3.10 \(\int \frac {5 x^2+3 \sqrt [3]{e^x+x}+e^x (3 x+2 x^2)}{x \sqrt [3]{e^x+x}} \, dx\)

Optimal. Leaf size=17 \[ 3 \left (x+e^x\right )^{2/3} x+3 \log (x) \]

[Out]

3*x*(exp(x)+x)^(2/3)+3*ln(x)

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Rubi [A]  time = 0.61, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6742, 2261, 2273, 2262} \[ 3 \left (x+e^x\right )^{2/3} x+3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(5*x^2 + 3*(E^x + x)^(1/3) + E^x*(3*x + 2*x^2))/(x*(E^x + x)^(1/3)),x]

[Out]

3*x*(E^x + x)^(2/3) + 3*Log[x]

Rule 2261

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*((b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))) + (a_.)*(x_)^(n_.))^(p_.), x_Sy
mbol] :> Simp[(a*x^n + b*F^(e*(c + d*x)))^(p + 1)/(b*d*e*(p + 1)*Log[F]), x] - Dist[(a*n)/(b*d*e*Log[F]), Int[
x^(n - 1)*(a*x^n + b*F^(e*(c + d*x)))^p, x], x] /; FreeQ[{F, a, b, c, d, e, n, p}, x] && NeQ[p, -1]

Rule 2262

Int[(F_)^((e_.)*((c_.) + (d_.)*(x_)))*(x_)^(m_.)*((b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))) + (a_.)*(x_)^(n_.))^
(p_.), x_Symbol] :> Simp[(x^m*(a*x^n + b*F^(e*(c + d*x)))^(p + 1))/(b*d*e*(p + 1)*Log[F]), x] + (-Dist[m/(b*d*
e*(p + 1)*Log[F]), Int[x^(m - 1)*(a*x^n + b*F^(e*(c + d*x)))^(p + 1), x], x] - Dist[(a*n)/(b*d*e*Log[F]), Int[
x^(m + n - 1)*(a*x^n + b*F^(e*(c + d*x)))^p, x], x]) /; FreeQ[{F, a, b, c, d, e, m, n, p}, x] && NeQ[p, -1]

Rule 2273

Int[(x_)^(m_.)*(E^(x_) + (x_)^(m_.))^(n_), x_Symbol] :> -Simp[(E^x + x^m)^(n + 1)/(n + 1), x] + (Dist[m, Int[x
^(m - 1)*(E^x + x^m)^n, x], x] + Int[(E^x + x^m)^(n + 1), x]) /; RationalQ[m, n] && GtQ[m, 0] && LtQ[n, 0] &&
NeQ[n, -1]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {5 x^2+3 \sqrt [3]{e^x+x}+e^x \left (3 x+2 x^2\right )}{x \sqrt [3]{e^x+x}} \, dx &=\int \left (\frac {3}{x}+\frac {3 e^x}{\sqrt [3]{e^x+x}}+\frac {\left (5+2 e^x\right ) x}{\sqrt [3]{e^x+x}}\right ) \, dx\\ &=3 \log (x)+3 \int \frac {e^x}{\sqrt [3]{e^x+x}} \, dx+\int \frac {\left (5+2 e^x\right ) x}{\sqrt [3]{e^x+x}} \, dx\\ &=\frac {9}{2} \left (e^x+x\right )^{2/3}+3 \log (x)-3 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx+\int \left (\frac {5 x}{\sqrt [3]{e^x+x}}+\frac {2 e^x x}{\sqrt [3]{e^x+x}}\right ) \, dx\\ &=\frac {9}{2} \left (e^x+x\right )^{2/3}+3 \log (x)+2 \int \frac {e^x x}{\sqrt [3]{e^x+x}} \, dx-3 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx+5 \int \frac {x}{\sqrt [3]{e^x+x}} \, dx\\ &=-3 \left (e^x+x\right )^{2/3}+3 x \left (e^x+x\right )^{2/3}+3 \log (x)-2 \int \frac {x}{\sqrt [3]{e^x+x}} \, dx-3 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx-3 \int \left (e^x+x\right )^{2/3} \, dx+5 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx+5 \int \left (e^x+x\right )^{2/3} \, dx\\ &=3 x \left (e^x+x\right )^{2/3}+3 \log (x)-2 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx-2 \int \left (e^x+x\right )^{2/3} \, dx-3 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx-3 \int \left (e^x+x\right )^{2/3} \, dx+5 \int \frac {1}{\sqrt [3]{e^x+x}} \, dx+5 \int \left (e^x+x\right )^{2/3} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 17, normalized size = 1.00 \[ 3 \left (x+e^x\right )^{2/3} x+3 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(5*x^2 + 3*(E^x + x)^(1/3) + E^x*(3*x + 2*x^2))/(x*(E^x + x)^(1/3)),x]

[Out]

3*x*(E^x + x)^(2/3) + 3*Log[x]

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*(exp(x)+x)^(1/3)+exp(x)*(2*x^2+3*x))/x/(exp(x)+x)^(1/3),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {5 \, x^{2} + {\left (2 \, x^{2} + 3 \, x\right )} e^{x} + 3 \, {\left (x + e^{x}\right )}^{\frac {1}{3}}}{{\left (x + e^{x}\right )}^{\frac {1}{3}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*(exp(x)+x)^(1/3)+exp(x)*(2*x^2+3*x))/x/(exp(x)+x)^(1/3),x, algorithm="giac")

[Out]

integrate((5*x^2 + (2*x^2 + 3*x)*e^x + 3*(x + e^x)^(1/3))/((x + e^x)^(1/3)*x), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {5 x^{2}+\left (2 x^{2}+3 x \right ) {\mathrm e}^{x}+3 \left (x +{\mathrm e}^{x}\right )^{\frac {1}{3}}}{\left (x +{\mathrm e}^{x}\right )^{\frac {1}{3}} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*(exp(x)+x)^(1/3)+exp(x)*(2*x^2+3*x))/x/(exp(x)+x)^(1/3),x)

[Out]

int((5*x^2+3*(exp(x)+x)^(1/3)+exp(x)*(2*x^2+3*x))/x/(exp(x)+x)^(1/3),x)

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maxima [A]  time = 0.57, size = 21, normalized size = 1.24 \[ \frac {3 \, {\left (x^{2} + x e^{x}\right )}}{{\left (x + e^{x}\right )}^{\frac {1}{3}}} + 3 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*(exp(x)+x)^(1/3)+exp(x)*(2*x^2+3*x))/x/(exp(x)+x)^(1/3),x, algorithm="maxima")

[Out]

3*(x^2 + x*e^x)/(x + e^x)^(1/3) + 3*log(x)

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mupad [B]  time = 0.27, size = 14, normalized size = 0.82 \[ 3\,\ln \relax (x)+3\,x\,{\left (x+{\mathrm {e}}^x\right )}^{2/3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*(x + exp(x))^(1/3) + exp(x)*(3*x + 2*x^2) + 5*x^2)/(x*(x + exp(x))^(1/3)),x)

[Out]

3*log(x) + 3*x*(x + exp(x))^(2/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 x^{2} e^{x} + 5 x^{2} + 3 x e^{x} + 3 \sqrt [3]{x + e^{x}}}{x \sqrt [3]{x + e^{x}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*(exp(x)+x)**(1/3)+exp(x)*(2*x**2+3*x))/x/(exp(x)+x)**(1/3),x)

[Out]

Integral((2*x**2*exp(x) + 5*x**2 + 3*x*exp(x) + 3*(x + exp(x))**(1/3))/(x*(x + exp(x))**(1/3)), x)

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