3.86.62 \(\int \frac {1}{2} (15 e^4+30 e^6+15 e^8) \, dx\)

Optimal. Leaf size=23 \[ 3 \left (e^2+e^4\right )^2 \left (-2+\frac {1}{2} (-2+x)+2 x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.65, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {8} \begin {gather*} \frac {15}{2} e^4 \left (1+e^2\right )^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15*E^4 + 30*E^6 + 15*E^8)/2,x]

[Out]

(15*E^4*(1 + E^2)^2*x)/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {15}{2} e^4 \left (1+e^2\right )^2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.00 \begin {gather*} \frac {15 e^4 x}{2}+15 e^6 x+\frac {15 e^8 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*E^4 + 30*E^6 + 15*E^8)/2,x]

[Out]

(15*E^4*x)/2 + 15*E^6*x + (15*E^8*x)/2

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fricas [A]  time = 0.49, size = 16, normalized size = 0.70 \begin {gather*} \frac {15}{2} \, x e^{8} + 15 \, x e^{6} + \frac {15}{2} \, x e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(15/2*exp(4)^2+15*exp(2)*exp(4)+15/2*exp(2)^2,x, algorithm="fricas")

[Out]

15/2*x*e^8 + 15*x*e^6 + 15/2*x*e^4

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giac [A]  time = 0.13, size = 12, normalized size = 0.52 \begin {gather*} \frac {15}{2} \, x {\left (e^{8} + 2 \, e^{6} + e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(15/2*exp(4)^2+15*exp(2)*exp(4)+15/2*exp(2)^2,x, algorithm="giac")

[Out]

15/2*x*(e^8 + 2*e^6 + e^4)

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maple [A]  time = 0.02, size = 17, normalized size = 0.74

method	result	size
risch	\(\frac {15 x \,{\mathrm e}^{8}}{2}+15 x \,{\mathrm e}^{6}+\frac {15 x \,{\mathrm e}^{4}}{2}\)	\(17\)
default	\(\left (\frac {15 \,{\mathrm e}^{8}}{2}+15 \,{\mathrm e}^{2} {\mathrm e}^{4}+\frac {15 \,{\mathrm e}^{4}}{2}\right ) x\)	\(22\)
norman	\(\left (\frac {15 \,{\mathrm e}^{8}}{2}+15 \,{\mathrm e}^{2} {\mathrm e}^{4}+\frac {15 \,{\mathrm e}^{4}}{2}\right ) x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(15/2*exp(4)^2+15*exp(2)*exp(4)+15/2*exp(2)^2,x,method=_RETURNVERBOSE)

[Out]

15/2*x*exp(8)+15*x*exp(6)+15/2*x*exp(4)

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maxima [A]  time = 0.46, size = 12, normalized size = 0.52 \begin {gather*} \frac {15}{2} \, x {\left (e^{8} + 2 \, e^{6} + e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(15/2*exp(4)^2+15*exp(2)*exp(4)+15/2*exp(2)^2,x, algorithm="maxima")

[Out]

15/2*x*(e^8 + 2*e^6 + e^4)

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mupad [B]  time = 0.00, size = 15, normalized size = 0.65 \begin {gather*} x\,\left (\frac {15\,{\mathrm {e}}^4}{2}+15\,{\mathrm {e}}^6+\frac {15\,{\mathrm {e}}^8}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*exp(4))/2 + 15*exp(6) + (15*exp(8))/2,x)

[Out]

x*((15*exp(4))/2 + 15*exp(6) + (15*exp(8))/2)

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sympy [A]  time = 0.05, size = 19, normalized size = 0.83 \begin {gather*} x \left (\frac {15 e^{4}}{2} + 15 e^{6} + \frac {15 e^{8}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(15/2*exp(4)**2+15*exp(2)*exp(4)+15/2*exp(2)**2,x)

[Out]

x*(15*exp(4)/2 + 15*exp(6) + 15*exp(8)/2)

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