3.73.1 \(\int \frac {\frac {5 x (-4+5 x^2)}{e^4 (16+20 x^2)}+e^x (4 x+5 x^3)}{4 x+5 x^3} \, dx\)

Optimal. Leaf size=22 \[ e^x-\frac {1}{4 e^4 \left (\frac {4}{5 x}+x\right )} \]

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Rubi [A]  time = 0.33, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1593, 6688, 2194, 383} \begin {gather*} e^x-\frac {5 x}{4 e^4 \left (5 x^2+4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((5*x*(-4 + 5*x^2))/(E^4*(16 + 20*x^2)) + E^x*(4*x + 5*x^3))/(4*x + 5*x^3),x]

[Out]

E^x - (5*x)/(4*E^4*(4 + 5*x^2))

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 1593

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

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {5 x \left (-4+5 x^2\right )}{e^4 \left (16+20 x^2\right )}+e^x \left (4 x+5 x^3\right )}{x \left (4+5 x^2\right )} \, dx\\ &=\int \left (e^x+\frac {5 \left (-4+5 x^2\right )}{4 e^4 \left (4+5 x^2\right )^2}\right ) \, dx\\ &=\frac {5 \int \frac {-4+5 x^2}{\left (4+5 x^2\right )^2} \, dx}{4 e^4}+\int e^x \, dx\\ &=e^x-\frac {5 x}{4 e^4 \left (4+5 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.95 \begin {gather*} e^x-\frac {5 x}{4 e^4 \left (4+5 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((5*x*(-4 + 5*x^2))/(E^4*(16 + 20*x^2)) + E^x*(4*x + 5*x^3))/(4*x + 5*x^3),x]

[Out]

E^x - (5*x)/(4*E^4*(4 + 5*x^2))

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fricas [A]  time = 1.03, size = 30, normalized size = 1.36 \begin {gather*} \frac {{\left (4 \, {\left (5 \, x^{2} + 4\right )} e^{\left (x + 4\right )} - 5 \, x\right )} e^{\left (-4\right )}}{4 \, {\left (5 \, x^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-4)*exp(log(5*x/(20*x^2+16))-4)+(5*x^3+4*x)*exp(x))/(5*x^3+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 35, normalized size = 1.59 \begin {gather*} \frac {20 \, x^{2} e^{\left (x + 4\right )} - 5 \, x + 16 \, e^{\left (x + 4\right )}}{4 \, {\left (5 \, x^{2} e^{4} + 4 \, e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-4)*exp(log(5*x/(20*x^2+16))-4)+(5*x^3+4*x)*exp(x))/(5*x^3+4*x),x, algorithm="giac")

[Out]

1/4*(20*x^2*e^(x + 4) - 5*x + 16*e^(x + 4))/(5*x^2*e^4 + 4*e^4)

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maple [A]  time = 0.13, size = 16, normalized size = 0.73

method	result	size
default	\(-\frac {{\mathrm e}^{-4} x}{4 \left (x^{2}+\frac {4}{5}\right )}+{\mathrm e}^{x}\)	\(16\)
norman	\(\frac {-\frac {5 \,{\mathrm e}^{-4} x}{4}+5 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x}}{5 x^{2}+4}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2-4)*exp(ln(5*x/(20*x^2+16))-4)+(5*x^3+4*x)*exp(x))/(5*x^3+4*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 20, normalized size = 0.91 \begin {gather*} -\frac {5 \, x}{4 \, {\left (5 \, x^{2} e^{4} + 4 \, e^{4}\right )}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-4)*exp(log(5*x/(20*x^2+16))-4)+(5*x^3+4*x)*exp(x))/(5*x^3+4*x),x, algorithm="maxima")

[Out]

-5/4*x/(5*x^2*e^4 + 4*e^4) + e^x

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mupad [B]  time = 4.57, size = 17, normalized size = 0.77 \begin {gather*} {\mathrm {e}}^x-\frac {5\,x\,{\mathrm {e}}^{-4}}{4\,\left (5\,x^2+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(4*x + 5*x^3) + exp(log((5*x)/(20*x^2 + 16)) - 4)*(5*x^2 - 4))/(4*x + 5*x^3),x)

[Out]

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

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sympy [A]  time = 0.26, size = 19, normalized size = 0.86 \begin {gather*} - \frac {5 x}{20 x^{2} e^{4} + 16 e^{4}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2-4)*exp(ln(5*x/(20*x**2+16))-4)+(5*x**3+4*x)*exp(x))/(5*x**3+4*x),x)

[Out]

-5*x/(20*x**2*exp(4) + 16*exp(4)) + exp(x)

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