3.88.19 \(\int \frac {-8+16 x+e^{\frac {1}{16} (36-36 x+9 x^2)} (-72+306 x-81 x^2)}{8 x+72 e^{\frac {1}{16} (36-36 x+9 x^2)} x} \, dx\)

Optimal. Leaf size=27 \[ 2 x+\log (5)-\log \left (x+9 e^{\frac {9}{16} (2-x)^2} x\right ) \]

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Rubi [F]  time = 2.15, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{8 x+72 e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + 16*x + E^((36 - 36*x + 9*x^2)/16)*(-72 + 306*x - 81*x^2))/(8*x + 72*E^((36 - 36*x + 9*x^2)/16)*x),x]

[Out]

(17*x)/4 - Defer[Int][1/((1 + 9*E^((9*(-2 + x)^2)/16))*x), x] - 9*Defer[Int][E^((9*(-2 + x)^2)/16)/((1 + 9*E^(
(9*(-2 + x)^2)/16))*x), x] - (81*Defer[Int][(E^((9*(-2 + x)^2)/16)*x)/(1 + 9*E^((9*(-2 + x)^2)/16)), x])/8 - (
9*Defer[Subst][Defer[Int][(1 + 9*E^((9*x^2)/16))^(-1), x], x, -2 + x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{8 \left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {1}{8} \int \frac {-8+16 x+e^{\frac {1}{16} \left (36-36 x+9 x^2\right )} \left (-72+306 x-81 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {1}{8} \int \left (\frac {8 (-1+2 x)}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}-\frac {9 e^{\frac {9}{16} (-2+x)^2} \left (8-34 x+9 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}\right ) \, dx\\ &=-\left (\frac {9}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} \left (8-34 x+9 x^2\right )}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\right )+\int \frac {-1+2 x}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=-\left (\frac {9}{8} \int \left (-\frac {34 e^{\frac {9}{16} (-2+x)^2}}{1+9 e^{\frac {9}{16} (-2+x)^2}}+\frac {8 e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}+\frac {9 e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}}\right ) \, dx\right )+\int \left (\frac {2}{1+9 e^{\frac {9}{16} (-2+x)^2}}-\frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x}\right ) \, dx\\ &=2 \int \frac {1}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \int \frac {e^{\frac {9}{16} (-2+x)^2}}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \operatorname {Subst}\left (\int \frac {e^{\frac {9 x^2}{16}}}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx+\frac {153}{4} \operatorname {Subst}\left (\int \left (\frac {1}{9}-\frac {1}{9 \left (1+9 e^{\frac {9 x^2}{16}}\right )}\right ) \, dx,x,-2+x\right )-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ &=\frac {17 x}{4}+2 \operatorname {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-\frac {17}{4} \operatorname {Subst}\left (\int \frac {1}{1+9 e^{\frac {9 x^2}{16}}} \, dx,x,-2+x\right )-9 \int \frac {e^{\frac {9}{16} (-2+x)^2}}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx-\frac {81}{8} \int \frac {e^{\frac {9}{16} (-2+x)^2} x}{1+9 e^{\frac {9}{16} (-2+x)^2}} \, dx-\int \frac {1}{\left (1+9 e^{\frac {9}{16} (-2+x)^2}\right ) x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 30, normalized size = 1.11 \begin {gather*} \frac {1}{8} \left (16 x-8 \log \left (1+9 e^{\frac {9}{16} (-2+x)^2}\right )-8 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 16*x + E^((36 - 36*x + 9*x^2)/16)*(-72 + 306*x - 81*x^2))/(8*x + 72*E^((36 - 36*x + 9*x^2)/16)
*x),x]

[Out]

(16*x - 8*Log[1 + 9*E^((9*(-2 + x)^2)/16)] - 8*Log[x])/8

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fricas [A]  time = 0.52, size = 26, normalized size = 0.96 \begin {gather*} 2 \, x - \log \relax (x) - \log \left (9 \, e^{\left (\frac {9}{16} \, x^{2} - \frac {9}{4} \, x + \frac {9}{4}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-81*x^2+306*x-72)*exp(9/16*x^2-9/4*x+9/4)+16*x-8)/(72*x*exp(9/16*x^2-9/4*x+9/4)+8*x),x, algorithm=
"fricas")

