3.44.28 \(\int \frac {88 e^5+e (33+88 x)}{16 e} \, dx\)

Optimal. Leaf size=15 \[ \frac {11}{4} x \left (\frac {3}{4}+2 e^4+x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.33, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12} \begin {gather*} \frac {11}{256} (8 x+3)^2+\frac {11 e^4 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(88*E^5 + E*(33 + 88*x))/(16*E),x]

[Out]

(11*E^4*x)/2 + (11*(3 + 8*x)^2)/256

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 {\int \left (88 e^5+e (33+88 x)\right ) \, dx}{16 e}\\ &=\frac {11 e^4 x}{2}+\frac {11}{256} (3+8 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.27 \begin {gather*} \frac {11}{16} \left (3 x+8 e^4 x+4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(88*E^5 + E*(33 + 88*x))/(16*E),x]

[Out]

(11*(3*x + 8*E^4*x + 4*x^2))/16

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fricas [A]  time = 0.71, size = 14, normalized size = 0.93 \begin {gather*} \frac {11}{4} \, x^{2} + \frac {11}{2} \, x e^{4} + \frac {33}{16} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(88*exp(5)+(88*x+33)*exp(1))/exp(1),x, algorithm="fricas")

[Out]

11/4*x^2 + 11/2*x*e^4 + 33/16*x

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giac [A]  time = 0.12, size = 22, normalized size = 1.47 \begin {gather*} \frac {11}{16} \, {\left (8 \, x e^{5} + {\left (4 \, x^{2} + 3 \, x\right )} e\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(88*exp(5)+(88*x+33)*exp(1))/exp(1),x, algorithm="giac")

[Out]

11/16*(8*x*e^5 + (4*x^2 + 3*x)*e)*e^(-1)

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maple [A]  time = 0.02, size = 15, normalized size = 1.00

method	result	size
risch	\(\frac {11 x \,{\mathrm e}^{4}}{2}+\frac {11 x^{2}}{4}+\frac {33 x}{16}\)	\(15\)
gosper	\(\frac {11 x \left (4 x \,{\mathrm e}+8 \,{\mathrm e}^{5}+3 \,{\mathrm e}\right ) {\mathrm e}^{-1}}{16}\)	\(22\)
norman	\(\frac {11 x^{2}}{4}+\frac {11 \left (8 \,{\mathrm e}^{5}+3 \,{\mathrm e}\right ) {\mathrm e}^{-1} x}{16}\)	\(23\)
default	\(\frac {{\mathrm e}^{-1} \left (88 x \,{\mathrm e}^{5}+{\mathrm e} \left (44 x^{2}+33 x \right )\right )}{16}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(88*exp(5)+(88*x+33)*exp(1))/exp(1),x,method=_RETURNVERBOSE)

[Out]

11/2*x*exp(4)+11/4*x^2+33/16*x

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maxima [A]  time = 0.35, size = 22, normalized size = 1.47 \begin {gather*} \frac {11}{16} \, {\left (8 \, x e^{5} + {\left (4 \, x^{2} + 3 \, x\right )} e\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(88*exp(5)+(88*x+33)*exp(1))/exp(1),x, algorithm="maxima")

[Out]

11/16*(8*x*e^5 + (4*x^2 + 3*x)*e)*e^(-1)

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mupad [B]  time = 0.43, size = 13, normalized size = 0.87 \begin {gather*} \frac {11\,{\left (8\,x+8\,{\mathrm {e}}^4+3\right )}^2}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-1)*((11*exp(5))/2 + (exp(1)*(88*x + 33))/16),x)

[Out]

(11*(8*x + 8*exp(4) + 3)^2)/256

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sympy [A]  time = 0.05, size = 17, normalized size = 1.13 \begin {gather*} \frac {11 x^{2}}{4} + x \left (\frac {33}{16} + \frac {11 e^{4}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(88*exp(5)+(88*x+33)*exp(1))/exp(1),x)

[Out]

11*x**2/4 + x*(33/16 + 11*exp(4)/2)

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