3.72.81 \(\int (e (-13+9 x^2)+e \log (16)+e^x (-13+6 x+3 x^2+\log (16))) \, dx\)

Optimal. Leaf size=19 \[ \left (e^x+e x\right ) \left (-4+3 \left (-3+x^2\right )+\log (16)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.84, number of steps used = 10, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2196, 2176, 2194} \begin {gather*} 3 e x^3+3 e^x x^2-13 e x+e x \log (16)-e^x (13-\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E*(-13 + 9*x^2) + E*Log[16] + E^x*(-13 + 6*x + 3*x^2 + Log[16]),x]

[Out]

-13*E*x + 3*E^x*x^2 + 3*E*x^3 - E^x*(13 - Log[16]) + E*x*Log[16]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e x \log (16)+e \int \left (-13+9 x^2\right ) \, dx+\int e^x \left (-13+6 x+3 x^2+\log (16)\right ) \, dx\\ &=-13 e x+3 e x^3+e x \log (16)+\int \left (6 e^x x+3 e^x x^2-13 e^x \left (1-\frac {4 \log (2)}{13}\right )\right ) \, dx\\ &=-13 e x+3 e x^3+e x \log (16)+3 \int e^x x^2 \, dx+6 \int e^x x \, dx-(13-4 \log (2)) \int e^x \, dx\\ &=-13 e x+6 e^x x+3 e^x x^2+3 e x^3-e^x (13-4 \log (2))+e x \log (16)-6 \int e^x \, dx-6 \int e^x x \, dx\\ &=-6 e^x-13 e x+3 e^x x^2+3 e x^3-e^x (13-4 \log (2))+e x \log (16)+6 \int e^x \, dx\\ &=-13 e x+3 e^x x^2+3 e x^3-e^x (13-4 \log (2))+e x \log (16)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 27, normalized size = 1.42 \begin {gather*} 3 e x^3+e x (-13+\log (16))+e^x \left (-13+3 x^2+\log (16)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E*(-13 + 9*x^2) + E*Log[16] + E^x*(-13 + 6*x + 3*x^2 + Log[16]),x]

[Out]

3*E*x^3 + E*x*(-13 + Log[16]) + E^x*(-13 + 3*x^2 + Log[16])

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 34, normalized size = 1.79 \begin {gather*} 4 \, x e \log \relax (2) + {\left (3 \, x^{3} - 13 \, x\right )} e + {\left (3 \, x^{2} + 4 \, \log \relax (2) - 13\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+3*x^2+6*x-13)*exp(x)+4*exp(1)*log(2)+(9*x^2-13)*exp(1),x, algorithm="fricas")

[Out]

4*x*e*log(2) + (3*x^3 - 13*x)*e + (3*x^2 + 4*log(2) - 13)*e^x

________________________________________________________________________________________

giac [A]  time = 0.21, size = 34, normalized size = 1.79 \begin {gather*} 4 \, x e \log \relax (2) + {\left (3 \, x^{3} - 13 \, x\right )} e + {\left (3 \, x^{2} + 4 \, \log \relax (2) - 13\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+3*x^2+6*x-13)*exp(x)+4*exp(1)*log(2)+(9*x^2-13)*exp(1),x, algorithm="giac")

[Out]

4*x*e*log(2) + (3*x^3 - 13*x)*e + (3*x^2 + 4*log(2) - 13)*e^x

________________________________________________________________________________________

maple [A]  time = 0.03, size = 35, normalized size = 1.84

method	result	size
risch	\(\left (4 \ln \relax (2)+3 x^{2}-13\right ) {\mathrm e}^{x}+4 x \,{\mathrm e} \ln \relax (2)+3 x^{3} {\mathrm e}-13 x \,{\mathrm e}\)	\(35\)
default	\(3 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} \ln \relax (2)-13 \,{\mathrm e}^{x}+{\mathrm e} \left (3 x^{3}-13 x \right )+4 x \,{\mathrm e} \ln \relax (2)\)	\(38\)
norman	\(\left (4 \,{\mathrm e} \ln \relax (2)-13 \,{\mathrm e}\right ) x +\left (4 \ln \relax (2)-13\right ) {\mathrm e}^{x}+3 x^{3} {\mathrm e}+3 \,{\mathrm e}^{x} x^{2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(2)+3*x^2+6*x-13)*exp(x)+4*exp(1)*ln(2)+(9*x^2-13)*exp(1),x,method=_RETURNVERBOSE)

[Out]

(4*ln(2)+3*x^2-13)*exp(x)+4*x*exp(1)*ln(2)+3*x^3*exp(1)-13*x*exp(1)

________________________________________________________________________________________

maxima [A]  time = 0.43, size = 34, normalized size = 1.79 \begin {gather*} 4 \, x e \log \relax (2) + {\left (3 \, x^{3} - 13 \, x\right )} e + {\left (3 \, x^{2} + 4 \, \log \relax (2) - 13\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(2)+3*x^2+6*x-13)*exp(x)+4*exp(1)*log(2)+(9*x^2-13)*exp(1),x, algorithm="maxima")

[Out]

4*x*e*log(2) + (3*x^3 - 13*x)*e + (3*x^2 + 4*log(2) - 13)*e^x

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 36, normalized size = 1.89 \begin {gather*} {\mathrm {e}}^x\,\left (\ln \left (16\right )-13\right )+3\,x^2\,{\mathrm {e}}^x-x\,\left (13\,\mathrm {e}-4\,\mathrm {e}\,\ln \relax (2)\right )+3\,x^3\,\mathrm {e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*exp(1)*log(2) + exp(1)*(9*x^2 - 13) + exp(x)*(6*x + 4*log(2) + 3*x^2 - 13),x)

[Out]

exp(x)*(log(16) - 13) + 3*x^2*exp(x) - x*(13*exp(1) - 4*exp(1)*log(2)) + 3*x^3*exp(1)

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 37, normalized size = 1.95 \begin {gather*} 3 e x^{3} + x \left (- 13 e + 4 e \log {\relax (2 )}\right ) + \left (3 x^{2} - 13 + 4 \log {\relax (2 )}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(2)+3*x**2+6*x-13)*exp(x)+4*exp(1)*ln(2)+(9*x**2-13)*exp(1),x)

[Out]

3*E*x**3 + x*(-13*E + 4*E*log(2)) + (3*x**2 - 13 + 4*log(2))*exp(x)

________________________________________________________________________________________