3.96.97 \(\int e^{-1+e^{4 x} x^2} x^4 (8 x^3+e^{4 x} x^2 (2 x^3+4 x^4)) \, dx\)

Optimal. Leaf size=17 \[ e^{-1+e^{4 x} x^2} x^8 \]

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Rubi [B]  time = 0.08, antiderivative size = 49, normalized size of antiderivative = 2.88, number of steps used = 1, number of rules used = 1, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2288} \begin {gather*} \frac {e^{e^{4 x} x^2+4 x-1} x^6 \left (2 x^4+x^3\right )}{2 e^{4 x} x^2+e^{4 x} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-1 + E^(4*x)*x^2)*x^4*(8*x^3 + E^(4*x)*x^2*(2*x^3 + 4*x^4)),x]

[Out]

(E^(-1 + 4*x + E^(4*x)*x^2)*x^6*(x^3 + 2*x^4))/(E^(4*x)*x + 2*E^(4*x)*x^2)

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} &=\frac {e^{-1+4 x+e^{4 x} x^2} x^6 \left (x^3+2 x^4\right )}{e^{4 x} x+2 e^{4 x} x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 1.00 \begin {gather*} e^{-1+e^{4 x} x^2} x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-1 + E^(4*x)*x^2)*x^4*(8*x^3 + E^(4*x)*x^2*(2*x^3 + 4*x^4)),x]

[Out]

E^(-1 + E^(4*x)*x^2)*x^8

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fricas [A]  time = 0.51, size = 20, normalized size = 1.18 \begin {gather*} x^{4} e^{\left (e^{\left (4 \, x + 2 \, \log \relax (x)\right )} + 4 \, \log \relax (x) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+2*x^3)*exp(log(exp(x)^2*x^2)+2*x)+8*x^3)*exp(1/4*exp(log(exp(x)^2*x^2)+2*x)+log(x)-1/4)^4,x,
algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 2 \, {\left (4 \, x^{3} + {\left (2 \, x^{4} + x^{3}\right )} e^{\left (2 \, x + \log \left (x^{2} e^{\left (2 \, x\right )}\right )\right )}\right )} e^{\left (e^{\left (2 \, x + \log \left (x^{2} e^{\left (2 \, x\right )}\right )\right )} + 4 \, \log \relax (x) - 1\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+2*x^3)*exp(log(exp(x)^2*x^2)+2*x)+8*x^3)*exp(1/4*exp(log(exp(x)^2*x^2)+2*x)+log(x)-1/4)^4,x,
algorithm="giac")

[Out]

integrate(2*(4*x^3 + (2*x^4 + x^3)*e^(2*x + log(x^2*e^(2*x))))*e^(e^(2*x + log(x^2*e^(2*x))) + 4*log(x) - 1),
x)

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maple [C]  time = 0.53, size = 185, normalized size = 10.88

method	result	size
risch	\(x^{8} {\mathrm e}^{x^{2} {\mathrm e}^{4 x} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{3}}{2}}-1}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4+2*x^3)*exp(ln(exp(x)^2*x^2)+2*x)+8*x^3)*exp(1/4*exp(ln(exp(x)^2*x^2)+2*x)+ln(x)-1/4)^4,x,method=_R
ETURNVERBOSE)

[Out]

x^8*exp(x^2*exp(4*x)*exp(-1/2*I*Pi*csgn(I*exp(2*x))^3)*exp(-1/2*I*Pi*csgn(I*exp(2*x))*csgn(I*exp(x))^2)*exp(-1
/2*I*Pi*csgn(I*x^2)^3)*exp(-1/2*I*Pi*csgn(I*x^2)*csgn(I*x)^2)*exp(-1/2*I*Pi*csgn(I*exp(2*x))*csgn(I*x^2)*csgn(
I*x^2*exp(2*x)))*exp(1/2*I*Pi*csgn(I*exp(2*x))*csgn(I*x^2*exp(2*x))^2)*exp(1/2*I*Pi*csgn(I*x^2)*csgn(I*x^2*exp
(2*x))^2)*exp(-1/2*I*Pi*csgn(I*x^2*exp(2*x))^3)-1)

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maxima [A]  time = 0.43, size = 15, normalized size = 0.88 \begin {gather*} x^{8} e^{\left (x^{2} e^{\left (4 \, x\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+2*x^3)*exp(log(exp(x)^2*x^2)+2*x)+8*x^3)*exp(1/4*exp(log(exp(x)^2*x^2)+2*x)+log(x)-1/4)^4,x,
algorithm="maxima")

[Out]

x^8*e^(x^2*e^(4*x) - 1)

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mupad [B]  time = 7.59, size = 15, normalized size = 0.88 \begin {gather*} x^8\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(2*x + log(x^2*exp(2*x))) + 4*log(x) - 1)*(exp(2*x + log(x^2*exp(2*x)))*(2*x^3 + 4*x^4) + 8*x^3),x)

[Out]

x^8*exp(-1)*exp(x^2*exp(4*x))

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sympy [A]  time = 0.20, size = 14, normalized size = 0.82 \begin {gather*} x^{8} e^{x^{2} e^{4 x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4+2*x**3)*exp(ln(exp(x)**2*x**2)+2*x)+8*x**3)*exp(1/4*exp(ln(exp(x)**2*x**2)+2*x)+ln(x)-1/4)*
*4,x)

[Out]

x**8*exp(x**2*exp(4*x) - 1)

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