\(\int \frac {e^{x^3+e^x (-5 x^2-x^3)+e^x x^3 \log (64+32 x+4 x^2)} (12 x^2+3 x^3+e^x (-40 x-42 x^2-10 x^3-x^4)+e^x (12 x^2+7 x^3+x^4) \log (64+32 x+4 x^2))}{4+x} \, dx\) [3721]

Optimal result

Integrand size = 108, antiderivative size = 27 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{x^2 \left (x-e^x \left (5+x-x \log \left (4 (4+x)^2\right )\right )\right )} \]

[Out]

exp(x^2*(x-exp(x)*(5-ln(4*(4+x)^2)*x+x)))

Rubi [F]

\[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=\int \frac {\exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx \]

[In]

Int[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*(12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2 - 10*
x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4)*Log[64 + 32*x + 4*x^2]))/(4 + x),x]

[Out]

32*Defer[Int][E^(x + x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2]), x] - 18*Defer[Int][E^(x + x^3
+ E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*x, x] + 3*Defer[Int][E^(x^3 + E^x*(-5*x^2 - x^3) + E^x
*x^3*Log[64 + 32*x + 4*x^2])*x^2, x] - 6*Defer[Int][E^(x + x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x +
4*x^2])*x^2, x] - Defer[Int][E^(x + x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*x^3, x] - 128*D
efer[Int][E^(x + x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])/(4 + x), x] + 3*Defer[Int][E^(x +
x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*x^2*Log[4*(4 + x)^2], x] + Defer[Int][E^(x + x^3 +
E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*x^3*Log[4*(4 + x)^2], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (3 \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2+\frac {\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x \left (-40-42 x-10 x^2-x^3+12 x \log \left (4 (4+x)^2\right )+7 x^2 \log \left (4 (4+x)^2\right )+x^3 \log \left (4 (4+x)^2\right )\right )}{4+x}\right ) \, dx \\ & = 3 \int \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx+\int \frac {\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x \left (-40-42 x-10 x^2-x^3+12 x \log \left (4 (4+x)^2\right )+7 x^2 \log \left (4 (4+x)^2\right )+x^3 \log \left (4 (4+x)^2\right )\right )}{4+x} \, dx \\ & = 3 \int \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx+\int \left (-\frac {\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x \left (40+42 x+10 x^2+x^3\right )}{4+x}+\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 (3+x) \log \left (4 (4+x)^2\right )\right ) \, dx \\ & = 3 \int \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx-\int \frac {\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x \left (40+42 x+10 x^2+x^3\right )}{4+x} \, dx+\int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 (3+x) \log \left (4 (4+x)^2\right ) \, dx \\ & = 3 \int \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx-\int \left (-32 \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right )+18 \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x+6 \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2+\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^3+\frac {128 \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right )}{4+x}\right ) \, dx+\int \left (3 \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \log \left (4 (4+x)^2\right )+\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^3 \log \left (4 (4+x)^2\right )\right ) \, dx \\ & = 3 \int \exp \left (x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx+3 \int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \log \left (4 (4+x)^2\right ) \, dx-6 \int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^2 \, dx-18 \int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x \, dx+32 \int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) \, dx-128 \int \frac {\exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx-\int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^3 \, dx+\int \exp \left (x+x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )\right ) x^3 \log \left (4 (4+x)^2\right ) \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=\int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx \]

[In]

Integrate[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*(12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2
- 10*x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4)*Log[64 + 32*x + 4*x^2]))/(4 + x),x]

[Out]

Integrate[(E^(x^3 + E^x*(-5*x^2 - x^3) + E^x*x^3*Log[64 + 32*x + 4*x^2])*(12*x^2 + 3*x^3 + E^x*(-40*x - 42*x^2
- 10*x^3 - x^4) + E^x*(12*x^2 + 7*x^3 + x^4)*Log[64 + 32*x + 4*x^2]))/(4 + x), x]

Maple [A] (verified)

Time = 5.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\({\mathrm e}^{x^{2} \left (x \,{\mathrm e}^{x} \ln \left (4 x^{2}+32 x +64\right )-{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}+x \right )}\)	\(32\)
risch	\(4^{{\mathrm e}^{x} x^{3}} \left (4+x \right )^{2 \,{\mathrm e}^{x} x^{3}} {\mathrm e}^{-\frac {x^{2} \left (i {\mathrm e}^{x} x \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i {\mathrm e}^{x} x \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )\right )+i {\mathrm e}^{x} x \pi \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i \left (4+x \right )\right )^{2}+2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}-2 x \right )}{2}}\)	\(108\)

