3.87.6 \(\int \frac {e^{\frac {1}{64} (16 e^x x+4 x^2+(4 e^x x+x^2) \log (1-x))} (-8 x+9 x^2+e^x (-16+4 x+16 x^2)+(-2 x+2 x^2+e^x (-4+4 x^2)) \log (1-x))}{-64+64 x} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {1}{16} \left (e^x+\frac {x}{4}\right ) x (4+\log (1-x))} \]

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Rubi [F]  time = 7.61, 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 {\exp \left (\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )\right ) \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((16*E^x*x + 4*x^2 + (4*E^x*x + x^2)*Log[1 - x])/64)*(-8*x + 9*x^2 + E^x*(-16 + 4*x + 16*x^2) + (-2*x +
2*x^2 + E^x*(-4 + 4*x^2))*Log[1 - x]))/(-64 + 64*x),x]

[Out]

Defer[Int][E^((x*(4*E^x + x)*(4 + Log[1 - x]))/64), x]/64 + (5*Defer[Int][E^(x + (x*(4*E^x + x)*(4 + Log[1 - x
]))/64), x])/16 + Defer[Int][E^((x*(4*E^x + x)*(4 + Log[1 - x]))/64)/(-1 + x), x]/64 + Defer[Int][E^(x + (x*(4
*E^x + x)*(4 + Log[1 - x]))/64)/(-1 + x), x]/16 + (9*Defer[Int][E^((x*(4*E^x + x)*(4 + Log[1 - x]))/64)*x, x])
/64 + Defer[Int][E^(x + (x*(4*E^x + x)*(4 + Log[1 - x]))/64)*x, x]/4 + Defer[Int][E^(x + (x*(4*E^x + x)*(4 + L
og[1 - x]))/64)*Log[1 - x], x]/16 + Defer[Int][E^((x*(4*E^x + x)*(4 + Log[1 - x]))/64)*x*Log[1 - x], x]/32 + D
efer[Int][E^(x + (x*(4*E^x + x)*(4 + Log[1 - x]))/64)*x*Log[1 - x], x]/16

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \left (8 x-9 x^2-e^x \left (-16+4 x+16 x^2\right )-\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{64-64 x} \, dx\\ &=\int \left (\frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x (-8+9 x-2 \log (1-x)+2 x \log (1-x))}{64 (-1+x)}+\frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \left (-4+x+4 x^2-\log (1-x)+x^2 \log (1-x)\right )}{16 (-1+x)}\right ) \, dx\\ &=\frac {1}{64} \int \frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x (-8+9 x-2 \log (1-x)+2 x \log (1-x))}{-1+x} \, dx+\frac {1}{16} \int \frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \left (-4+x+4 x^2-\log (1-x)+x^2 \log (1-x)\right )}{-1+x} \, dx\\ &=\frac {1}{64} \int \left (\frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x (-8+9 x)}{-1+x}+2 e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x)\right ) \, dx+\frac {1}{16} \int \left (\frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \left (-4+x+4 x^2\right )}{-1+x}+e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} (1+x) \log (1-x)\right ) \, dx\\ &=\frac {1}{64} \int \frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x (-8+9 x)}{-1+x} \, dx+\frac {1}{32} \int e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x) \, dx+\frac {1}{16} \int \frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \left (-4+x+4 x^2\right )}{-1+x} \, dx+\frac {1}{16} \int e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} (1+x) \log (1-x) \, dx\\ &=\frac {1}{64} \int \left (e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}+\frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}}{-1+x}+9 e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x\right ) \, dx+\frac {1}{32} \int e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x) \, dx+\frac {1}{16} \int \left (5 e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}+\frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}}{-1+x}+4 e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x\right ) \, dx+\frac {1}{16} \int \left (e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \log (1-x)+e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x)\right ) \, dx\\ &=\frac {1}{64} \int e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \, dx+\frac {1}{64} \int \frac {e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}}{-1+x} \, dx+\frac {1}{32} \int e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x) \, dx+\frac {1}{16} \int \frac {e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))}}{-1+x} \, dx+\frac {1}{16} \int e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \log (1-x) \, dx+\frac {1}{16} \int e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \log (1-x) \, dx+\frac {9}{64} \int e^{\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \, dx+\frac {1}{4} \int e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} x \, dx+\frac {5}{16} \int e^{x+\frac {1}{64} x \left (4 e^x+x\right ) (4+\log (1-x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 33, normalized size = 1.38 \begin {gather*} e^{\frac {1}{16} x \left (4 e^x+x\right )} (1-x)^{\frac {1}{64} x \left (4 e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((16*E^x*x + 4*x^2 + (4*E^x*x + x^2)*Log[1 - x])/64)*(-8*x + 9*x^2 + E^x*(-16 + 4*x + 16*x^2) + (
-2*x + 2*x^2 + E^x*(-4 + 4*x^2))*Log[1 - x]))/(-64 + 64*x),x]

