3.71.46 \(\int \frac {1}{25} e^{\frac {1}{50} (25+6 x+6 x^2+e^5 (6 x+6 x^2))} (3+6 x+e^5 (3+6 x)) \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {1}{2}+\frac {3}{25} \left (1+e^5\right ) x (1+x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 28, normalized size of antiderivative = 1.47, number of steps used = 3, number of rules used = 3, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {12, 2244, 2236} \begin {gather*} e^{\frac {3}{25} \left (1+e^5\right ) x^2+\frac {3}{25} \left (1+e^5\right ) x+\frac {1}{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((25 + 6*x + 6*x^2 + E^5*(6*x + 6*x^2))/50)*(3 + 6*x + E^5*(3 + 6*x)))/25,x]

[Out]

E^(1/2 + (3*(1 + E^5)*x)/25 + (3*(1 + E^5)*x^2)/25)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \exp \left (\frac {1}{50} \left (25+6 x+6 x^2+e^5 \left (6 x+6 x^2\right )\right )\right ) \left (3+6 x+e^5 (3+6 x)\right ) \, dx\\ &=\frac {1}{25} \int e^{\frac {1}{2}+\frac {3}{25} \left (1+e^5\right ) x+\frac {3}{25} \left (1+e^5\right ) x^2} \left (3 \left (1+e^5\right )+6 \left (1+e^5\right ) x\right ) \, dx\\ &=e^{\frac {1}{2}+\frac {3}{25} \left (1+e^5\right ) x+\frac {3}{25} \left (1+e^5\right ) x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 28, normalized size = 1.47 \begin {gather*} e^{\frac {1}{2}+\frac {3}{25} \left (1+e^5\right ) x+\frac {3}{25} \left (1+e^5\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((25 + 6*x + 6*x^2 + E^5*(6*x + 6*x^2))/50)*(3 + 6*x + E^5*(3 + 6*x)))/25,x]

[Out]

E^(1/2 + (3*(1 + E^5)*x)/25 + (3*(1 + E^5)*x^2)/25)

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fricas [A]  time = 0.59, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (\frac {3}{25} \, x^{2} + \frac {3}{25} \, {\left (x^{2} + x\right )} e^{5} + \frac {3}{25} \, x + \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((6*x+3)*exp(5)+6*x+3)*exp(1/50*(6*x^2+6*x)*exp(5)+3/25*x^2+3/25*x+1/2),x, algorithm="fricas")

[Out]

e^(3/25*x^2 + 3/25*(x^2 + x)*e^5 + 3/25*x + 1/2)

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giac [A]  time = 0.21, size = 23, normalized size = 1.21 \begin {gather*} e^{\left (\frac {3}{25} \, x^{2} e^{5} + \frac {3}{25} \, x^{2} + \frac {3}{25} \, x e^{5} + \frac {3}{25} \, x + \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((6*x+3)*exp(5)+6*x+3)*exp(1/50*(6*x^2+6*x)*exp(5)+3/25*x^2+3/25*x+1/2),x, algorithm="giac")

[Out]

e^(3/25*x^2*e^5 + 3/25*x^2 + 3/25*x*e^5 + 3/25*x + 1/2)

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maple [A]  time = 0.06, size = 24, normalized size = 1.26

method	result	size
gosper	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{5}}{25}+\frac {3 x \,{\mathrm e}^{5}}{25}+\frac {3 x^{2}}{25}+\frac {3 x}{25}+\frac {1}{2}}\)	\(24\)
risch	\({\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{5}}{25}+\frac {3 x \,{\mathrm e}^{5}}{25}+\frac {3 x^{2}}{25}+\frac {3 x}{25}+\frac {1}{2}}\)	\(24\)
norman	\({\mathrm e}^{\frac {\left (6 x^{2}+6 x \right ) {\mathrm e}^{5}}{50}+\frac {3 x^{2}}{25}+\frac {3 x}{25}+\frac {1}{2}}\)	\(25\)
default	\(\frac {3 \,{\mathrm e}^{\frac {1}{2}+\left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right ) x^{2}+\left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right ) x}}{25 \left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right )}+\frac {3 \,{\mathrm e}^{\frac {11}{2}+\left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right ) x^{2}+\left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right ) x}}{25 \left (\frac {3 \,{\mathrm e}^{5}}{25}+\frac {3}{25}\right )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((6*x+3)*exp(5)+6*x+3)*exp(1/50*(6*x^2+6*x)*exp(5)+3/25*x^2+3/25*x+1/2),x,method=_RETURNVERBOSE)

[Out]

exp(3/25*x^2*exp(5)+3/25*x*exp(5)+3/25*x^2+3/25*x+1/2)

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maxima [A]  time = 0.38, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (\frac {3}{25} \, x^{2} + \frac {3}{25} \, {\left (x^{2} + x\right )} e^{5} + \frac {3}{25} \, x + \frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((6*x+3)*exp(5)+6*x+3)*exp(1/50*(6*x^2+6*x)*exp(5)+3/25*x^2+3/25*x+1/2),x, algorithm="maxima")

[Out]

e^(3/25*x^2 + 3/25*(x^2 + x)*e^5 + 3/25*x + 1/2)

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mupad [B]  time = 4.25, size = 27, normalized size = 1.42 \begin {gather*} {\mathrm {e}}^{\frac {3\,x^2\,{\mathrm {e}}^5}{25}}\,{\mathrm {e}}^{\frac {3\,x}{25}}\,\sqrt {\mathrm {e}}\,{\mathrm {e}}^{\frac {3\,x^2}{25}}\,{\mathrm {e}}^{\frac {3\,x\,{\mathrm {e}}^5}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((3*x)/25 + (exp(5)*(6*x + 6*x^2))/50 + (3*x^2)/25 + 1/2)*(6*x + exp(5)*(6*x + 3) + 3))/25,x)

[Out]

exp((3*x^2*exp(5))/25)*exp((3*x)/25)*exp(1/2)*exp((3*x^2)/25)*exp((3*x*exp(5))/25)

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sympy [A]  time = 0.14, size = 31, normalized size = 1.63 \begin {gather*} e^{\frac {3 x^{2}}{25} + \frac {3 x}{25} + \left (\frac {3 x^{2}}{25} + \frac {3 x}{25}\right ) e^{5} + \frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((6*x+3)*exp(5)+6*x+3)*exp(1/50*(6*x**2+6*x)*exp(5)+3/25*x**2+3/25*x+1/2),x)

[Out]

exp(3*x**2/25 + 3*x/25 + (3*x**2/25 + 3*x/25)*exp(5) + 1/2)

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