3.100.7 \(\int (-160+e (10-4 x)+64 x) \, dx\)

Optimal. Leaf size=11 \[ 2 (-16+e) (5-x) x \]

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Rubi [A]  time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.91, number of steps used = 1, number of rules used = 0, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} 32 x^2-\frac {1}{2} e (5-2 x)^2-160 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-160 + E*(10 - 4*x) + 64*x,x]

[Out]

-1/2*(E*(5 - 2*x)^2) - 160*x + 32*x^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {1}{2} e (5-2 x)^2-160 x+32 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.09 \begin {gather*} -2 (-16+e) \left (-5 x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-160 + E*(10 - 4*x) + 64*x,x]

[Out]

-2*(-16 + E)*(-5*x + x^2)

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fricas [A]  time = 0.64, size = 20, normalized size = 1.82 \begin {gather*} 32 \, x^{2} - 2 \, {\left (x^{2} - 5 \, x\right )} e - 160 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10-4*x)*exp(1)+64*x-160,x, algorithm="fricas")

[Out]

32*x^2 - 2*(x^2 - 5*x)*e - 160*x

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giac [A]  time = 0.25, size = 20, normalized size = 1.82 \begin {gather*} 32 \, x^{2} - 2 \, {\left (x^{2} - 5 \, x\right )} e - 160 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10-4*x)*exp(1)+64*x-160,x, algorithm="giac")

[Out]

32*x^2 - 2*(x^2 - 5*x)*e - 160*x

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maple [A]  time = 0.02, size = 11, normalized size = 1.00

method	result	size
gosper	\(-2 \left ({\mathrm e}-16\right ) \left (x -5\right ) x\)	\(11\)
norman	\(\left (-2 \,{\mathrm e}+32\right ) x^{2}+\left (10 \,{\mathrm e}-160\right ) x\)	\(20\)
default	\({\mathrm e} \left (-2 x^{2}+10 x \right )+32 x^{2}-160 x\)	\(22\)
risch	\(-2 x^{2} {\mathrm e}+10 x \,{\mathrm e}+32 x^{2}-160 x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10-4*x)*exp(1)+64*x-160,x,method=_RETURNVERBOSE)

[Out]

-2*(exp(1)-16)*(x-5)*x

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maxima [A]  time = 0.36, size = 20, normalized size = 1.82 \begin {gather*} 32 \, x^{2} - 2 \, {\left (x^{2} - 5 \, x\right )} e - 160 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10-4*x)*exp(1)+64*x-160,x, algorithm="maxima")

[Out]

32*x^2 - 2*(x^2 - 5*x)*e - 160*x

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mupad [B]  time = 5.64, size = 10, normalized size = 0.91 \begin {gather*} -2\,x\,\left (\mathrm {e}-16\right )\,\left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(64*x - exp(1)*(4*x - 10) - 160,x)

[Out]

-2*x*(exp(1) - 16)*(x - 5)

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sympy [A]  time = 0.05, size = 17, normalized size = 1.55 \begin {gather*} x^{2} \left (32 - 2 e\right ) + x \left (-160 + 10 e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10-4*x)*exp(1)+64*x-160,x)

[Out]

x**2*(32 - 2*E) + x*(-160 + 10*E)

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