3.103.61 \(\int \frac {1}{4} e^{\frac {1}{4} (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2)))} (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))) \, dx\)

Optimal. Leaf size=29 \[ 1+e^{\frac {1}{4} \left (x+\left (4+4 \left (3-e^x\right )\right ) x\right ) (3-\log (2))} \]

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Rubi [A]  time = 0.56, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {12, 6741, 6706} \begin {gather*} 2^{-\frac {1}{4} \left (17-4 e^x\right ) x} e^{\frac {3}{4} \left (17-4 e^x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((51*x - 17*x*Log[2] + E^x*(-12*x + 4*x*Log[2]))/4)*(51 - 17*Log[2] + E^x*(-12 - 12*x + (4 + 4*x)*Log[2
])))/4,x]

[Out]

E^((3*(17 - 4*E^x)*x)/4)/2^(((17 - 4*E^x)*x)/4)

Rule 12

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \exp \left (\frac {1}{4} \left (51 x-17 x \log (2)+e^x (-12 x+4 x \log (2))\right )\right ) \left (51-17 \log (2)+e^x (-12-12 x+(4+4 x) \log (2))\right ) \, dx\\ &=\frac {1}{4} \int e^{\frac {1}{4} \left (-17+4 e^x\right ) x (-3+\log (2))} \left (17-4 e^x-4 e^x x\right ) (3-\log (2)) \, dx\\ &=\frac {1}{4} (3-\log (2)) \int e^{\frac {1}{4} \left (-17+4 e^x\right ) x (-3+\log (2))} \left (17-4 e^x-4 e^x x\right ) \, dx\\ &=2^{-\frac {1}{4} \left (17-4 e^x\right ) x} e^{\frac {3}{4} \left (17-4 e^x\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.50, size = 25, normalized size = 0.86 \begin {gather*} 2^{-17 x/4} e^{\frac {51 x}{4}+e^x x (-3+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((51*x - 17*x*Log[2] + E^x*(-12*x + 4*x*Log[2]))/4)*(51 - 17*Log[2] + E^x*(-12 - 12*x + (4 + 4*x)
*Log[2])))/4,x]

[Out]

E^((51*x)/4 + E^x*x*(-3 + Log[2]))/2^((17*x)/4)

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fricas [A]  time = 0.73, size = 21, normalized size = 0.72 \begin {gather*} e^{\left ({\left (x \log \relax (2) - 3 \, x\right )} e^{x} - \frac {17}{4} \, x \log \relax (2) + \frac {51}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((4*x+4)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x*log(2)-12*x)*exp(x)-17/4*x*log(2)+51
/4*x),x, algorithm="fricas")

[Out]

e^((x*log(2) - 3*x)*e^x - 17/4*x*log(2) + 51/4*x)

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giac [A]  time = 0.17, size = 21, normalized size = 0.72 \begin {gather*} e^{\left (x e^{x} \log \relax (2) - 3 \, x e^{x} - \frac {17}{4} \, x \log \relax (2) + \frac {51}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((4*x+4)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x*log(2)-12*x)*exp(x)-17/4*x*log(2)+51
/4*x),x, algorithm="giac")

[Out]

e^(x*e^x*log(2) - 3*x*e^x - 17/4*x*log(2) + 51/4*x)

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maple [A]  time = 0.09, size = 23, normalized size = 0.79

method	result	size
risch	\(2^{-\frac {17 x}{4}} 2^{{\mathrm e}^{x} x} {\mathrm e}^{-\frac {3 x \left (4 \,{\mathrm e}^{x}-17\right )}{4}}\)	\(23\)
norman	\({\mathrm e}^{\frac {\left (4 x \ln \relax (2)-12 x \right ) {\mathrm e}^{x}}{4}-\frac {17 x \ln \relax (2)}{4}+\frac {51 x}{4}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(((4*x+4)*ln(2)-12*x-12)*exp(x)-17*ln(2)+51)*exp(1/4*(4*x*ln(2)-12*x)*exp(x)-17/4*x*ln(2)+51/4*x),x,me
thod=_RETURNVERBOSE)

[Out]

2^(-17/4*x)*2^(exp(x)*x)*exp(-3/4*x*(4*exp(x)-17))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{4} \, \int {\left (4 \, {\left ({\left (x + 1\right )} \log \relax (2) - 3 \, x - 3\right )} e^{x} - 17 \, \log \relax (2) + 51\right )} e^{\left ({\left (x \log \relax (2) - 3 \, x\right )} e^{x} - \frac {17}{4} \, x \log \relax (2) + \frac {51}{4} \, x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((4*x+4)*log(2)-12*x-12)*exp(x)-17*log(2)+51)*exp(1/4*(4*x*log(2)-12*x)*exp(x)-17/4*x*log(2)+51
/4*x),x, algorithm="maxima")

[Out]

1/4*integrate((4*((x + 1)*log(2) - 3*x - 3)*e^x - 17*log(2) + 51)*e^((x*log(2) - 3*x)*e^x - 17/4*x*log(2) + 51
/4*x), x)

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mupad [B]  time = 0.18, size = 21, normalized size = 0.72 \begin {gather*} 2^{x\,{\mathrm {e}}^x-\frac {17\,x}{4}}\,{\mathrm {e}}^{-3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {51\,x}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((51*x)/4 - (17*x*log(2))/4 - (exp(x)*(12*x - 4*x*log(2)))/4)*(17*log(2) + exp(x)*(12*x - log(2)*(4*x
+ 4) + 12) - 51))/4,x)

[Out]

2^(x*exp(x) - (17*x)/4)*exp(-3*x*exp(x))*exp((51*x)/4)

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sympy [A]  time = 0.28, size = 26, normalized size = 0.90 \begin {gather*} e^{- \frac {17 x \log {\relax (2 )}}{4} + \frac {51 x}{4} + \left (- 3 x + x \log {\relax (2 )}\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((4*x+4)*ln(2)-12*x-12)*exp(x)-17*ln(2)+51)*exp(1/4*(4*x*ln(2)-12*x)*exp(x)-17/4*x*ln(2)+51/4*x
),x)

[Out]

exp(-17*x*log(2)/4 + 51*x/4 + (-3*x + x*log(2))*exp(x))

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