3.19.13 \(\int \frac {68 x+32 x^2+4 x^3+e^{2 x} (-4-71 x-20 x^2)+e^{2 x} (-7 x-2 x^2) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx\)

Optimal. Leaf size=27 \[ x-\frac {-x+e^{2 x} (10+\log (-2 x))}{16+4 x} \]

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Rubi [A]  time = 1.54, antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 20, number of rules used = 9, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1594, 27, 12, 6742, 683, 2178, 2177, 2197, 2554} \begin {gather*} x-\frac {5 e^{2 x}}{2 (x+4)}-\frac {1}{x+4}-\frac {e^{2 x} \log (-2 x)}{4 (x+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(68*x + 32*x^2 + 4*x^3 + E^(2*x)*(-4 - 71*x - 20*x^2) + E^(2*x)*(-7*x - 2*x^2)*Log[-2*x])/(64*x + 32*x^2 +
4*x^3),x]

[Out]

x - (4 + x)^(-1) - (5*E^(2*x))/(2*(4 + x)) - (E^(2*x)*Log[-2*x])/(4*(4 + 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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 1594

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

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, 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 \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{x \left (64+32 x+4 x^2\right )} \, dx\\ &=\int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{4 x (4+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{x (4+x)^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {4 \left (17+8 x+x^2\right )}{(4+x)^2}-\frac {e^{2 x} \left (4+71 x+20 x^2+7 x \log (-2 x)+2 x^2 \log (-2 x)\right )}{x (4+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{2 x} \left (4+71 x+20 x^2+7 x \log (-2 x)+2 x^2 \log (-2 x)\right )}{x (4+x)^2} \, dx\right )+\int \frac {17+8 x+x^2}{(4+x)^2} \, dx\\ &=-\left (\frac {1}{4} \int \left (\frac {e^{2 x} \left (4+71 x+20 x^2\right )}{x (4+x)^2}+\frac {e^{2 x} (7+2 x) \log (-2 x)}{(4+x)^2}\right ) \, dx\right )+\int \left (1+\frac {1}{(4+x)^2}\right ) \, dx\\ &=x-\frac {1}{4+x}-\frac {1}{4} \int \frac {e^{2 x} \left (4+71 x+20 x^2\right )}{x (4+x)^2} \, dx-\frac {1}{4} \int \frac {e^{2 x} (7+2 x) \log (-2 x)}{(4+x)^2} \, dx\\ &=x-\frac {1}{4+x}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}+\frac {1}{4} \int \frac {e^{2 x}}{x (4+x)} \, dx-\frac {1}{4} \int \left (\frac {e^{2 x}}{4 x}-\frac {10 e^{2 x}}{(4+x)^2}+\frac {79 e^{2 x}}{4 (4+x)}\right ) \, dx\\ &=x-\frac {1}{4+x}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}-\frac {1}{16} \int \frac {e^{2 x}}{x} \, dx+\frac {1}{4} \int \left (\frac {e^{2 x}}{4 x}-\frac {e^{2 x}}{4 (4+x)}\right ) \, dx+\frac {5}{2} \int \frac {e^{2 x}}{(4+x)^2} \, dx-\frac {79}{16} \int \frac {e^{2 x}}{4+x} \, dx\\ &=x-\frac {1}{4+x}-\frac {5 e^{2 x}}{2 (4+x)}-\frac {\text {Ei}(2 x)}{16}-\frac {79 \text {Ei}(2 (4+x))}{16 e^8}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}+\frac {1}{16} \int \frac {e^{2 x}}{x} \, dx-\frac {1}{16} \int \frac {e^{2 x}}{4+x} \, dx+5 \int \frac {e^{2 x}}{4+x} \, dx\\ &=x-\frac {1}{4+x}-\frac {5 e^{2 x}}{2 (4+x)}-\frac {e^{2 x} \log (-2 x)}{4 (4+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.37, size = 36, normalized size = 1.33 \begin {gather*} -\frac {4+10 e^{2 x}-16 x-4 x^2+e^{2 x} \log (-2 x)}{4 (4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(68*x + 32*x^2 + 4*x^3 + E^(2*x)*(-4 - 71*x - 20*x^2) + E^(2*x)*(-7*x - 2*x^2)*Log[-2*x])/(64*x + 32
*x^2 + 4*x^3),x]

[Out]

-1/4*(4 + 10*E^(2*x) - 16*x - 4*x^2 + E^(2*x)*Log[-2*x])/(4 + x)

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fricas [A]  time = 0.84, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^2+68*x)/(4*x^3+32*x^2+64*x),x,
algorithm="fricas")

[Out]

1/4*(4*x^2 - e^(2*x)*log(-2*x) + 16*x - 10*e^(2*x) - 4)/(x + 4)

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giac [A]  time = 0.31, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^2+68*x)/(4*x^3+32*x^2+64*x),x,
algorithm="giac")

[Out]

1/4*(4*x^2 - e^(2*x)*log(-2*x) + 16*x - 10*e^(2*x) - 4)/(x + 4)

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maple [A]  time = 0.07, size = 28, normalized size = 1.04

method	result	size
norman	\(\frac {x^{2}-17-\frac {\ln \left (-2 x \right ) {\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{2 x}}{2}}{4+x}\)	\(28\)
default	\(\frac {-\ln \left (-2 x \right ) {\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}}{4 x +16}+x -\frac {1}{4+x}\)	\(34\)
risch	\(-\frac {{\mathrm e}^{2 x} \ln \left (-2 x \right )}{4 \left (4+x \right )}+\frac {2 x^{2}+8 x -5 \,{\mathrm e}^{2 x}-2}{2 x +8}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-7*x)*exp(2*x)*ln(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^2+68*x)/(4*x^3+32*x^2+64*x),x,method=
_RETURNVERBOSE)

[Out]

(x^2-17-1/4*ln(-2*x)*exp(2*x)-5/2*exp(2*x))/(4+x)

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maxima [A]  time = 0.95, size = 35, normalized size = 1.30 \begin {gather*} x - \frac {{\left (\log \relax (2) + 10\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (-x\right )}{4 \, {\left (x + 4\right )}} - \frac {1}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^2+68*x)/(4*x^3+32*x^2+64*x),x,
algorithm="maxima")

[Out]

x - 1/4*((log(2) + 10)*e^(2*x) + e^(2*x)*log(-x))/(x + 4) - 1/(x + 4)

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mupad [B]  time = 1.28, size = 33, normalized size = 1.22 \begin {gather*} \frac {17\,x-10\,{\mathrm {e}}^{2\,x}+4\,x^2-\ln \left (-2\,x\right )\,{\mathrm {e}}^{2\,x}}{4\,\left (x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((68*x - exp(2*x)*(71*x + 20*x^2 + 4) + 32*x^2 + 4*x^3 - log(-2*x)*exp(2*x)*(7*x + 2*x^2))/(64*x + 32*x^2 +
4*x^3),x)

[Out]

(17*x - 10*exp(2*x) + 4*x^2 - log(-2*x)*exp(2*x))/(4*(x + 4))

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sympy [A]  time = 0.33, size = 26, normalized size = 0.96 \begin {gather*} x + \frac {\left (- \log {\left (- 2 x \right )} - 10\right ) e^{2 x}}{4 x + 16} - \frac {1}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-7*x)*exp(2*x)*ln(-2*x)+(-20*x**2-71*x-4)*exp(2*x)+4*x**3+32*x**2+68*x)/(4*x**3+32*x**2+64*
x),x)

[Out]

x + (-log(-2*x) - 10)*exp(2*x)/(4*x + 16) - 1/(x + 4)

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