3.15.57 \(\int \frac {16 x^2+6 x^3+e^x (75+90 x+27 x^2)}{-10 x^2-6 x^3+e^x (-75 x+310 x^2+453 x^3+144 x^4)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (16+\frac {2 e^{-x}}{-5-3 x}-\frac {3}{x}\right ) \]

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Rubi [F]  time = 2.07, 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 {16 x^2+6 x^3+e^x \left (75+90 x+27 x^2\right )}{-10 x^2-6 x^3+e^x \left (-75 x+310 x^2+453 x^3+144 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16*x^2 + 6*x^3 + E^x*(75 + 90*x + 27*x^2))/(-10*x^2 - 6*x^3 + E^x*(-75*x + 310*x^2 + 453*x^3 + 144*x^4)),
x]

[Out]

Log[3 - 16*x] - Log[x] + 2*Defer[Int][(-15*E^x - 2*x + 71*E^x*x + 48*E^x*x^2)^(-1), x] + 2*Defer[Int][x/(-15*E
^x - 2*x + 71*E^x*x + 48*E^x*x^2), x] - 10*Defer[Int][1/((5 + 3*x)*(-15*E^x - 2*x + 71*E^x*x + 48*E^x*x^2)), x
] + 6*Defer[Int][1/((-3 + 16*x)*(-15*E^x - 2*x + 71*E^x*x + 48*E^x*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 x^2-6 x^3-e^x \left (75+90 x+27 x^2\right )}{x (5+3 x) \left (15 e^x+2 x-71 e^x x-48 e^x x^2\right )} \, dx\\ &=\int \left (\frac {3}{x (-3+16 x)}+\frac {2 \left (15-15 x+119 x^2+48 x^3\right )}{(5+3 x) (-3+16 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )}\right ) \, dx\\ &=2 \int \frac {15-15 x+119 x^2+48 x^3}{(5+3 x) (-3+16 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )} \, dx+3 \int \frac {1}{x (-3+16 x)} \, dx\\ &=2 \int \left (\frac {1}{-15 e^x-2 x+71 e^x x+48 e^x x^2}+\frac {x}{-15 e^x-2 x+71 e^x x+48 e^x x^2}-\frac {5}{(5+3 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )}+\frac {3}{(-3+16 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )}\right ) \, dx+16 \int \frac {1}{-3+16 x} \, dx-\int \frac {1}{x} \, dx\\ &=\log (3-16 x)-\log (x)+2 \int \frac {1}{-15 e^x-2 x+71 e^x x+48 e^x x^2} \, dx+2 \int \frac {x}{-15 e^x-2 x+71 e^x x+48 e^x x^2} \, dx+6 \int \frac {1}{(-3+16 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )} \, dx-10 \int \frac {1}{(5+3 x) \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.29, size = 52, normalized size = 2.36 \begin {gather*} -x+\log (3-16 x)-\log (x)-\log ((3-16 x) (5+3 x))+\log \left (-15 e^x-2 x+71 e^x x+48 e^x x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^2 + 6*x^3 + E^x*(75 + 90*x + 27*x^2))/(-10*x^2 - 6*x^3 + E^x*(-75*x + 310*x^2 + 453*x^3 + 144*
x^4)),x]

[Out]

-x + Log[3 - 16*x] - Log[x] - Log[(3 - 16*x)*(5 + 3*x)] + Log[-15*E^x - 2*x + 71*E^x*x + 48*E^x*x^2]

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fricas [B]  time = 0.67, size = 45, normalized size = 2.05 \begin {gather*} -x + \log \left (16 \, x - 3\right ) - \log \relax (x) + \log \left (\frac {{\left (48 \, x^{2} + 71 \, x - 15\right )} e^{x} - 2 \, x}{48 \, x^{2} + 71 \, x - 15}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((27*x^2+90*x+75)*exp(x)+6*x^3+16*x^2)/((144*x^4+453*x^3+310*x^2-75*x)*exp(x)-6*x^3-10*x^2),x, algor
ithm="fricas")

