3.35.86 \(\int \frac {1}{2} (50+e^{\frac {1}{2} (23 x+x^2+4 x \log (x))} (2+27 x+2 x^2+4 x \log (x))) \, dx\) [3486]

Optimal. Leaf size=22 \[ \left (25+e^{\frac {1}{2} x (15+x)+2 x (2+\log (x))}\right ) x \]

[Out]

(25+exp(1/4*(x+15)*x+(ln(x)+2)*x)^2)*x

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(22)=44\).
time = 0.07, antiderivative size = 48, normalized size of antiderivative = 2.18, number of steps used = 3, number of rules used = 2, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12, 2326} \begin {gather*} \frac {e^{\frac {1}{2} \left (x^2+23 x\right )} x^{2 x} \left (2 x^2+27 x+4 x \log (x)\right )}{2 x+4 \log (x)+27}+25 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50 + E^((23*x + x^2 + 4*x*Log[x])/2)*(2 + 27*x + 2*x^2 + 4*x*Log[x]))/2,x]

[Out]

25*x + (E^((23*x + x^2)/2)*x^(2*x)*(27*x + 2*x^2 + 4*x*Log[x]))/(27 + 2*x + 4*Log[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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (50+e^{\frac {1}{2} \left (23 x+x^2+4 x \log (x)\right )} \left (2+27 x+2 x^2+4 x \log (x)\right )\right ) \, dx\\ &=25 x+\frac {1}{2} \int e^{\frac {1}{2} \left (23 x+x^2+4 x \log (x)\right )} \left (2+27 x+2 x^2+4 x \log (x)\right ) \, dx\\ &=25 x+\frac {e^{\frac {1}{2} \left (23 x+x^2\right )} x^{2 x} \left (27 x+2 x^2+4 x \log (x)\right )}{27+2 x+4 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.12, size = 27, normalized size = 1.23 \begin {gather*} 25 x+e^{\frac {23 x}{2}+\frac {x^2}{2}} x^{1+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50 + E^((23*x + x^2 + 4*x*Log[x])/2)*(2 + 27*x + 2*x^2 + 4*x*Log[x]))/2,x]

[Out]

25*x + E^((23*x)/2 + x^2/2)*x^(1 + 2*x)

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Maple [A]
time = 0.21, size = 23, normalized size = 1.05

method	result	size
risch	\(25 x +x^{2 x} {\mathrm e}^{\frac {x \left (x +23\right )}{2}} x\)	\(19\)
default	\(25 x +{\mathrm e}^{2 x \ln \left (x \right )+\frac {x^{2}}{2}+\frac {23 x}{2}} x\)	\(23\)
norman	\(25 x +{\mathrm e}^{2 x \ln \left (x \right )+\frac {x^{2}}{2}+\frac {23 x}{2}} x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(4*x*ln(x)+2*x^2+27*x+2)*exp(x*ln(x)+1/4*x^2+23/4*x)^2+25,x,method=_RETURNVERBOSE)

[Out]

25*x+exp(x*ln(x)+1/4*x^2+23/4*x)^2*x

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Maxima [A]
time = 0.32, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (\frac {1}{2} \, x^{2} + 2 \, x \log \left (x\right ) + \frac {23}{2} \, x\right )} + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(4*x*log(x)+2*x^2+27*x+2)*exp(x*log(x)+1/4*x^2+23/4*x)^2+25,x, algorithm="maxima")

[Out]

x*e^(1/2*x^2 + 2*x*log(x) + 23/2*x) + 25*x

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Fricas [A]
time = 0.37, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (\frac {1}{2} \, x^{2} + 2 \, x \log \left (x\right ) + \frac {23}{2} \, x\right )} + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(4*x*log(x)+2*x^2+27*x+2)*exp(x*log(x)+1/4*x^2+23/4*x)^2+25,x, algorithm="fricas")

[Out]

x*e^(1/2*x^2 + 2*x*log(x) + 23/2*x) + 25*x

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Sympy [A]
time = 0.08, size = 22, normalized size = 1.00 \begin {gather*} x e^{\frac {x^{2}}{2} + 2 x \log {\left (x \right )} + \frac {23 x}{2}} + 25 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(4*x*ln(x)+2*x**2+27*x+2)*exp(x*ln(x)+1/4*x**2+23/4*x)**2+25,x)

[Out]

x*exp(x**2/2 + 2*x*log(x) + 23*x/2) + 25*x

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Giac [A]
time = 0.42, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (\frac {1}{2} \, x^{2} + 2 \, x \log \left (x\right ) + \frac {23}{2} \, x\right )} + 25 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(4*x*log(x)+2*x^2+27*x+2)*exp(x*log(x)+1/4*x^2+23/4*x)^2+25,x, algorithm="giac")

[Out]

x*e^(1/2*x^2 + 2*x*log(x) + 23/2*x) + 25*x

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Mupad [B]
time = 2.13, size = 20, normalized size = 0.91 \begin {gather*} x\,\left (x^{2\,x}\,{\mathrm {e}}^{\frac {x^2}{2}+\frac {23\,x}{2}}+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((23*x)/2 + 2*x*log(x) + x^2/2)*(27*x + 4*x*log(x) + 2*x^2 + 2))/2 + 25,x)

[Out]

x*(x^(2*x)*exp((23*x)/2 + x^2/2) + 25)

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