \,{\mathrm e}^{\frac {x}{5}}}{125 \left (\frac {x}{5}-\frac {4}{25}\right )}-\frac {32 \,{\mathrm e}^{\frac {x}{5}}}{5}-8 x \,{\mathrm e}^{\frac {x}{5}}\)	\(45\)
risch	\(-\frac {8 \left (25 x^{2}-20 x +16\right ) \ln \relax (x )}{5 \left (5 x -4\right )}-\frac {8 \left (25 \,{\mathrm e}^{\frac {x}{5}} x^{2}+20 x \ln \relax (x )-16 \ln \relax (x )\right )}{5 \left (5 x -4\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-200*x^2+320*x)*ln(x)+(-40*x^3-168*x^2+320*x)*exp(1/5*x)-200*x^2+160*x)/(25*x^2-40*x+16),x,method=_RETUR
NVERBOSE)

[Out]

(-40*x^2*ln(x)-40*exp(1/5*x)*x^2)/(5*x-4)

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maxima [B]  time = 0.40, size = 48, normalized size = 2.00 \begin {gather*} -8 \, x - \frac {8 \, {\left (25 \, x^{2} e^{\left (\frac {1}{5} \, x\right )} - 25 \, x^{2} + {\left (25 \, x^{2} - 20 \, x + 16\right )} \log \relax (x) + 20 \, x\right )}}{5 \, {\left (5 \, x - 4\right )}} - \frac {32}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x^2+320*x)*log(x)+(-40*x^3-168*x^2+320*x)*exp(1/5*x)-200*x^2+160*x)/(25*x^2-40*x+16),x, algor
ithm="maxima")

[Out]

-8*x - 8/5*(25*x^2*e^(1/5*x) - 25*x^2 + (25*x^2 - 20*x + 16)*log(x) + 20*x)/(5*x - 4) - 32/5*log(x)

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mupad [B]  time = 5.40, size = 19, normalized size = 0.79 \begin {gather*} -\frac {40\,x^2\,\left ({\mathrm {e}}^{x/5}+\ln \relax (x)\right )}{5\,x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((160*x - exp(x/5)*(168*x^2 - 320*x + 40*x^3) + log(x)*(320*x - 200*x^2) - 200*x^2)/(25*x^2 - 40*x + 16),x)

[Out]

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

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sympy [B]  time = 0.42, size = 39, normalized size = 1.62 \begin {gather*} - \frac {40 x^{2} e^{\frac {x}{5}}}{5 x - 4} - \frac {32 \log {\relax (x )}}{5} + \frac {\left (- 200 x^{2} + 160 x - 128\right ) \log {\relax (x )}}{25 x - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x**2+320*x)*ln(x)+(-40*x**3-168*x**2+320*x)*exp(1/5*x)-200*x**2+160*x)/(25*x**2-40*x+16),x)

[Out]

-40*x**2*exp(x/5)/(5*x - 4) - 32*log(x)/5 + (-200*x**2 + 160*x - 128)*log(x)/(25*x - 20)

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