3.90.78 \(\int (5+e^8 (40-20 x)+e^{16} (-40 x+15 x^2)) \, dx\)

Optimal. Leaf size=15 \[ 5 (-4+x) \left (1-e^8 x\right )^2 \]

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Rubi [B]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 2.13, number of steps used = 2, number of rules used = 0, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} 5 e^{16} x^3-20 e^{16} x^2+5 x-10 e^8 (2-x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[5 + E^8*(40 - 20*x) + E^16*(-40*x + 15*x^2),x]

[Out]

-10*E^8*(2 - x)^2 + 5*x - 20*E^16*x^2 + 5*E^16*x^3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-10 e^8 (2-x)^2+5 x+e^{16} \int \left (-40 x+15 x^2\right ) \, dx\\ &=-10 e^8 (2-x)^2+5 x-20 e^{16} x^2+5 e^{16} x^3\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.00, size = 34, normalized size = 2.27 \begin {gather*} 5 x+40 e^8 x-10 e^8 x^2-20 e^{16} x^2+5 e^{16} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[5 + E^8*(40 - 20*x) + E^16*(-40*x + 15*x^2),x]

[Out]

5*x + 40*E^8*x - 10*E^8*x^2 - 20*E^16*x^2 + 5*E^16*x^3

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fricas [B]  time = 0.47, size = 28, normalized size = 1.87 \begin {gather*} 5 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{16} - 10 \, {\left (x^{2} - 4 \, x\right )} e^{8} + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2-40*x)*exp(4)^4+(-20*x+40)*exp(4)^2+5,x, algorithm="fricas")

[Out]

5*(x^3 - 4*x^2)*e^16 - 10*(x^2 - 4*x)*e^8 + 5*x

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giac [B]  time = 0.13, size = 28, normalized size = 1.87 \begin {gather*} 5 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{16} - 10 \, {\left (x^{2} - 4 \, x\right )} e^{8} + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2-40*x)*exp(4)^4+(-20*x+40)*exp(4)^2+5,x, algorithm="giac")

[Out]

5*(x^3 - 4*x^2)*e^16 - 10*(x^2 - 4*x)*e^8 + 5*x

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maple [A]  time = 0.03, size = 31, normalized size = 2.07

method	result	size
risch	\(5 x^{3} {\mathrm e}^{16}-20 x^{2} {\mathrm e}^{16}-10 x^{2} {\mathrm e}^{8}+40 x \,{\mathrm e}^{8}+5 x\)	\(31\)
gosper	\(5 x \left (x^{2} {\mathrm e}^{16}-4 x \,{\mathrm e}^{16}-2 x \,{\mathrm e}^{8}+8 \,{\mathrm e}^{8}+1\right )\)	\(34\)
default	\({\mathrm e}^{16} \left (5 x^{3}-20 x^{2}\right )+{\mathrm e}^{8} \left (-10 x^{2}+40 x \right )+5 x\)	\(35\)
norman	\(\left (40 \,{\mathrm e}^{8}+5\right ) x +\left (-20 \,{\mathrm e}^{16}-10 \,{\mathrm e}^{8}\right ) x^{2}+5 x^{3} {\mathrm e}^{16}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x^2-40*x)*exp(4)^4+(-20*x+40)*exp(4)^2+5,x,method=_RETURNVERBOSE)

[Out]

5*x^3*exp(16)-20*x^2*exp(16)-10*x^2*exp(8)+40*x*exp(8)+5*x

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maxima [B]  time = 0.34, size = 28, normalized size = 1.87 \begin {gather*} 5 \, {\left (x^{3} - 4 \, x^{2}\right )} e^{16} - 10 \, {\left (x^{2} - 4 \, x\right )} e^{8} + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^2-40*x)*exp(4)^4+(-20*x+40)*exp(4)^2+5,x, algorithm="maxima")

[Out]

5*(x^3 - 4*x^2)*e^16 - 10*(x^2 - 4*x)*e^8 + 5*x

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mupad [B]  time = 0.05, size = 30, normalized size = 2.00 \begin {gather*} 5\,{\mathrm {e}}^{16}\,x^3+\left (-10\,{\mathrm {e}}^8-20\,{\mathrm {e}}^{16}\right )\,x^2+\left (40\,{\mathrm {e}}^8+5\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5 - exp(8)*(20*x - 40) - exp(16)*(40*x - 15*x^2),x)

[Out]

5*x^3*exp(16) - x^2*(10*exp(8) + 20*exp(16)) + x*(40*exp(8) + 5)

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sympy [B]  time = 0.06, size = 31, normalized size = 2.07 \begin {gather*} 5 x^{3} e^{16} + x^{2} \left (- 20 e^{16} - 10 e^{8}\right ) + x \left (5 + 40 e^{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x**2-40*x)*exp(4)**4+(-20*x+40)*exp(4)**2+5,x)

[Out]

5*x**3*exp(16) + x**2*(-20*exp(16) - 10*exp(8)) + x*(5 + 40*exp(8))

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