3.23.37 \(\int \frac {32-12 x-2 x^2+e^x (-320-400 x-85 x^2-5 x^3)+16 \log (5 e^{-x} x^2)}{320+80 x+5 x^2} \, dx\)

Optimal. Leaf size=30 \[ -e^x x+\frac {4 x \log \left (5 e^{-x} x^2\right )}{5 (16+2 x)} \]

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Rubi [A]  time = 0.24, antiderivative size = 43, normalized size of antiderivative = 1.43, number of steps used = 13, number of rules used = 8, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {27, 12, 6742, 2176, 2194, 43, 2551, 72} \begin {gather*} -\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (x+8)}-\frac {2 x}{5}+e^x-e^x (x+1)+\frac {4 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 12*x - 2*x^2 + E^x*(-320 - 400*x - 85*x^2 - 5*x^3) + 16*Log[(5*x^2)/E^x])/(320 + 80*x + 5*x^2),x]

[Out]

E^x - (2*x)/5 - E^x*(1 + x) + (4*Log[x])/5 - (16*Log[(5*x^2)/E^x])/(5*(8 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

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 {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{(8+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-5 e^x (1+x)+\frac {32}{(8+x)^2}-\frac {12 x}{(8+x)^2}-\frac {2 x^2}{(8+x)^2}+\frac {16 \log \left (5 e^{-x} x^2\right )}{(8+x)^2}\right ) \, dx\\ &=-\frac {32}{5 (8+x)}-\frac {2}{5} \int \frac {x^2}{(8+x)^2} \, dx-\frac {12}{5} \int \frac {x}{(8+x)^2} \, dx+\frac {16}{5} \int \frac {\log \left (5 e^{-x} x^2\right )}{(8+x)^2} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)-\frac {32}{5 (8+x)}-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}-\frac {2}{5} \int \left (1+\frac {64}{(8+x)^2}-\frac {16}{8+x}\right ) \, dx-\frac {12}{5} \int \left (-\frac {8}{(8+x)^2}+\frac {1}{8+x}\right ) \, dx+\frac {16}{5} \int \frac {2-x}{x (8+x)} \, dx+\int e^x \, dx\\ &=e^x-\frac {2 x}{5}-e^x (1+x)-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}+4 \log (8+x)+\frac {16}{5} \int \left (\frac {1}{4 x}-\frac {5}{4 (8+x)}\right ) \, dx\\ &=e^x-\frac {2 x}{5}-e^x (1+x)+\frac {4 \log (x)}{5}-\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (8+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 36, normalized size = 1.20 \begin {gather*} \frac {1}{5} \left (-2 x-5 e^x x+4 \log (x)-\frac {16 \log \left (5 e^{-x} x^2\right )}{8+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 12*x - 2*x^2 + E^x*(-320 - 400*x - 85*x^2 - 5*x^3) + 16*Log[(5*x^2)/E^x])/(320 + 80*x + 5*x^2)
,x]

[Out]

(-2*x - 5*E^x*x + 4*Log[x] - (16*Log[(5*x^2)/E^x])/(8 + x))/5

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fricas [A]  time = 0.71, size = 32, normalized size = 1.07 \begin {gather*} -\frac {5 \, {\left (x^{2} + 8 \, x\right )} e^{x} - 2 \, x \log \left (5 \, x^{2} e^{\left (-x\right )}\right )}{5 \, {\left (x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/(5*x^2+80*x+320),x, algorithm=
"fricas")

[Out]

-1/5*(5*(x^2 + 8*x)*e^x - 2*x*log(5*x^2*e^(-x)))/(x + 8)

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giac [A]  time = 0.24, size = 46, normalized size = 1.53 \begin {gather*} -\frac {5 \, x^{2} e^{x} + 2 \, x^{2} + 40 \, x e^{x} - 4 \, x \log \relax (x) + 16 \, x + 16 \, \log \left (5 \, x^{2}\right ) - 32 \, \log \relax (x) + 128}{5 \, {\left (x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/(5*x^2+80*x+320),x, algorithm=
"giac")

