3.15.83 \(\int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} (25+20 x+x^2)+e^x (25+35 x+11 x^2+x^3))}{25+10 x+x^2} \, dx\)

Optimal. Leaf size=24 \[ e^{1+\left (1+e^x+e^{\frac {2 x}{5+x}}-3 x\right ) x} \]

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Rubi [A]  time = 3.84, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {27, 6706} \begin {gather*} e^{-3 x^2+e^x x+e^{\frac {2 x}{x+5}} x+x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)*(25 - 140*x - 59*x^2 - 6*x^3 + E^((2*x)/(5 + x))*(25 + 20
*x + x^2) + E^x*(25 + 35*x + 11*x^2 + x^3)))/(25 + 10*x + x^2),x]

[Out]

E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)

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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2} \left (25-140 x-59 x^2-6 x^3+e^{\frac {2 x}{5+x}} \left (25+20 x+x^2\right )+e^x \left (25+35 x+11 x^2+x^3\right )\right )}{(5+x)^2} \, dx\\ &=e^{1+x+e^x x+e^{\frac {2 x}{5+x}} x-3 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.29, size = 28, normalized size = 1.17 \begin {gather*} e^{1+x+e^x x+e^{2-\frac {10}{5+x}} x-3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 + x + E^x*x + E^((2*x)/(5 + x))*x - 3*x^2)*(25 - 140*x - 59*x^2 - 6*x^3 + E^((2*x)/(5 + x))*(2
5 + 20*x + x^2) + E^x*(25 + 35*x + 11*x^2 + x^3)))/(25 + 10*x + x^2),x]

[Out]

E^(1 + x + E^x*x + E^(2 - 10/(5 + x))*x - 3*x^2)

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fricas [A]  time = 0.78, size = 24, normalized size = 1.00 \begin {gather*} e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (\frac {2 \, x}{x + 5}\right )} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2-140*x+25)*exp(exp(x)*x+x*exp(
2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.55, size = 24, normalized size = 1.00 \begin {gather*} e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (\frac {2 \, x}{x + 5}\right )} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2-140*x+25)*exp(exp(x)*x+x*exp(
2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 25, normalized size = 1.04

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {2 x}{5+x}}-3 x^{2}+x +1}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2-140*x+25)*exp(exp(x)*x+x*exp(2*x/(5
+x))-3*x^2+x+1)/(x^2+10*x+25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.60, size = 25, normalized size = 1.04 \begin {gather*} e^{\left (-3 \, x^{2} + x e^{x} + x e^{\left (-\frac {10}{x + 5} + 2\right )} + x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+11*x^2+35*x+25)*exp(x)+(x^2+20*x+25)*exp(2*x/(5+x))-6*x^3-59*x^2-140*x+25)*exp(exp(x)*x+x*exp(
2*x/(5+x))-3*x^2+x+1)/(x^2+10*x+25),x, algorithm="maxima")

[Out]

e^(-3*x^2 + x*e^x + x*e^(-10/(x + 5) + 2) + x + 1)

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mupad [B]  time = 1.33, size = 28, normalized size = 1.17 \begin {gather*} {\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\mathrm {e}\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{\frac {2\,x}{x+5}}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + x*exp((2*x)/(x + 5)) + x*exp(x) - 3*x^2 + 1)*(140*x - exp(x)*(35*x + 11*x^2 + x^3 + 25) + 59*x^2
+ 6*x^3 - exp((2*x)/(x + 5))*(20*x + x^2 + 25) - 25))/(10*x + x^2 + 25),x)

[Out]

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

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sympy [A]  time = 0.73, size = 24, normalized size = 1.00 \begin {gather*} e^{- 3 x^{2} + x e^{x} + x e^{\frac {2 x}{x + 5}} + x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+11*x**2+35*x+25)*exp(x)+(x**2+20*x+25)*exp(2*x/(5+x))-6*x**3-59*x**2-140*x+25)*exp(exp(x)*x+x
*exp(2*x/(5+x))-3*x**2+x+1)/(x**2+10*x+25),x)

[Out]

exp(-3*x**2 + x*exp(x) + x*exp(2*x/(x + 5)) + x + 1)

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