3.51.49 \(\int \frac {1}{5} (1+e^3 (5+90 x)) \, dx\)

Optimal. Leaf size=17 \[ \frac {x}{5}+e^3 \left (x+9 x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12} \begin {gather*} \frac {1}{36} e^3 (18 x+1)^2+\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^3*(5 + 90*x))/5,x]

[Out]

x/5 + (E^3*(1 + 18*x)^2)/36

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (1+e^3 (5+90 x)\right ) \, dx\\ &=\frac {x}{5}+\frac {1}{36} e^3 (1+18 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.12 \begin {gather*} \frac {x}{5}+e^3 x+9 e^3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^3*(5 + 90*x))/5,x]

[Out]

x/5 + E^3*x + 9*E^3*x^2

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fricas [A]  time = 0.84, size = 14, normalized size = 0.82 \begin {gather*} {\left (9 \, x^{2} + x\right )} e^{3} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(90*x+5)*exp(3)+1/5,x, algorithm="fricas")

[Out]

(9*x^2 + x)*e^3 + 1/5*x

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giac [A]  time = 0.13, size = 14, normalized size = 0.82 \begin {gather*} {\left (9 \, x^{2} + x\right )} e^{3} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(90*x+5)*exp(3)+1/5,x, algorithm="giac")

[Out]

(9*x^2 + x)*e^3 + 1/5*x

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maple [A]  time = 0.02, size = 15, normalized size = 0.88

method	result	size
gosper	\(\frac {x \left (45 x \,{\mathrm e}^{3}+5 \,{\mathrm e}^{3}+1\right )}{5}\)	\(15\)
norman	\(\left ({\mathrm e}^{3}+\frac {1}{5}\right ) x +9 x^{2} {\mathrm e}^{3}\)	\(15\)
risch	\(9 x^{2} {\mathrm e}^{3}+x \,{\mathrm e}^{3}+\frac {x}{5}\)	\(16\)
default	\(\frac {{\mathrm e}^{3} \left (45 x^{2}+5 x \right )}{5}+\frac {x}{5}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(90*x+5)*exp(3)+1/5,x,method=_RETURNVERBOSE)

[Out]

1/5*x*(45*x*exp(3)+5*exp(3)+1)

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maxima [A]  time = 0.34, size = 14, normalized size = 0.82 \begin {gather*} {\left (9 \, x^{2} + x\right )} e^{3} + \frac {1}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(90*x+5)*exp(3)+1/5,x, algorithm="maxima")

[Out]

(9*x^2 + x)*e^3 + 1/5*x

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mupad [B]  time = 0.07, size = 17, normalized size = 1.00 \begin {gather*} \frac {\left (90\,x+5\right )\,\left ({\mathrm {e}}^3\,\left (90\,x+5\right )+2\right )}{900} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(90*x + 5))/5 + 1/5,x)

[Out]

((90*x + 5)*(exp(3)*(90*x + 5) + 2))/900

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sympy [A]  time = 0.05, size = 15, normalized size = 0.88 \begin {gather*} 9 x^{2} e^{3} + x \left (\frac {1}{5} + e^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(90*x+5)*exp(3)+1/5,x)

[Out]

9*x**2*exp(3) + x*(1/5 + exp(3))

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