3.88.82 \(\int \frac {1-e^{-4+3 x-x^4}+e^{-4+3 x-x^4} (15+3 x-20 x^3-4 x^4) \log (5+x)+(5+x) \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx\)

Optimal. Leaf size=23 \[ x+\frac {-1+e^{-4+3 x-x^4}}{\log (5+x)} \]

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Rubi [B]  time = 0.96, antiderivative size = 75, normalized size of antiderivative = 3.26, number of steps used = 8, number of rules used = 6, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6742, 2288, 6688, 2390, 2302, 30} \begin {gather*} \frac {e^{-x^4+3 x-4} \left (-4 x^4 \log (x+5)-20 x^3 \log (x+5)+3 x \log (x+5)+15 \log (x+5)\right )}{(x+5) \left (3-4 x^3\right ) \log ^2(x+5)}+x-\frac {1}{\log (x+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - E^(-4 + 3*x - x^4) + E^(-4 + 3*x - x^4)*(15 + 3*x - 20*x^3 - 4*x^4)*Log[5 + x] + (5 + x)*Log[5 + x]^2
)/((5 + x)*Log[5 + x]^2),x]

[Out]

x - Log[5 + x]^(-1) + (E^(-4 + 3*x - x^4)*(15*Log[5 + x] + 3*x*Log[5 + x] - 20*x^3*Log[5 + x] - 4*x^4*Log[5 +
x]))/((5 + x)*(3 - 4*x^3)*Log[5 + x]^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 \left (-\frac {e^{-4+3 x-x^4} \left (1-15 \log (5+x)-3 x \log (5+x)+20 x^3 \log (5+x)+4 x^4 \log (5+x)\right )}{(5+x) \log ^2(5+x)}+\frac {1+5 \log ^2(5+x)+x \log ^2(5+x)}{(5+x) \log ^2(5+x)}\right ) \, dx\\ &=-\int \frac {e^{-4+3 x-x^4} \left (1-15 \log (5+x)-3 x \log (5+x)+20 x^3 \log (5+x)+4 x^4 \log (5+x)\right )}{(5+x) \log ^2(5+x)} \, dx+\int \frac {1+5 \log ^2(5+x)+x \log ^2(5+x)}{(5+x) \log ^2(5+x)} \, dx\\ &=\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\int \left (1+\frac {1}{(5+x) \log ^2(5+x)}\right ) \, dx\\ &=x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\int \frac {1}{(5+x) \log ^2(5+x)} \, dx\\ &=x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5+x\right )\\ &=x+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}+\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (5+x)\right )\\ &=x-\frac {1}{\log (5+x)}+\frac {e^{-4+3 x-x^4} \left (15 \log (5+x)+3 x \log (5+x)-20 x^3 \log (5+x)-4 x^4 \log (5+x)\right )}{(5+x) \left (3-4 x^3\right ) \log ^2(5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 27, normalized size = 1.17 \begin {gather*} \frac {-1+e^{-4+3 x-x^4}+x \log (5+x)}{\log (5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - E^(-4 + 3*x - x^4) + E^(-4 + 3*x - x^4)*(15 + 3*x - 20*x^3 - 4*x^4)*Log[5 + x] + (5 + x)*Log[5
+ x]^2)/((5 + x)*Log[5 + x]^2),x]

[Out]

(-1 + E^(-4 + 3*x - x^4) + x*Log[5 + x])/Log[5 + x]

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fricas [A]  time = 0.69, size = 26, normalized size = 1.13 \begin {gather*} \frac {x \log \left (x + 5\right ) + e^{\left (-x^{4} + 3 \, x - 4\right )} - 1}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*log(x + 5) + e^(-x^4 + 3*x - 4) - 1)/log(x + 5)

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giac [A]  time = 0.16, size = 32, normalized size = 1.39 \begin {gather*} \frac {{\left (x e^{4} \log \left (x + 5\right ) - e^{4} + e^{\left (-x^{4} + 3 \, x\right )}\right )} e^{\left (-4\right )}}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.42, size = 23, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-x^{4}+3 x -4}-1}{\ln \left (5+x \right )}+x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 37, normalized size = 1.61 \begin {gather*} \frac {{\left (x e^{4} \log \left (x + 5\right ) + e^{\left (-x^{4} + 3 \, x\right )}\right )} e^{\left (-4\right )}}{\log \left (x + 5\right )} - \frac {1}{\log \left (x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^4*log(x + 5) + e^(-x^4 + 3*x))*e^(-4)/log(x + 5) - 1/log(x + 5)

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mupad [B]  time = 5.56, size = 29, normalized size = 1.26 \begin {gather*} x-\frac {1}{\ln \left (x+5\right )}+\frac {{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-x^4}}{\ln \left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 5)^2*(x + 5) - exp(3*x - x^4 - 4) + log(x + 5)*exp(3*x - x^4 - 4)*(3*x - 20*x^3 - 4*x^4 + 15) + 1
)/(log(x + 5)^2*(x + 5)),x)

[Out]

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

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sympy [A]  time = 0.33, size = 22, normalized size = 0.96 \begin {gather*} x + \frac {e^{- x^{4} + 3 x - 4}}{\log {\left (x + 5 \right )}} - \frac {1}{\log {\left (x + 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(-x**4 + 3*x - 4)/log(x + 5) - 1/log(x + 5)

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