[In]

int(((x^4+7*x^3+12*x^2)*exp(x)*ln(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3*exp(x)
*ln(4*x^2+32*x+64)+(-x^3-5*x^2)*exp(x)+x^3)/(4+x),x,method=_RETURNVERBOSE)

[Out]

exp(x^2*(x*exp(x)*ln(4*x^2+32*x+64)-exp(x)*x-5*exp(x)+x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (x^{3} e^{x} \log \left (4 \, x^{2} + 32 \, x + 64\right ) + x^{3} - {\left (x^{3} + 5 \, x^{2}\right )} e^{x}\right )} \]

[In]

integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3
*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2)*exp(x)+x^3)/(4+x),x, algorithm="fricas")

[Out]

e^(x^3*e^x*log(4*x^2 + 32*x + 64) + x^3 - (x^3 + 5*x^2)*e^x)

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{x^{3} e^{x} \log {\left (4 x^{2} + 32 x + 64 \right )} + x^{3} + \left (- x^{3} - 5 x^{2}\right ) e^{x}} \]

[In]

integrate(((x**4+7*x**3+12*x**2)*exp(x)*ln(4*x**2+32*x+64)+(-x**4-10*x**3-42*x**2-40*x)*exp(x)+3*x**3+12*x**2)
*exp(x**3*exp(x)*ln(4*x**2+32*x+64)+(-x**3-5*x**2)*exp(x)+x**3)/(4+x),x)

[Out]

exp(x**3*exp(x)*log(4*x**2 + 32*x + 64) + x**3 + (-x**3 - 5*x**2)*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (2 \, x^{3} e^{x} \log \left (2\right ) + 2 \, x^{3} e^{x} \log \left (x + 4\right ) - x^{3} e^{x} + x^{3} - 5 \, x^{2} e^{x}\right )} \]

[In]

integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3
*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2)*exp(x)+x^3)/(4+x),x, algorithm="maxima")

[Out]

e^(2*x^3*e^x*log(2) + 2*x^3*e^x*log(x + 4) - x^3*e^x + x^3 - 5*x^2*e^x)

Giac [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx=e^{\left (x^{3} e^{x} \log \left (4 \, x^{2} + 32 \, x + 64\right ) - x^{3} e^{x} + x^{3} - 5 \, x^{2} e^{x}\right )} \]

[In]

integrate(((x^4+7*x^3+12*x^2)*exp(x)*log(4*x^2+32*x+64)+(-x^4-10*x^3-42*x^2-40*x)*exp(x)+3*x^3+12*x^2)*exp(x^3
*exp(x)*log(4*x^2+32*x+64)+(-x^3-5*x^2)*exp(x)+x^3)/(4+x),x, algorithm="giac")

[Out]

e^(x^3*e^x*log(4*x^2 + 32*x + 64) - x^3*e^x + x^3 - 5*x^2*e^x)

Mupad [B] (verification not implemented)

Time = 8.71 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x^3+e^x \left (-5 x^2-x^3\right )+e^x x^3 \log \left (64+32 x+4 x^2\right )} \left (12 x^2+3 x^3+e^x \left (-40 x-42 x^2-10 x^3-x^4\right )+e^x \left (12 x^2+7 x^3+x^4\right ) \log \left (64+32 x+4 x^2\right )\right )}{4+x} \, dx={\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-5\,x^2\,{\mathrm {e}}^x}\,{\left (4\,x^2+32\,x+64\right )}^{x^3\,{\mathrm {e}}^x} \]

[In]

int((exp(x^3 - exp(x)*(5*x^2 + x^3) + x^3*exp(x)*log(32*x + 4*x^2 + 64))*(12*x^2 - exp(x)*(40*x + 42*x^2 + 10*
x^3 + x^4) + 3*x^3 + exp(x)*log(32*x + 4*x^2 + 64)*(12*x^2 + 7*x^3 + x^4)))/(x + 4),x)

[Out]

exp(x^3)*exp(-x^3*exp(x))*exp(-5*x^2*exp(x))*(32*x + 4*x^2 + 64)^(x^3*exp(x))