[Out]

-x + log(16*x - 3) - log(x) + log(((48*x^2 + 71*x - 15)*e^x - 2*x)/(48*x^2 + 71*x - 15))

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giac [B]  time = 0.33, size = 48, normalized size = 2.18 \begin {gather*} -x + \log \left (48 \, x^{2} e^{x} + 71 \, x e^{x} - 2 \, x - 15 \, e^{x}\right ) - \log \left (48 \, x^{2} + 71 \, x - 15\right ) + \log \left (16 \, x - 3\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((27*x^2+90*x+75)*exp(x)+6*x^3+16*x^2)/((144*x^4+453*x^3+310*x^2-75*x)*exp(x)-6*x^3-10*x^2),x, algor
ithm="giac")

[Out]

-x + log(48*x^2*e^x + 71*x*e^x - 2*x - 15*e^x) - log(48*x^2 + 71*x - 15) + log(16*x - 3) - log(x)

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maple [A]  time = 0.07, size = 34, normalized size = 1.55

method	result	size
risch	\(-\ln \relax (x )+\ln \left (16 x -3\right )-x +\ln \left ({\mathrm e}^{x}-\frac {2 x}{48 x^{2}+71 x -15}\right )\)	\(34\)
norman	\(-x -\ln \relax (x )-\ln \left (3 x +5\right )+\ln \left (48 \,{\mathrm e}^{x} x^{2}+71 \,{\mathrm e}^{x} x -2 x -15 \,{\mathrm e}^{x}\right )\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((27*x^2+90*x+75)*exp(x)+6*x^3+16*x^2)/((144*x^4+453*x^3+310*x^2-75*x)*exp(x)-6*x^3-10*x^2),x,method=_RETU
RNVERBOSE)

[Out]

-ln(x)+ln(16*x-3)-x+ln(exp(x)-2*x/(48*x^2+71*x-15))

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maxima [B]  time = 0.62, size = 45, normalized size = 2.05 \begin {gather*} -x + \log \left (16 \, x - 3\right ) - \log \relax (x) + \log \left (\frac {{\left (48 \, x^{2} + 71 \, x - 15\right )} e^{x} - 2 \, x}{48 \, x^{2} + 71 \, x - 15}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((27*x^2+90*x+75)*exp(x)+6*x^3+16*x^2)/((144*x^4+453*x^3+310*x^2-75*x)*exp(x)-6*x^3-10*x^2),x, algor
ithm="maxima")

[Out]

-x + log(16*x - 3) - log(x) + log(((48*x^2 + 71*x - 15)*e^x - 2*x)/(48*x^2 + 71*x - 15))

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mupad [B]  time = 1.07, size = 35, normalized size = 1.59 \begin {gather*} \ln \left (48\,x^2\,{\mathrm {e}}^x-15\,{\mathrm {e}}^x-2\,x+71\,x\,{\mathrm {e}}^x\right )-\ln \left (x+\frac {5}{3}\right )-x-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(90*x + 27*x^2 + 75) + 16*x^2 + 6*x^3)/(10*x^2 - exp(x)*(310*x^2 - 75*x + 453*x^3 + 144*x^4) + 6*
x^3),x)

[Out]

log(48*x^2*exp(x) - 15*exp(x) - 2*x + 71*x*exp(x)) - log(x + 5/3) - x - log(x)

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sympy [A]  time = 0.44, size = 29, normalized size = 1.32 \begin {gather*} - x - \log {\relax (x )} + \log {\left (x - \frac {3}{16} \right )} + \log {\left (- \frac {2 x}{48 x^{2} + 71 x - 15} + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((27*x**2+90*x+75)*exp(x)+6*x**3+16*x**2)/((144*x**4+453*x**3+310*x**2-75*x)*exp(x)-6*x**3-10*x**2),
x)

[Out]

-x - log(x) + log(x - 3/16) + log(-2*x/(48*x**2 + 71*x - 15) + exp(x))

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