3.49.89 \(\int \frac {10 x-6 x^5+e^x (-5 x+4 x^4-x^5)+e^{\log ^2(x)} (4 x^4+(-10-2 x^4) \log (x))}{25 x+10 x^5+x^9} \, dx\)

Optimal. Leaf size=25 \[ \frac {-e^x-e^{\log ^2(x)}+2 x}{5+x^4} \]

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Rubi [B]  time = 0.89, antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 9, number of rules used = 5, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1594, 28, 6742, 2288, 383} \begin {gather*} \frac {2 x}{x^4+5}-\frac {e^x}{x^4+5}-\frac {e^{\log ^2(x)} \left (x^4 \log (x)+5 \log (x)\right )}{\left (x^4+5\right )^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x - 6*x^5 + E^x*(-5*x + 4*x^4 - x^5) + E^Log[x]^2*(4*x^4 + (-10 - 2*x^4)*Log[x]))/(25*x + 10*x^5 + x^9
),x]

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*x*(a + b*x^n)^(p + 1))/a, x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a*d - b*c*(n*(p + 1) + 1), 0]

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 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 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 {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{x \left (25+10 x^4+x^8\right )} \, dx\\ &=\int \frac {10 x-6 x^5+e^x \left (-5 x+4 x^4-x^5\right )+e^{\log ^2(x)} \left (4 x^4+\left (-10-2 x^4\right ) \log (x)\right )}{x \left (5+x^4\right )^2} \, dx\\ &=\int \left (-\frac {-10+5 e^x-4 e^x x^3+6 x^4+e^x x^4}{\left (5+x^4\right )^2}-\frac {2 e^{\log ^2(x)} \left (-2 x^4+5 \log (x)+x^4 \log (x)\right )}{x \left (5+x^4\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{\log ^2(x)} \left (-2 x^4+5 \log (x)+x^4 \log (x)\right )}{x \left (5+x^4\right )^2} \, dx\right )-\int \frac {-10+5 e^x-4 e^x x^3+6 x^4+e^x x^4}{\left (5+x^4\right )^2} \, dx\\ &=-\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)}-\int \left (\frac {e^x \left (5-4 x^3+x^4\right )}{\left (5+x^4\right )^2}+\frac {2 \left (-5+3 x^4\right )}{\left (5+x^4\right )^2}\right ) \, dx\\ &=-\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)}-2 \int \frac {-5+3 x^4}{\left (5+x^4\right )^2} \, dx-\int \frac {e^x \left (5-4 x^3+x^4\right )}{\left (5+x^4\right )^2} \, dx\\ &=-\frac {e^x}{5+x^4}+\frac {2 x}{5+x^4}-\frac {e^{\log ^2(x)} \left (5 \log (x)+x^4 \log (x)\right )}{\left (5+x^4\right )^2 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.67, size = 22, normalized size = 0.88 \begin {gather*} -\frac {e^x+e^{\log ^2(x)}-2 x}{5+x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x - 6*x^5 + E^x*(-5*x + 4*x^4 - x^5) + E^Log[x]^2*(4*x^4 + (-10 - 2*x^4)*Log[x]))/(25*x + 10*x^5
+ x^9),x]

[Out]

-((E^x + E^Log[x]^2 - 2*x)/(5 + x^4))

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fricas [A]  time = 1.01, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 \, x - e^{\left (\log \relax (x)^{2}\right )} - e^{x}}{x^{4} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^4-10)*log(x)+4*x^4)*exp(log(x)^2)+(-x^5+4*x^4-5*x)*exp(x)-6*x^5+10*x)/(x^9+10*x^5+25*x),x, a
lgorithm="fricas")

[Out]

(2*x - e^(log(x)^2) - e^x)/(x^4 + 5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {6 \, x^{5} - 2 \, {\left (2 \, x^{4} - {\left (x^{4} + 5\right )} \log \relax (x)\right )} e^{\left (\log \relax (x)^{2}\right )} + {\left (x^{5} - 4 \, x^{4} + 5 \, x\right )} e^{x} - 10 \, x}{x^{9} + 10 \, x^{5} + 25 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^4-10)*log(x)+4*x^4)*exp(log(x)^2)+(-x^5+4*x^4-5*x)*exp(x)-6*x^5+10*x)/(x^9+10*x^5+25*x),x, a
lgorithm="giac")

[Out]

integrate(-(6*x^5 - 2*(2*x^4 - (x^4 + 5)*log(x))*e^(log(x)^2) + (x^5 - 4*x^4 + 5*x)*e^x - 10*x)/(x^9 + 10*x^5
+ 25*x), x)

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maple [A]  time = 0.04, size = 32, normalized size = 1.28

method	result	size
risch	\(\frac {2 x -{\mathrm e}^{x}}{x^{4}+5}-\frac {{\mathrm e}^{\ln \relax (x )^{2}}}{x^{4}+5}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^4-10)*ln(x)+4*x^4)*exp(ln(x)^2)+(-x^5+4*x^4-5*x)*exp(x)-6*x^5+10*x)/(x^9+10*x^5+25*x),x,method=_RE
TURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 28, normalized size = 1.12 \begin {gather*} \frac {2 \, x}{x^{4} + 5} - \frac {e^{\left (\log \relax (x)^{2}\right )} + e^{x}}{x^{4} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^4-10)*log(x)+4*x^4)*exp(log(x)^2)+(-x^5+4*x^4-5*x)*exp(x)-6*x^5+10*x)/(x^9+10*x^5+25*x),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 3.45, size = 20, normalized size = 0.80 \begin {gather*} -\frac {{\mathrm {e}}^{{\ln \relax (x)}^2}-2\,x+{\mathrm {e}}^x}{x^4+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + exp(log(x)^2)*(4*x^4 - log(x)*(2*x^4 + 10)) - exp(x)*(5*x - 4*x^4 + x^5) - 6*x^5)/(25*x + 10*x^5 +
x^9),x)

[Out]

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

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sympy [A]  time = 0.49, size = 27, normalized size = 1.08 \begin {gather*} \frac {2 x}{x^{4} + 5} - \frac {e^{x}}{x^{4} + 5} - \frac {e^{\log {\relax (x )}^{2}}}{x^{4} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**4-10)*ln(x)+4*x**4)*exp(ln(x)**2)+(-x**5+4*x**4-5*x)*exp(x)-6*x**5+10*x)/(x**9+10*x**5+25*x
),x)

[Out]

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

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