[Out]

E^((x*(4*E^x + x))/16)*(1 - x)^((x*(4*E^x + x))/64)

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fricas [A]  time = 0.53, size = 29, normalized size = 1.21 \begin {gather*} e^{\left (\frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x} + \frac {1}{64} \, {\left (x^{2} + 4 \, x e^{x}\right )} \log \left (-x + 1\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(-x+1)+(16*x^2+4*x-16)*exp(x)+9*x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*
log(-x+1)+1/4*exp(x)*x+1/16*x^2)/(64*x-64),x, algorithm="fricas")

[Out]

e^(1/16*x^2 + 1/4*x*e^x + 1/64*(x^2 + 4*x*e^x)*log(-x + 1))

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giac [A]  time = 0.24, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (\frac {1}{64} \, x^{2} \log \left (-x + 1\right ) + \frac {1}{16} \, x e^{x} \log \left (-x + 1\right ) + \frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(-x+1)+(16*x^2+4*x-16)*exp(x)+9*x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*
log(-x+1)+1/4*exp(x)*x+1/16*x^2)/(64*x-64),x, algorithm="giac")

[Out]

e^(1/64*x^2*log(-x + 1) + 1/16*x*e^x*log(-x + 1) + 1/16*x^2 + 1/4*x*e^x)

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maple [A]  time = 0.39, size = 27, normalized size = 1.12

method	result	size
risch	\(\left (1-x \right )^{\frac {x \left (4 \,{\mathrm e}^{x}+x \right )}{64}} {\mathrm e}^{\frac {x \left (4 \,{\mathrm e}^{x}+x \right )}{16}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2-4)*exp(x)+2*x^2-2*x)*ln(1-x)+(16*x^2+4*x-16)*exp(x)+9*x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*ln(1-x)+
1/4*exp(x)*x+1/16*x^2)/(64*x-64),x,method=_RETURNVERBOSE)

[Out]

(1-x)^(1/64*x*(4*exp(x)+x))*exp(1/16*x*(4*exp(x)+x))

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maxima [A]  time = 0.47, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (\frac {1}{64} \, x^{2} \log \left (-x + 1\right ) + \frac {1}{16} \, x e^{x} \log \left (-x + 1\right ) + \frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(-x+1)+(16*x^2+4*x-16)*exp(x)+9*x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*
log(-x+1)+1/4*exp(x)*x+1/16*x^2)/(64*x-64),x, algorithm="maxima")

[Out]

e^(1/64*x^2*log(-x + 1) + 1/16*x*e^x*log(-x + 1) + 1/16*x^2 + 1/4*x*e^x)

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mupad [B]  time = 5.84, size = 30, normalized size = 1.25 \begin {gather*} {\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{4}+\frac {x^2}{16}}\,{\left (1-x\right )}^{\frac {x\,{\mathrm {e}}^x}{16}+\frac {x^2}{64}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*exp(x))/4 + x^2/16 + (log(1 - x)*(4*x*exp(x) + x^2))/64)*(log(1 - x)*(exp(x)*(4*x^2 - 4) - 2*x + 2
*x^2) - 8*x + exp(x)*(4*x + 16*x^2 - 16) + 9*x^2))/(64*x - 64),x)

[Out]

exp((x*exp(x))/4 + x^2/16)*(1 - x)^((x*exp(x))/16 + x^2/64)

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sympy [A]  time = 0.73, size = 29, normalized size = 1.21 \begin {gather*} e^{\frac {x^{2}}{16} + \frac {x e^{x}}{4} + \left (\frac {x^{2}}{64} + \frac {x e^{x}}{16}\right ) \log {\left (1 - x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2-4)*exp(x)+2*x**2-2*x)*ln(-x+1)+(16*x**2+4*x-16)*exp(x)+9*x**2-8*x)*exp(1/64*(4*exp(x)*x+x*
*2)*ln(-x+1)+1/4*exp(x)*x+1/16*x**2)/(64*x-64),x)

[Out]

exp(x**2/16 + x*exp(x)/4 + (x**2/64 + x*exp(x)/16)*log(1 - x))

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