[Out]

2*x - log(x) - log(9*e^(9/16*x^2 - 9/4*x + 9/4) + 1)

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giac [A]  time = 0.25, size = 26, normalized size = 0.96 \begin {gather*} 2 \, x - \log \relax (x) - \log \left (9 \, e^{\left (\frac {9}{16} \, x^{2} - \frac {9}{4} \, x + \frac {9}{4}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-81*x^2+306*x-72)*exp(9/16*x^2-9/4*x+9/4)+16*x-8)/(72*x*exp(9/16*x^2-9/4*x+9/4)+8*x),x, algorithm=
"giac")

[Out]

2*x - log(x) - log(9*e^(9/16*x^2 - 9/4*x + 9/4) + 1)

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maple [A]  time = 0.07, size = 23, normalized size = 0.85

method	result	size
risch	\(2 x -\ln \relax (x )+\frac {9}{4}-\ln \left ({\mathrm e}^{\frac {9 \left (x -2\right )^{2}}{16}}+\frac {1}{9}\right )\)	\(23\)
norman	\(2 x -\ln \relax (x )-\ln \left (9 \,{\mathrm e}^{\frac {9}{16} x^{2}-\frac {9}{4} x +\frac {9}{4}}+1\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-81*x^2+306*x-72)*exp(9/16*x^2-9/4*x+9/4)+16*x-8)/(72*x*exp(9/16*x^2-9/4*x+9/4)+8*x),x,method=_RETURNVER
BOSE)

[Out]

2*x-ln(x)+9/4-ln(exp(9/16*(x-2)^2)+1/9)

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maxima [A]  time = 0.40, size = 30, normalized size = 1.11 \begin {gather*} \frac {17}{4} \, x - \log \left (\frac {1}{9} \, {\left (9 \, e^{\left (\frac {9}{16} \, x^{2} + \frac {9}{4}\right )} + e^{\left (\frac {9}{4} \, x\right )}\right )} e^{\left (-\frac {9}{4}\right )}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-81*x^2+306*x-72)*exp(9/16*x^2-9/4*x+9/4)+16*x-8)/(72*x*exp(9/16*x^2-9/4*x+9/4)+8*x),x, algorithm=
"maxima")

[Out]

17/4*x - log(1/9*(9*e^(9/16*x^2 + 9/4) + e^(9/4*x))*e^(-9/4)) - log(x)

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mupad [B]  time = 0.15, size = 25, normalized size = 0.93 \begin {gather*} \frac {17\,x}{4}-\ln \left (x\,{\left ({\mathrm {e}}^x\right )}^{9/4}+9\,x\,{\left ({\mathrm {e}}^{x^2}\right )}^{9/16}\,{\mathrm {e}}^{9/4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((9*x^2)/16 - (9*x)/4 + 9/4)*(81*x^2 - 306*x + 72) - 16*x + 8)/(8*x + 72*x*exp((9*x^2)/16 - (9*x)/4 +
9/4)),x)

[Out]

(17*x)/4 - log(x*exp(x)^(9/4) + 9*x*exp(x^2)^(9/16)*exp(9/4))

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sympy [A]  time = 0.16, size = 27, normalized size = 1.00 \begin {gather*} 2 x - \log {\relax (x )} - \log {\left (e^{\frac {9 x^{2}}{16} - \frac {9 x}{4} + \frac {9}{4}} + \frac {1}{9} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-81*x**2+306*x-72)*exp(9/16*x**2-9/4*x+9/4)+16*x-8)/(72*x*exp(9/16*x**2-9/4*x+9/4)+8*x),x)

[Out]

2*x - log(x) - log(exp(9*x**2/16 - 9*x/4 + 9/4) + 1/9)

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