\(\int e^{4 x} (10 x+17 x^2-4 x^3) \log ^2(3) \, dx\) [3209]

Optimal result

Integrand size = 24, antiderivative size = 20 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=1+e^{4 x} (5-x) x^2 \log ^2(3) \]

[Out]

5*(1-1/5*x)*exp(x)^4*ln(3)^2*x^2+1

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {12, 1608, 2227, 2207, 2225} \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=5 e^{4 x} x^2 \log ^2(3)-e^{4 x} x^3 \log ^2(3) \]

[In]

Int[E^(4*x)*(10*x + 17*x^2 - 4*x^3)*Log[3]^2,x]

[Out]

5*E^(4*x)*x^2*Log[3]^2 - E^(4*x)*x^3*Log[3]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2227

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] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \log ^2(3) \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \, dx \\ & = \log ^2(3) \int e^{4 x} x \left (10+17 x-4 x^2\right ) \, dx \\ & = \log ^2(3) \int \left (10 e^{4 x} x+17 e^{4 x} x^2-4 e^{4 x} x^3\right ) \, dx \\ & = -\left (\left (4 \log ^2(3)\right ) \int e^{4 x} x^3 \, dx\right )+\left (10 \log ^2(3)\right ) \int e^{4 x} x \, dx+\left (17 \log ^2(3)\right ) \int e^{4 x} x^2 \, dx \\ & = \frac {5}{2} e^{4 x} x \log ^2(3)+\frac {17}{4} e^{4 x} x^2 \log ^2(3)-e^{4 x} x^3 \log ^2(3)-\frac {1}{2} \left (5 \log ^2(3)\right ) \int e^{4 x} \, dx+\left (3 \log ^2(3)\right ) \int e^{4 x} x^2 \, dx-\frac {1}{2} \left (17 \log ^2(3)\right ) \int e^{4 x} x \, dx \\ & = -\frac {5}{8} e^{4 x} \log ^2(3)+\frac {3}{8} e^{4 x} x \log ^2(3)+5 e^{4 x} x^2 \log ^2(3)-e^{4 x} x^3 \log ^2(3)-\frac {1}{2} \left (3 \log ^2(3)\right ) \int e^{4 x} x \, dx+\frac {1}{8} \left (17 \log ^2(3)\right ) \int e^{4 x} \, dx \\ & = -\frac {3}{32} e^{4 x} \log ^2(3)+5 e^{4 x} x^2 \log ^2(3)-e^{4 x} x^3 \log ^2(3)+\frac {1}{8} \left (3 \log ^2(3)\right ) \int e^{4 x} \, dx \\ & = 5 e^{4 x} x^2 \log ^2(3)-e^{4 x} x^3 \log ^2(3) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=-e^{4 x} (-5+x) x^2 \log ^2(3) \]

[In]

Integrate[E^(4*x)*(10*x + 17*x^2 - 4*x^3)*Log[3]^2,x]

[Out]

-(E^(4*x)*(-5 + x)*x^2*Log[3]^2)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
gosper	\(-{\mathrm e}^{4 x} \ln \left (3\right )^{2} x^{2} \left (-5+x \right )\)	\(17\)
risch	\(\ln \left (3\right )^{2} \left (-x^{3}+5 x^{2}\right ) {\mathrm e}^{4 x}\)	\(21\)
default	\(\ln \left (3\right )^{2} \left (5 x^{2} {\mathrm e}^{4 x}-x^{3} {\mathrm e}^{4 x}\right )\)	\(25\)
parallelrisch	\(\ln \left (3\right )^{2} \left (5 x^{2} {\mathrm e}^{4 x}-x^{3} {\mathrm e}^{4 x}\right )\)	\(25\)
norman	\(5 x^{2} \ln \left (3\right )^{2} {\mathrm e}^{4 x}-x^{3} \ln \left (3\right )^{2} {\mathrm e}^{4 x}\)	\(28\)
meijerg	\(-\frac {\ln \left (3\right )^{2} \left (6-\frac {\left (-256 x^{3}+192 x^{2}-96 x +24\right ) {\mathrm e}^{4 x}}{4}\right )}{64}-\frac {17 \ln \left (3\right )^{2} \left (2-\frac {\left (48 x^{2}-24 x +6\right ) {\mathrm e}^{4 x}}{3}\right )}{64}+\frac {5 \ln \left (3\right )^{2} \left (1-\frac {\left (-8 x +2\right ) {\mathrm e}^{4 x}}{2}\right )}{8}\)	\(74\)

[In]

int((-4*x^3+17*x^2+10*x)*ln(3)^2*exp(x)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=-{\left (x^{3} - 5 \, x^{2}\right )} e^{\left (4 \, x\right )} \log \left (3\right )^{2} \]

[In]

integrate((-4*x^3+17*x^2+10*x)*log(3)^2*exp(x)^4,x, algorithm="fricas")

[Out]

-(x^3 - 5*x^2)*e^(4*x)*log(3)^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=\left (- x^{3} \log {\left (3 \right )}^{2} + 5 x^{2} \log {\left (3 \right )}^{2}\right ) e^{4 x} \]

[In]

integrate((-4*x**3+17*x**2+10*x)*ln(3)**2*exp(x)**4,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (18) = 36\).

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.70 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=-\frac {1}{32} \, {\left ({\left (32 \, x^{3} - 24 \, x^{2} + 12 \, x - 3\right )} e^{\left (4 \, x\right )} - 17 \, {\left (8 \, x^{2} - 4 \, x + 1\right )} e^{\left (4 \, x\right )} - 20 \, {\left (4 \, x - 1\right )} e^{\left (4 \, x\right )}\right )} \log \left (3\right )^{2} \]

[In]

integrate((-4*x^3+17*x^2+10*x)*log(3)^2*exp(x)^4,x, algorithm="maxima")

[Out]

-1/32*((32*x^3 - 24*x^2 + 12*x - 3)*e^(4*x) - 17*(8*x^2 - 4*x + 1)*e^(4*x) - 20*(4*x - 1)*e^(4*x))*log(3)^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=-{\left (x^{3} - 5 \, x^{2}\right )} e^{\left (4 \, x\right )} \log \left (3\right )^{2} \]

[In]

integrate((-4*x^3+17*x^2+10*x)*log(3)^2*exp(x)^4,x, algorithm="giac")

[Out]

-(x^3 - 5*x^2)*e^(4*x)*log(3)^2

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int e^{4 x} \left (10 x+17 x^2-4 x^3\right ) \log ^2(3) \, dx=-x^2\,{\mathrm {e}}^{4\,x}\,{\ln \left (3\right )}^2\,\left (x-5\right ) \]

[In]

int(exp(4*x)*log(3)^2*(10*x + 17*x^2 - 4*x^3),x)

[Out]

-x^2*exp(4*x)*log(3)^2*(x - 5)