3.55.36 \(\int e^5 (5+5 x) \, dx\)

Optimal. Leaf size=18 \[ -\frac {5}{4} e^5 \left (-4+\frac {3}{x}-2 x\right ) x \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.67, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {9} \begin {gather*} \frac {5}{2} e^5 (x+1)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^5*(5 + 5*x),x]

[Out]

(5*E^5*(1 + x)^2)/2

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {5}{2} e^5 (1+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.78 \begin {gather*} 5 e^5 \left (x+\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^5*(5 + 5*x),x]

[Out]

5*E^5*(x + x^2/2)

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fricas [A]  time = 1.61, size = 11, normalized size = 0.61 \begin {gather*} \frac {5}{2} \, {\left (x^{2} + 2 \, x\right )} e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x+5)*exp(5),x, algorithm="fricas")

[Out]

5/2*(x^2 + 2*x)*e^5

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giac [A]  time = 0.15, size = 11, normalized size = 0.61 \begin {gather*} \frac {5}{2} \, {\left (x^{2} + 2 \, x\right )} e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x+5)*exp(5),x, algorithm="giac")

[Out]

5/2*(x^2 + 2*x)*e^5

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maple [A]  time = 0.03, size = 9, normalized size = 0.50

method	result	size
gosper	\(\frac {5 \,{\mathrm e}^{5} x \left (2+x \right )}{2}\)	\(9\)
default	\({\mathrm e}^{5} \left (\frac {5}{2} x^{2}+5 x \right )\)	\(13\)
norman	\(5 x \,{\mathrm e}^{5}+\frac {5 x^{2} {\mathrm e}^{5}}{2}\)	\(14\)
risch	\(5 x \,{\mathrm e}^{5}+\frac {5 x^{2} {\mathrm e}^{5}}{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x+5)*exp(5),x,method=_RETURNVERBOSE)

[Out]

5/2*exp(5)*x*(2+x)

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maxima [A]  time = 0.38, size = 11, normalized size = 0.61 \begin {gather*} \frac {5}{2} \, {\left (x^{2} + 2 \, x\right )} e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x+5)*exp(5),x, algorithm="maxima")

[Out]

5/2*(x^2 + 2*x)*e^5

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mupad [B]  time = 0.06, size = 9, normalized size = 0.50 \begin {gather*} \frac {5\,{\mathrm {e}}^5\,{\left (x+1\right )}^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5)*(5*x + 5),x)

[Out]

(5*exp(5)*(x + 1)^2)/2

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sympy [A]  time = 0.07, size = 15, normalized size = 0.83 \begin {gather*} \frac {5 x^{2} e^{5}}{2} + 5 x e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x+5)*exp(5),x)

[Out]

5*x**2*exp(5)/2 + 5*x*exp(5)

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