3.43.35 \(\int \frac {1}{4} (-125+75 x-15 x^2+x^3+29^{2 x} (8 x+8 x^2 \log (29))+29^x (50-40 x+6 x^2+(50 x-20 x^2+2 x^3) \log (29))) \, dx\)

Optimal. Leaf size=19 \[ \left (\frac {1}{4} (5-x)^2+29^x x\right )^2 \]

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Rubi [B]  time = 0.23, antiderivative size = 62, normalized size of antiderivative = 3.26, number of steps used = 29, number of rules used = 5, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {x^4}{16}+\frac {29^x x^3}{2}-\frac {5 x^3}{4}-5\ 29^x x^2+29^{2 x} x^2+\frac {75 x^2}{8}+\frac {25\ 29^x x}{2}-\frac {125 x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-125 + 75*x - 15*x^2 + x^3 + 29^(2*x)*(8*x + 8*x^2*Log[29]) + 29^x*(50 - 40*x + 6*x^2 + (50*x - 20*x^2 +
2*x^3)*Log[29]))/4,x]

[Out]

(-125*x)/4 + (25*29^x*x)/2 + (75*x^2)/8 - 5*29^x*x^2 + 29^(2*x)*x^2 - (5*x^3)/4 + (29^x*x^3)/2 + x^4/16

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (-125+75 x-15 x^2+x^3+29^{2 x} \left (8 x+8 x^2 \log (29)\right )+29^x \left (50-40 x+6 x^2+\left (50 x-20 x^2+2 x^3\right ) \log (29)\right )\right ) \, dx\\ &=-\frac {125 x}{4}+\frac {75 x^2}{8}-\frac {5 x^3}{4}+\frac {x^4}{16}+\frac {1}{4} \int 29^{2 x} \left (8 x+8 x^2 \log (29)\right ) \, dx+\frac {1}{4} \int 29^x \left (50-40 x+6 x^2+\left (50 x-20 x^2+2 x^3\right ) \log (29)\right ) \, dx\\ &=-\frac {125 x}{4}+\frac {75 x^2}{8}-\frac {5 x^3}{4}+\frac {x^4}{16}+\frac {1}{4} \int 29^{2 x} x (8+8 x \log (29)) \, dx+\frac {1}{4} \int \left (50\ 29^x-40\ 29^x x+6\ 29^x x^2+2\ 29^x (-5+x)^2 x \log (29)\right ) \, dx\\ &=-\frac {125 x}{4}+\frac {75 x^2}{8}-\frac {5 x^3}{4}+\frac {x^4}{16}+\frac {1}{4} \int \left (8\ 29^{2 x} x+8\ 29^{2 x} x^2 \log (29)\right ) \, dx+\frac {3}{2} \int 29^x x^2 \, dx-10 \int 29^x x \, dx+\frac {25 \int 29^x \, dx}{2}+\frac {1}{2} \log (29) \int 29^x (-5+x)^2 x \, dx\\ &=-\frac {125 x}{4}+\frac {75 x^2}{8}-\frac {5 x^3}{4}+\frac {x^4}{16}+\frac {25\ 29^x}{2 \log (29)}-\frac {10\ 29^x x}{\log (29)}+\frac {3\ 29^x x^2}{2 \log (29)}+2 \int 29^{2 x} x \, dx-\frac {3 \int 29^x x \, dx}{\log (29)}+\frac {10 \int 29^x \, dx}{\log (29)}+\frac {1}{2} \log (29) \int \left (25\ 29^x x-10\ 29^x x^2+29^x x^3\right ) \, dx+(2 \log (29)) \int 29^{2 x} x^2 \, dx\\ &=-\frac {125 x}{4}+\frac {75 x^2}{8}+29^{2 x} x^2-\frac {5 x^3}{4}+\frac {x^4}{16}+\frac {10\ 29^x}{\log ^2(29)}-\frac {3\ 29^x x}{\log ^2(29)}+\frac {25\ 29^x}{2 \log (29)}-\frac {10\ 29^x x}{\log (29)}+\frac {29^{2 x} x}{\log (29)}+\frac {3\ 29^x x^2}{2 \log (29)}-2 \int 29^{2 x} x \, dx+\frac {3 \int 29^x \, dx}{\log ^2(29)}-\frac {\int 29^{2 x} \, dx}{\log (29)}+\frac {1}{2} \log (29) \int 29^x x^3 \, dx-(5 \log (29)) \int 29^x x^2 \, dx+\frac {1}{2} (25 \log (29)) \int 29^x x \, dx\\ &=-\frac {125 x}{4}+\frac {25\ 29^x x}{2}+\frac {75 x^2}{8}-5\ 29^x x^2+29^{2 x} x^2-\frac {5 x^3}{4}+\frac {29^x x^3}{2}+\frac {x^4}{16}+\frac {3\ 29^x}{\log ^3(29)}+\frac {10\ 29^x}{\log ^2(29)}-\frac {29^{2 x}}{2 \log ^2(29)}-\frac {3\ 29^x x}{\log ^2(29)}+\frac {25\ 29^x}{2 \log (29)}-\frac {10\ 29^x x}{\log (29)}+\frac {3\ 29^x x^2}{2 \log (29)}-\frac {3}{2} \int 29^x x^2 \, dx+10 \int 29^x x \, dx-\frac {25 \int 29^x \, dx}{2}+\frac {\int 29^{2 x} \, dx}{\log (29)}\\ &=-\frac {125 x}{4}+\frac {25\ 29^x x}{2}+\frac {75 x^2}{8}-5\ 29^x x^2+29^{2 x} x^2-\frac {5 x^3}{4}+\frac {29^x x^3}{2}+\frac {x^4}{16}+\frac {3\ 29^x}{\log ^3(29)}+\frac {10\ 29^x}{\log ^2(29)}-\frac {3\ 29^x x}{\log ^2(29)}+\frac {3 \int 29^x x \, dx}{\log (29)}-\frac {10 \int 29^x \, dx}{\log (29)}\\ &=-\frac {125 x}{4}+\frac {25\ 29^x x}{2}+\frac {75 x^2}{8}-5\ 29^x x^2+29^{2 x} x^2-\frac {5 x^3}{4}+\frac {29^x x^3}{2}+\frac {x^4}{16}+\frac {3\ 29^x}{\log ^3(29)}-\frac {3 \int 29^x \, dx}{\log ^2(29)}\\ &=-\frac {125 x}{4}+\frac {25\ 29^x x}{2}+\frac {75 x^2}{8}-5\ 29^x x^2+29^{2 x} x^2-\frac {5 x^3}{4}+\frac {29^x x^3}{2}+\frac {x^4}{16}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.11 \begin {gather*} \frac {1}{16} \left (25+2 \left (-5+2\ 29^x\right ) x+x^2\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-125 + 75*x - 15*x^2 + x^3 + 29^(2*x)*(8*x + 8*x^2*Log[29]) + 29^x*(50 - 40*x + 6*x^2 + (50*x - 20*
x^2 + 2*x^3)*Log[29]))/4,x]

