3.17.92 \(\int 4 e^{-3+3 x+25 x^2} (3+50 x) \, dx\)

Optimal. Leaf size=14 \[ 4 e^{-3+3 x+25 x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2236} \begin {gather*} 4 e^{25 x^2+3 x-3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[4*E^(-3 + 3*x + 25*x^2)*(3 + 50*x),x]

[Out]

4*E^(-3 + 3*x + 25*x^2)

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \int e^{-3+3 x+25 x^2} (3+50 x) \, dx\\ &=4 e^{-3+3 x+25 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 14, normalized size = 1.00 \begin {gather*} 4 e^{-3+3 x+25 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[4*E^(-3 + 3*x + 25*x^2)*(3 + 50*x),x]

[Out]

4*E^(-3 + 3*x + 25*x^2)

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fricas [A]  time = 0.97, size = 15, normalized size = 1.07 \begin {gather*} e^{\left (25 \, x^{2} + 3 \, x + 2 \, \log \relax (2) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x+3)*exp(2*log(2)+25*x^2+3*x-3),x, algorithm="fricas")

[Out]

e^(25*x^2 + 3*x + 2*log(2) - 3)

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giac [A]  time = 0.25, size = 15, normalized size = 1.07 \begin {gather*} e^{\left (25 \, x^{2} + 3 \, x + 2 \, \log \relax (2) - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x+3)*exp(2*log(2)+25*x^2+3*x-3),x, algorithm="giac")

[Out]

e^(25*x^2 + 3*x + 2*log(2) - 3)

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maple [A]  time = 0.02, size = 14, normalized size = 1.00

method	result	size
risch	\(4 \,{\mathrm e}^{25 x^{2}+3 x -3}\)	\(14\)
gosper	\({\mathrm e}^{2 \ln \relax (2)+25 x^{2}+3 x -3}\)	\(16\)
derivativedivides	\({\mathrm e}^{2 \ln \relax (2)+25 x^{2}+3 x -3}\)	\(16\)
default	\({\mathrm e}^{2 \ln \relax (2)+25 x^{2}+3 x -3}\)	\(16\)
norman	\({\mathrm e}^{2 \ln \relax (2)+25 x^{2}+3 x -3}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x+3)*exp(2*ln(2)+25*x^2+3*x-3),x,method=_RETURNVERBOSE)

[Out]

4*exp(25*x^2+3*x-3)

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maxima [A]  time = 0.36, size = 13, normalized size = 0.93 \begin {gather*} 4 \, e^{\left (25 \, x^{2} + 3 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x+3)*exp(2*log(2)+25*x^2+3*x-3),x, algorithm="maxima")

[Out]

4*e^(25*x^2 + 3*x - 3)

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mupad [B]  time = 0.07, size = 14, normalized size = 1.00 \begin {gather*} 4\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{25\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x + 2*log(2) + 25*x^2 - 3)*(50*x + 3),x)

[Out]

4*exp(3*x)*exp(-3)*exp(25*x^2)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.86 \begin {gather*} 4 e^{25 x^{2} + 3 x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x+3)*exp(2*ln(2)+25*x**2+3*x-3),x)

[Out]

4*exp(25*x**2 + 3*x - 3)

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