[Out]

-1/5*(5*x^2*e^x + 2*x^2 + 40*x*e^x - 4*x*log(x) + 16*x + 16*log(5*x^2) - 32*log(x) + 128)/(x + 8)

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maple [A]  time = 0.49, size = 31, normalized size = 1.03

method	result	size
default	\(-{\mathrm e}^{x} x -\frac {2 x}{5}-\frac {16 \ln \left (5 x^{2} {\mathrm e}^{-x}\right )}{5 \left (x +8\right )}+\frac {4 \ln \relax (x )}{5}\)	\(31\)
norman	\(\frac {-8 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}+\frac {2 \ln \left (5 x^{2} {\mathrm e}^{-x}\right ) x}{5}}{x +8}\)	\(33\)
risch	\(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{5 \left (x +8\right )}-\frac {-8 i \pi \,\mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{3}+8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i x^{2}\right )+16 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-8 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+8 i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-8 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+5 \,{\mathrm e}^{x} x^{2}-4 x \ln \relax (x )+2 x^{2}+40 \,{\mathrm e}^{x} x +16 \ln \relax (5)+16 x}{5 \left (x +8\right )}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*ln(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/(5*x^2+80*x+320),x,method=_RETURNVERB
OSE)

[Out]

-exp(x)*x-2/5*x-16/5/(x+8)*ln(5*x^2/exp(x))+4/5*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {2}{5} \, x + \frac {64 \, e^{\left (-8\right )} E_{2}\left (-x - 8\right )}{x + 8} + \frac {4 \, {\left (x \log \relax (x) - 4 \, \log \relax (5) - 32\right )}}{5 \, {\left (x + 8\right )}} - \frac {1}{5} \, \int \frac {5 \, {\left (x^{3} + 17 \, x^{2} + 80 \, x\right )} e^{x}}{x^{2} + 16 \, x + 64}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/(5*x^2+80*x+320),x, algorithm=
"maxima")

[Out]

-2/5*x + 64*e^(-8)*exp_integral_e(2, -x - 8)/(x + 8) + 4/5*(x*log(x) - 4*log(5) - 32)/(x + 8) - 1/5*integrate(
5*(x^3 + 17*x^2 + 80*x)*e^x/(x^2 + 16*x + 64), x)

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mupad [B]  time = 1.45, size = 47, normalized size = 1.57 \begin {gather*} \frac {4\,\ln \relax (x)}{5}-\frac {2\,x}{5}-\frac {16\,\ln \relax (5)}{5\,\left (x+8\right )}+\frac {16\,x}{5\,\left (x+8\right )}-x\,{\mathrm {e}}^x-\frac {16\,\ln \left (x^2\right )}{5\,\left (x+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x - 16*log(5*x^2*exp(-x)) + 2*x^2 + exp(x)*(400*x + 85*x^2 + 5*x^3 + 320) - 32)/(80*x + 5*x^2 + 320),
x)

[Out]

(4*log(x))/5 - (2*x)/5 - (16*log(5))/(5*(x + 8)) + (16*x)/(5*(x + 8)) - x*exp(x) - (16*log(x^2))/(5*(x + 8))

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sympy [A]  time = 0.26, size = 32, normalized size = 1.07 \begin {gather*} - x e^{x} - \frac {2 x}{5} + \frac {4 \log {\relax (x )}}{5} - \frac {16 \log {\left (5 x^{2} e^{- x} \right )}}{5 x + 40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*ln(5*x**2/exp(x))+(-5*x**3-85*x**2-400*x-320)*exp(x)-2*x**2-12*x+32)/(5*x**2+80*x+320),x)

[Out]

-x*exp(x) - 2*x/5 + 4*log(x)/5 - 16*log(5*x**2*exp(-x))/(5*x + 40)

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