[Out]

(25 + 2*(-5 + 2*29^x)*x + x^2)^2/16

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fricas [B]  time = 0.78, size = 45, normalized size = 2.37 \begin {gather*} \frac {1}{16} \, x^{4} + 29^{2 \, x} x^{2} - \frac {5}{4} \, x^{3} + \frac {1}{2} \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} 29^{x} + \frac {75}{8} \, x^{2} - \frac {125}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(8*x^2*log(29)+8*x)*exp(x*log(29))^2+1/4*((2*x^3-20*x^2+50*x)*log(29)+6*x^2-40*x+50)*exp(x*log(2
9))+1/4*x^3-15/4*x^2+75/4*x-125/4,x, algorithm="fricas")

[Out]

1/16*x^4 + 29^(2*x)*x^2 - 5/4*x^3 + 1/2*(x^3 - 10*x^2 + 25*x)*29^x + 75/8*x^2 - 125/4*x

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giac [B]  time = 0.17, size = 62, normalized size = 3.26 \begin {gather*} \frac {1}{16} \, x^{4} + 29^{2 \, x} x^{2} - \frac {5}{4} \, x^{3} + \frac {75}{8} \, x^{2} - \frac {125}{4} \, x + \frac {{\left (x^{3} \log \left (29\right )^{4} - 10 \, x^{2} \log \left (29\right )^{4} + 25 \, x \log \left (29\right )^{4}\right )} 29^{x}}{2 \, \log \left (29\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(8*x^2*log(29)+8*x)*exp(x*log(29))^2+1/4*((2*x^3-20*x^2+50*x)*log(29)+6*x^2-40*x+50)*exp(x*log(2
9))+1/4*x^3-15/4*x^2+75/4*x-125/4,x, algorithm="giac")

[Out]

1/16*x^4 + 29^(2*x)*x^2 - 5/4*x^3 + 75/8*x^2 - 125/4*x + 1/2*(x^3*log(29)^4 - 10*x^2*log(29)^4 + 25*x*log(29)^
4)*29^x/log(29)^4

