3.49.49 \(\int \frac {1}{2} (2 e^x+e^{\frac {1}{4} (-40 x+7 x^2+4 x \log (2 x))} (-18+7 x+2 \log (2 x))) \, dx\)

Optimal. Leaf size=19 \[ e^x+e^{x \left (-10+\frac {7 x}{4}+\log (2 x)\right )} \]

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Rubi [A]  time = 0.10, antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 4, number of rules used = 3, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {12, 2194, 6706} \begin {gather*} 2^x e^{\frac {1}{4} \left (7 x^2-40 x\right )} x^x+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^x + E^((-40*x + 7*x^2 + 4*x*Log[2*x])/4)*(-18 + 7*x + 2*Log[2*x]))/2,x]

[Out]

E^x + 2^x*E^((-40*x + 7*x^2)/4)*x^x

Rule 12

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (2 e^x+e^{\frac {1}{4} \left (-40 x+7 x^2+4 x \log (2 x)\right )} (-18+7 x+2 \log (2 x))\right ) \, dx\\ &=\frac {1}{2} \int e^{\frac {1}{4} \left (-40 x+7 x^2+4 x \log (2 x)\right )} (-18+7 x+2 \log (2 x)) \, dx+\int e^x \, dx\\ &=e^x+2^x e^{\frac {1}{4} \left (-40 x+7 x^2\right )} x^x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 24, normalized size = 1.26 \begin {gather*} e^x+2^x e^{-10 x+\frac {7 x^2}{4}} x^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^x + E^((-40*x + 7*x^2 + 4*x*Log[2*x])/4)*(-18 + 7*x + 2*Log[2*x]))/2,x]

[Out]

E^x + 2^x*E^(-10*x + (7*x^2)/4)*x^x

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fricas [A]  time = 0.80, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {7}{4} \, x^{2} + x \log \left (2 \, x\right ) - 10 \, x\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*log(2*x)+7*x-18)*exp(x*log(2*x)+7/4*x^2-10*x)+exp(x),x, algorithm="fricas")

[Out]

e^(7/4*x^2 + x*log(2*x) - 10*x) + e^x

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giac [A]  time = 0.35, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {7}{4} \, x^{2} + x \log \left (2 \, x\right ) - 10 \, x\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*log(2*x)+7*x-18)*exp(x*log(2*x)+7/4*x^2-10*x)+exp(x),x, algorithm="giac")

[Out]

e^(7/4*x^2 + x*log(2*x) - 10*x) + e^x

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maple [A]  time = 0.05, size = 19, normalized size = 1.00

method	result	size
risch	\(\left (2 x \right )^{x} {\mathrm e}^{\frac {x \left (7 x -40\right )}{4}}+{\mathrm e}^{x}\)	\(19\)
default	\({\mathrm e}^{x \ln \left (2 x \right )+\frac {7 x^{2}}{4}-10 x}+{\mathrm e}^{x}\)	\(20\)
norman	\({\mathrm e}^{x \ln \left (2 x \right )+\frac {7 x^{2}}{4}-10 x}+{\mathrm e}^{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*ln(2*x)+7*x-18)*exp(x*ln(2*x)+7/4*x^2-10*x)+exp(x),x,method=_RETURNVERBOSE)

[Out]

(2*x)^x*exp(1/4*x*(7*x-40))+exp(x)

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maxima [A]  time = 0.36, size = 19, normalized size = 1.00 \begin {gather*} e^{\left (\frac {7}{4} \, x^{2} + x \log \left (2 \, x\right ) - 10 \, x\right )} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*log(2*x)+7*x-18)*exp(x*log(2*x)+7/4*x^2-10*x)+exp(x),x, algorithm="maxima")

[Out]

e^(7/4*x^2 + x*log(2*x) - 10*x) + e^x

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mupad [B]  time = 3.34, size = 19, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^x+{\mathrm {e}}^{\frac {7\,x^2}{4}-10\,x}\,{\left (2\,x\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x) + (exp(x*log(2*x) - 10*x + (7*x^2)/4)*(7*x + 2*log(2*x) - 18))/2,x)

[Out]

exp(x) + exp((7*x^2)/4 - 10*x)*(2*x)^x

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sympy [A]  time = 0.38, size = 20, normalized size = 1.05 \begin {gather*} e^{x} + e^{\frac {7 x^{2}}{4} + x \log {\left (2 x \right )} - 10 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*ln(2*x)+7*x-18)*exp(x*ln(2*x)+7/4*x**2-10*x)+exp(x),x)

[Out]

exp(x) + exp(7*x**2/4 + x*log(2*x) - 10*x)

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