3.33.9 $\int e^{4x}(10x+17x^2-4x^3){\log }^2(3)dx$

Optimal. Leaf size=20 $$1+e^{4x}(5-x)x^2{\log }^2(3)$$

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Rubi [A]  time = 0.10, antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 13, number of rules used = 5, integrand size = 24, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.208, Rules used = {12, 1594, 2196, 2176, 2194} $$\begin{array}{c}5e^{4x}{x}^2{\log }^2(3)-e^{4x}{x}^3{\log }^2(3)\end{array}$$

Antiderivative was successfully verified.

[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 1594

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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& ={\log }^2(3)\int e^{4x}(10x+17x^2-4x^3)dx\\ & ={\log }^2(3)\int e^{4x}x(10+17x-4x^2)dx\\ & ={\log }^2(3)\int (10e^{4x}x+17e^{4x}{x}^2-4e^{4x}{x}^3)dx\\ & =-((4{\log }^2(3))\int e^{4x}{x}^{3}dx)+(10{\log }^2(3))\int e^{4x}xdx+(17{\log }^2(3))\int e^{4x}{x}^{2}dx\\ & =\frac{5}{2}{e}^{4x}x{\log }^2(3)+\frac{17}{4}{e}^{4x}{x}^2{\log }^2(3)-e^{4x}{x}^3{\log }^2(3)-\frac{1}{2}(5{\log }^2(3))\int e^{4x}dx+(3{\log }^2(3))\int e^{4x}{x}^{2}dx-\frac{1}{2}(17{\log }^2(3))\int e^{4x}xdx\\ & =-\frac{5}{8}{e}^{4x}{\log }^2(3)+\frac{3}{8}{e}^{4x}x{\log }^2(3)+5e^{4x}{x}^2{\log }^2(3)-e^{4x}{x}^3{\log }^2(3)-\frac{1}{2}(3{\log }^2(3))\int e^{4x}xdx+\frac{1}{8}(17{\log }^2(3))\int e^{4x}dx\\ & =-\frac{3}{32}{e}^{4x}{\log }^2(3)+5e^{4x}{x}^2{\log }^2(3)-e^{4x}{x}^3{\log }^2(3)+\frac{1}{8}(3{\log }^2(3))\int e^{4x}dx\\ & =5e^{4x}{x}^2{\log }^2(3)-e^{4x}{x}^3{\log }^2(3)\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.85 $$\begin{array}{c}-e^{4x}(-5+x)x^2{\log }^2(3)\end{array}$$

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.53, size = 19, normalized size = 0.95 $$\begin{array}{c}-(x^3-5x^2)e^{\left(4x\right)}\log (3{)}^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.17, size = 19, normalized size = 0.95 $$\begin{array}{c}-(x^3-5x^2)e^{\left(4x\right)}\log (3{)}^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.04, size = 17, normalized size = 0.85

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

Verification of antiderivative is not currently implemented for this CAS.

[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*(x-5)

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maxima [B]  time = 0.41, size = 54, normalized size = 2.70 $$\begin{array}{c}-\frac{1}{32}((32x^3-24x^2+12x-3)e^{\left(4x\right)}-17(8x^2-4x+1)e^{\left(4x\right)}-20(4x-1)e^{\left(4x\right)})\log (3{)}^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.85, size = 16, normalized size = 0.80 $$\begin{array}{c}-x^2{\mathrm{e}}^{4x}{\ln (3)}^2(x-5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 22, normalized size = 1.10 $$\begin{array}{c}(-x^3\log {(3)}^2+5x^2\log {(3)}^2)e^{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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