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maple [B]  time = 0.06, size = 48, normalized size = 2.53

method	result	size
risch	\(29^{2 x} x^{2}+\frac {\left (2 x^{3}-20 x^{2}+50 x \right ) 29^{x}}{4}+\frac {x^{4}}{16}-\frac {5 x^{3}}{4}+\frac {75 x^{2}}{8}-\frac {125 x}{4}\)	\(48\)
default	\(-\frac {125 x}{4}+\frac {75 x^{2}}{8}-\frac {5 x^{3}}{4}+\frac {x^{4}}{16}+{\mathrm e}^{2 x \ln \left (29\right )} x^{2}+\frac {{\mathrm e}^{x \ln \left (29\right )} x^{3}}{2}-5 \,{\mathrm e}^{x \ln \left (29\right )} x^{2}+\frac {25 \,{\mathrm e}^{x \ln \left (29\right )} x}{2}\)	\(59\)
norman	\(-\frac {125 x}{4}+\frac {75 x^{2}}{8}-\frac {5 x^{3}}{4}+\frac {x^{4}}{16}+{\mathrm e}^{2 x \ln \left (29\right )} x^{2}+\frac {{\mathrm e}^{x \ln \left (29\right )} x^{3}}{2}-5 \,{\mathrm e}^{x \ln \left (29\right )} x^{2}+\frac {25 \,{\mathrm e}^{x \ln \left (29\right )} x}{2}\)	\(59\)
derivativedivides	\(\frac {-125 x \ln \left (29\right )+\frac {x^{4} \ln \left (29\right )}{4}+4 \ln \left (29\right ) {\mathrm e}^{2 x \ln \left (29\right )} x^{2}+50 \ln \left (29\right ) {\mathrm e}^{x \ln \left (29\right )} x -20 \ln \left (29\right ) {\mathrm e}^{x \ln \left (29\right )} x^{2}+2 \ln \left (29\right ) {\mathrm e}^{x \ln \left (29\right )} x^{3}+\frac {75 x^{2} \ln \left (29\right )}{2}-5 x^{3} \ln \left (29\right )}{4 \ln \left (29\right )}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(8*x^2*ln(29)+8*x)*exp(x*ln(29))^2+1/4*((2*x^3-20*x^2+50*x)*ln(29)+6*x^2-40*x+50)*exp(x*ln(29))+1/4*x^
3-15/4*x^2+75/4*x-125/4,x,method=_RETURNVERBOSE)

[Out]

(29^x)^2*x^2+1/4*(2*x^3-20*x^2+50*x)*29^x+1/16*x^4-5/4*x^3+75/8*x^2-125/4*x

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maxima [B]  time = 0.43, size = 45, normalized size = 2.37 \begin {gather*} \frac {1}{16} \, x^{4} + 29^{2 \, x} x^{2} - \frac {5}{4} \, x^{3} + \frac {1}{2} \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} 29^{x} + \frac {75}{8} \, x^{2} - \frac {125}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(8*x^2*log(29)+8*x)*exp(x*log(29))^2+1/4*((2*x^3-20*x^2+50*x)*log(29)+6*x^2-40*x+50)*exp(x*log(2
9))+1/4*x^3-15/4*x^2+75/4*x-125/4,x, algorithm="maxima")

[Out]

1/16*x^4 + 29^(2*x)*x^2 - 5/4*x^3 + 1/2*(x^3 - 10*x^2 + 25*x)*29^x + 75/8*x^2 - 125/4*x

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mupad [B]  time = 2.90, size = 43, normalized size = 2.26 \begin {gather*} \frac {x\,\left (150\,x+16\,{29}^{2\,x}\,x+8\,{29}^x\,x^2-80\,{29}^x\,x-20\,x^2+x^3+200\,{29}^x-500\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((75*x)/4 - (15*x^2)/4 + x^3/4 + (exp(2*x*log(29))*(8*x + 8*x^2*log(29)))/4 + (exp(x*log(29))*(log(29)*(50*
x - 20*x^2 + 2*x^3) - 40*x + 6*x^2 + 50))/4 - 125/4,x)

[Out]

(x*(150*x + 16*29^(2*x)*x + 8*29^x*x^2 - 80*29^x*x - 20*x^2 + x^3 + 200*29^x - 500))/16

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sympy [B]  time = 0.18, size = 54, normalized size = 2.84 \begin {gather*} \frac {x^{4}}{16} - \frac {5 x^{3}}{4} + x^{2} e^{2 x \log {\left (29 \right )}} + \frac {75 x^{2}}{8} - \frac {125 x}{4} + \frac {\left (x^{3} - 10 x^{2} + 25 x\right ) e^{x \log {\left (29 \right )}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(8*x**2*ln(29)+8*x)*exp(x*ln(29))**2+1/4*((2*x**3-20*x**2+50*x)*ln(29)+6*x**2-40*x+50)*exp(x*ln(
29))+1/4*x**3-15/4*x**2+75/4*x-125/4,x)

[Out]

x**4/16 - 5*x**3/4 + x**2*exp(2*x*log(29)) + 75*x**2/8 - 125*x/4 + (x**3 - 10*x**2 + 25*x)*exp(x*log(29))/2

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