3.52.76 \(\int (1+2 x-3 x^2+3 e^{x^3} x^2+e^x (-1+x+x^2-\log (25))+\log (25)) \, dx\)

Optimal. Leaf size=24 \[ e^{x^3}+x+\left (-e^x+x\right ) \left (x-x^2+\log (25)\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 10, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2209, 2196, 2176, 2194} \begin {gather*} -x^3+e^{x^3}+e^x x^2+x^2-e^x x+e^x+x (1+\log (25))-e^x (1+\log (25)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + 2*x - 3*x^2 + 3*E^x^3*x^2 + E^x*(-1 + x + x^2 - Log[25]) + Log[25],x]

[Out]

E^x + E^x^3 - E^x*x + x^2 + E^x*x^2 - x^3 - E^x*(1 + Log[25]) + x*(1 + Log[25])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^2-x^3+x (1+\log (25))+3 \int e^{x^3} x^2 \, dx+\int e^x \left (-1+x+x^2-\log (25)\right ) \, dx\\ &=e^{x^3}+x^2-x^3+x (1+\log (25))+\int \left (e^x x+e^x x^2+e^x (-1-\log (25))\right ) \, dx\\ &=e^{x^3}+x^2-x^3+x (1+\log (25))+(-1-\log (25)) \int e^x \, dx+\int e^x x \, dx+\int e^x x^2 \, dx\\ &=e^{x^3}+e^x x+x^2+e^x x^2-x^3-e^x (1+\log (25))+x (1+\log (25))-2 \int e^x x \, dx-\int e^x \, dx\\ &=-e^x+e^{x^3}-e^x x+x^2+e^x x^2-x^3-e^x (1+\log (25))+x (1+\log (25))+2 \int e^x \, dx\\ &=e^x+e^{x^3}-e^x x+x^2+e^x x^2-x^3-e^x (1+\log (25))+x (1+\log (25))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 34, normalized size = 1.42 \begin {gather*} e^{x^3}+x+x^2-x^3+e^x \left (-x+x^2-\log (25)\right )+x \log (25) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + 2*x - 3*x^2 + 3*E^x^3*x^2 + E^x*(-1 + x + x^2 - Log[25]) + Log[25],x]

[Out]

E^x^3 + x + x^2 - x^3 + E^x*(-x + x^2 - Log[25]) + x*Log[25]

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fricas [A]  time = 0.60, size = 33, normalized size = 1.38 \begin {gather*} -x^{3} + x^{2} + {\left (x^{2} - x - 2 \, \log \relax (5)\right )} e^{x} + 2 \, x \log \relax (5) + x + e^{\left (x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x^2*exp(x^3)+(-2*log(5)+x^2+x-1)*exp(x)+2*log(5)-3*x^2+2*x+1,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 33, normalized size = 1.38 \begin {gather*} -x^{3} + x^{2} + {\left (x^{2} - x - 2 \, \log \relax (5)\right )} e^{x} + 2 \, x \log \relax (5) + x + e^{\left (x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x^2*exp(x^3)+(-2*log(5)+x^2+x-1)*exp(x)+2*log(5)-3*x^2+2*x+1,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 34, normalized size = 1.42

method	result	size
risch	\({\mathrm e}^{x^{3}}+\left (x^{2}-2 \ln \relax (5)-x \right ) {\mathrm e}^{x}+2 x \ln \relax (5)-x^{3}+x^{2}+x\)	\(34\)
default	\(x -{\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} \ln \relax (5)+x^{2}-x^{3}+{\mathrm e}^{x^{3}}+2 x \ln \relax (5)\)	\(37\)
norman	\(x^{2}+\left (2 \ln \relax (5)+1\right ) x +{\mathrm e}^{x} x^{2}-x^{3}-{\mathrm e}^{x} x -2 \,{\mathrm e}^{x} \ln \relax (5)+{\mathrm e}^{x^{3}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*x^2*exp(x^3)+(-2*ln(5)+x^2+x-1)*exp(x)+2*ln(5)-3*x^2+2*x+1,x,method=_RETURNVERBOSE)

[Out]

exp(x^3)+(x^2-2*ln(5)-x)*exp(x)+2*x*ln(5)-x^3+x^2+x

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maxima [A]  time = 0.43, size = 33, normalized size = 1.38 \begin {gather*} -x^{3} + x^{2} + {\left (x^{2} - x - 2 \, \log \relax (5)\right )} e^{x} + 2 \, x \log \relax (5) + x + e^{\left (x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x^2*exp(x^3)+(-2*log(5)+x^2+x-1)*exp(x)+2*log(5)-3*x^2+2*x+1,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.23, size = 36, normalized size = 1.50 \begin {gather*} {\mathrm {e}}^{x^3}+x^2\,{\mathrm {e}}^x+x\,\left (\ln \left (25\right )+1\right )-{\mathrm {e}}^x\,\ln \left (25\right )-x\,{\mathrm {e}}^x+x^2-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + 2*log(5) + 3*x^2*exp(x^3) + exp(x)*(x - 2*log(5) + x^2 - 1) - 3*x^2 + 1,x)

[Out]

exp(x^3) + x^2*exp(x) + x*(log(25) + 1) - exp(x)*log(25) - x*exp(x) + x^2 - x^3

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sympy [A]  time = 0.15, size = 32, normalized size = 1.33 \begin {gather*} - x^{3} + x^{2} + x \left (1 + 2 \log {\relax (5 )}\right ) + \left (x^{2} - x - 2 \log {\relax (5 )}\right ) e^{x} + e^{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*x**2*exp(x**3)+(-2*ln(5)+x**2+x-1)*exp(x)+2*ln(5)-3*x**2+2*x+1,x)

[Out]

-x**3 + x**2 + x*(1 + 2*log(5)) + (x**2 - x - 2*log(5))*exp(x) + exp(x**3)

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