3.80.40 \(\int \frac {e^{\frac {1}{225} (x^2+6 x \log (x)+9 \log ^2(x))} (4950+132 x+50 x^2+2 x^3+(396+150 x+6 x^2) \log (x))}{108900+9900 x+225 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^{\frac {1}{25} \left (\frac {x}{3}+\log (x)\right )^2} x}{22+x} \]

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Rubi [B]  time = 0.17, antiderivative size = 69, normalized size of antiderivative = 3.00, number of steps used = 3, number of rules used = 3, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {27, 12, 2288} \begin {gather*} \frac {x^{2 x/75} e^{\frac {1}{225} \left (x^2+9 \log ^2(x)\right )} \left (x^3+25 x^2+3 \left (x^2+25 x+66\right ) \log (x)+66 x\right )}{(x+22)^2 \left (x+3 \log (x)+\frac {9 \log (x)}{x}+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x^2 + 6*x*Log[x] + 9*Log[x]^2)/225)*(4950 + 132*x + 50*x^2 + 2*x^3 + (396 + 150*x + 6*x^2)*Log[x]))/(
108900 + 9900*x + 225*x^2),x]

[Out]

(E^((x^2 + 9*Log[x]^2)/225)*x^((2*x)/75)*(66*x + 25*x^2 + x^3 + 3*(66 + 25*x + x^2)*Log[x]))/((22 + x)^2*(3 +
x + 3*Log[x] + (9*Log[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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2288

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} &=\int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{225 (22+x)^2} \, dx\\ &=\frac {1}{225} \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{(22+x)^2} \, dx\\ &=\frac {e^{\frac {1}{225} \left (x^2+9 \log ^2(x)\right )} x^{2 x/75} \left (66 x+25 x^2+x^3+3 \left (66+25 x+x^2\right ) \log (x)\right )}{(22+x)^2 \left (3+x+3 \log (x)+\frac {9 \log (x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.94, size = 33, normalized size = 1.43 \begin {gather*} \frac {e^{\frac {x^2}{225}+\frac {\log ^2(x)}{25}} x^{1+\frac {2 x}{75}}}{22+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2 + 6*x*Log[x] + 9*Log[x]^2)/225)*(4950 + 132*x + 50*x^2 + 2*x^3 + (396 + 150*x + 6*x^2)*Log[
x]))/(108900 + 9900*x + 225*x^2),x]

[Out]

(E^(x^2/225 + Log[x]^2/25)*x^(1 + (2*x)/75))/(22 + x)

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fricas [A]  time = 0.82, size = 25, normalized size = 1.09 \begin {gather*} \frac {x e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \relax (x) + \frac {1}{25} \, \log \relax (x)^{2}\right )}}{x + 22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+150*x+396)*log(x)+2*x^3+50*x^2+132*x+4950)*exp(1/25*log(x)^2+2/75*x*log(x)+1/225*x^2)/(225*x
^2+9900*x+108900),x, algorithm="fricas")

[Out]

x*e^(1/225*x^2 + 2/75*x*log(x) + 1/25*log(x)^2)/(x + 22)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{3} + 25 \, x^{2} + 3 \, {\left (x^{2} + 25 \, x + 66\right )} \log \relax (x) + 66 \, x + 2475\right )} e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \relax (x) + \frac {1}{25} \, \log \relax (x)^{2}\right )}}{225 \, {\left (x^{2} + 44 \, x + 484\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+150*x+396)*log(x)+2*x^3+50*x^2+132*x+4950)*exp(1/25*log(x)^2+2/75*x*log(x)+1/225*x^2)/(225*x
^2+9900*x+108900),x, algorithm="giac")

[Out]

integrate(2/225*(x^3 + 25*x^2 + 3*(x^2 + 25*x + 66)*log(x) + 66*x + 2475)*e^(1/225*x^2 + 2/75*x*log(x) + 1/25*
log(x)^2)/(x^2 + 44*x + 484), x)

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maple [A]  time = 0.25, size = 26, normalized size = 1.13

method	result	size
norman	\(\frac {x \,{\mathrm e}^{\frac {\ln \relax (x )^{2}}{25}+\frac {2 x \ln \relax (x )}{75}+\frac {x^{2}}{225}}}{22+x}\)	\(26\)
risch	\(\frac {x \,x^{\frac {2 x}{75}} {\mathrm e}^{\frac {\ln \relax (x )^{2}}{25}+\frac {x^{2}}{225}}}{22+x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2+150*x+396)*ln(x)+2*x^3+50*x^2+132*x+4950)*exp(1/25*ln(x)^2+2/75*x*ln(x)+1/225*x^2)/(225*x^2+9900*x
+108900),x,method=_RETURNVERBOSE)

[Out]

x*exp(1/25*ln(x)^2+2/75*x*ln(x)+1/225*x^2)/(22+x)

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maxima [A]  time = 0.45, size = 25, normalized size = 1.09 \begin {gather*} \frac {x e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \relax (x) + \frac {1}{25} \, \log \relax (x)^{2}\right )}}{x + 22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2+150*x+396)*log(x)+2*x^3+50*x^2+132*x+4950)*exp(1/25*log(x)^2+2/75*x*log(x)+1/225*x^2)/(225*x
^2+9900*x+108900),x, algorithm="maxima")

[Out]

x*e^(1/225*x^2 + 2/75*x*log(x) + 1/25*log(x)^2)/(x + 22)

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mupad [B]  time = 5.23, size = 25, normalized size = 1.09 \begin {gather*} \frac {x\,x^{\frac {2\,x}{75}}\,{\mathrm {e}}^{\frac {x^2}{225}+\frac {{\ln \relax (x)}^2}{25}}}{x+22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x)^2/25 + (2*x*log(x))/75 + x^2/225)*(132*x + log(x)*(150*x + 6*x^2 + 396) + 50*x^2 + 2*x^3 + 495
0))/(9900*x + 225*x^2 + 108900),x)

[Out]

(x*x^((2*x)/75)*exp(log(x)^2/25 + x^2/225))/(x + 22)

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sympy [A]  time = 0.33, size = 26, normalized size = 1.13 \begin {gather*} \frac {x e^{\frac {x^{2}}{225} + \frac {2 x \log {\relax (x )}}{75} + \frac {\log {\relax (x )}^{2}}{25}}}{x + 22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2+150*x+396)*ln(x)+2*x**3+50*x**2+132*x+4950)*exp(1/25*ln(x)**2+2/75*x*ln(x)+1/225*x**2)/(225
*x**2+9900*x+108900),x)

[Out]

x*exp(x**2/225 + 2*x*log(x)/75 + log(x)**2/25)/(x + 22)

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