3.32.25 \(\int e^{-25+e^{x^3}+25 x^2-5 x^4} (1+50 x^2+3 e^{x^3} x^3-20 x^4) \, dx\)

Optimal. Leaf size=23 \[ 5+e^{-25+e^{x^3}-5 x^2 \left (-5+x^2\right )} x \]

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Rubi [B]  time = 0.17, antiderivative size = 62, normalized size of antiderivative = 2.70, number of steps used = 1, number of rules used = 1, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2288} \begin {gather*} \frac {e^{-5 x^4+e^{x^3}+25 x^2-25} \left (-20 x^4+3 e^{x^3} x^3+50 x^2\right )}{-20 x^3+3 e^{x^3} x^2+50 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-25 + E^x^3 + 25*x^2 - 5*x^4)*(1 + 50*x^2 + 3*E^x^3*x^3 - 20*x^4),x]

[Out]

(E^(-25 + E^x^3 + 25*x^2 - 5*x^4)*(50*x^2 + 3*E^x^3*x^3 - 20*x^4))/(50*x + 3*E^x^3*x^2 - 20*x^3)

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^{-25+e^{x^3}+25 x^2-5 x^4} \left (50 x^2+3 e^{x^3} x^3-20 x^4\right )}{50 x+3 e^{x^3} x^2-20 x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 22, normalized size = 0.96 \begin {gather*} e^{e^{x^3}-5 \left (5-5 x^2+x^4\right )} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-25 + E^x^3 + 25*x^2 - 5*x^4)*(1 + 50*x^2 + 3*E^x^3*x^3 - 20*x^4),x]

[Out]

E^(E^x^3 - 5*(5 - 5*x^2 + x^4))*x

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fricas [A]  time = 0.93, size = 19, normalized size = 0.83 \begin {gather*} e^{\left (-5 \, x^{4} + 25 \, x^{2} + e^{\left (x^{3}\right )} + \log \relax (x) - 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3*exp(x^3)-20*x^4+50*x^2+1)*exp(log(x)+exp(x^3)-5*x^4+25*x^2-25)/x,x, algorithm="fricas")

[Out]

e^(-5*x^4 + 25*x^2 + e^(x^3) + log(x) - 25)

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giac [A]  time = 0.22, size = 19, normalized size = 0.83 \begin {gather*} e^{\left (-5 \, x^{4} + 25 \, x^{2} + e^{\left (x^{3}\right )} + \log \relax (x) - 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3*exp(x^3)-20*x^4+50*x^2+1)*exp(log(x)+exp(x^3)-5*x^4+25*x^2-25)/x,x, algorithm="giac")

[Out]

e^(-5*x^4 + 25*x^2 + e^(x^3) + log(x) - 25)

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maple [A]  time = 0.07, size = 20, normalized size = 0.87

method	result	size
risch	\(x \,{\mathrm e}^{-25+{\mathrm e}^{x^{3}}-5 x^{4}+25 x^{2}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^3*exp(x^3)-20*x^4+50*x^2+1)*exp(ln(x)+exp(x^3)-5*x^4+25*x^2-25)/x,x,method=_RETURNVERBOSE)

[Out]

x*exp(-25+exp(x^3)-5*x^4+25*x^2)

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maxima [A]  time = 0.72, size = 19, normalized size = 0.83 \begin {gather*} x e^{\left (-5 \, x^{4} + 25 \, x^{2} + e^{\left (x^{3}\right )} - 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3*exp(x^3)-20*x^4+50*x^2+1)*exp(log(x)+exp(x^3)-5*x^4+25*x^2-25)/x,x, algorithm="maxima")

[Out]

x*e^(-5*x^4 + 25*x^2 + e^(x^3) - 25)

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mupad [B]  time = 1.87, size = 21, normalized size = 0.91 \begin {gather*} x\,{\mathrm {e}}^{-25}\,{\mathrm {e}}^{-5\,x^4}\,{\mathrm {e}}^{25\,x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{x^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x^3) + log(x) + 25*x^2 - 5*x^4 - 25)*(3*x^3*exp(x^3) + 50*x^2 - 20*x^4 + 1))/x,x)

[Out]

x*exp(-25)*exp(-5*x^4)*exp(25*x^2)*exp(exp(x^3))

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sympy [A]  time = 0.72, size = 19, normalized size = 0.83 \begin {gather*} x e^{- 5 x^{4} + 25 x^{2} + e^{x^{3}} - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**3*exp(x**3)-20*x**4+50*x**2+1)*exp(ln(x)+exp(x**3)-5*x**4+25*x**2-25)/x,x)

[Out]

x*exp(-5*x**4 + 25*x**2 + exp(x**3) - 25)

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