\(\int (150 x^5+e^{-12+10 x} (100 x^3+250 x^4)+e^{-6+5 x} (250 x^4+250 x^5)) \, dx\) [701]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 44, antiderivative size = 26 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=2+25 x^4 \left (e^{-2 (3-2 x)+x}+x\right )^2-\log (4) \]

[Out]

2-2*ln(2)+25*x^4*(exp(5*x-6)+x)^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 27, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1607, 2227, 2207, 2225} \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25 x^6+50 e^{5 x-6} x^5+25 e^{10 x-12} x^4 \]

[In]

Int[150*x^5 + E^(-12 + 10*x)*(100*x^3 + 250*x^4) + E^(-6 + 5*x)*(250*x^4 + 250*x^5),x]

[Out]

25*E^(-12 + 10*x)*x^4 + 50*E^(-6 + 5*x)*x^5 + 25*x^6

Rule 1607

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 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}& = 25 x^6+\int e^{-12+10 x} \left (100 x^3+250 x^4\right ) \, dx+\int e^{-6+5 x} \left (250 x^4+250 x^5\right ) \, dx \\ & = 25 x^6+\int e^{-12+10 x} x^3 (100+250 x) \, dx+\int e^{-6+5 x} x^4 (250+250 x) \, dx \\ & = 25 x^6+\int \left (100 e^{-12+10 x} x^3+250 e^{-12+10 x} x^4\right ) \, dx+\int \left (250 e^{-6+5 x} x^4+250 e^{-6+5 x} x^5\right ) \, dx \\ & = 25 x^6+100 \int e^{-12+10 x} x^3 \, dx+250 \int e^{-6+5 x} x^4 \, dx+250 \int e^{-12+10 x} x^4 \, dx+250 \int e^{-6+5 x} x^5 \, dx \\ & = 10 e^{-12+10 x} x^3+50 e^{-6+5 x} x^4+25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6-30 \int e^{-12+10 x} x^2 \, dx-100 \int e^{-12+10 x} x^3 \, dx-200 \int e^{-6+5 x} x^3 \, dx-250 \int e^{-6+5 x} x^4 \, dx \\ & = -3 e^{-12+10 x} x^2-40 e^{-6+5 x} x^3+25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6+6 \int e^{-12+10 x} x \, dx+30 \int e^{-12+10 x} x^2 \, dx+120 \int e^{-6+5 x} x^2 \, dx+200 \int e^{-6+5 x} x^3 \, dx \\ & = \frac {3}{5} e^{-12+10 x} x+24 e^{-6+5 x} x^2+25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6-\frac {3}{5} \int e^{-12+10 x} \, dx-6 \int e^{-12+10 x} x \, dx-48 \int e^{-6+5 x} x \, dx-120 \int e^{-6+5 x} x^2 \, dx \\ & = -\frac {3}{50} e^{-12+10 x}-\frac {48}{5} e^{-6+5 x} x+25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6+\frac {3}{5} \int e^{-12+10 x} \, dx+\frac {48}{5} \int e^{-6+5 x} \, dx+48 \int e^{-6+5 x} x \, dx \\ & = \frac {48}{25} e^{-6+5 x}+25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6-\frac {48}{5} \int e^{-6+5 x} \, dx \\ & = 25 e^{-12+10 x} x^4+50 e^{-6+5 x} x^5+25 x^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=\frac {25 x^4 \left (e^{5 x}+e^6 x\right )^2}{e^{12}} \]

[In]

Integrate[150*x^5 + E^(-12 + 10*x)*(100*x^3 + 250*x^4) + E^(-6 + 5*x)*(250*x^4 + 250*x^5),x]

[Out]

(25*x^4*(E^(5*x) + E^6*x)^2)/E^12

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19

method result size
norman \(25 x^{6}+50 \,{\mathrm e}^{5 x -6} x^{5}+25 \,{\mathrm e}^{10 x -12} x^{4}\) \(31\)
risch \(25 x^{6}+50 \,{\mathrm e}^{5 x -6} x^{5}+25 \,{\mathrm e}^{10 x -12} x^{4}\) \(31\)
parallelrisch \(25 x^{6}+50 \,{\mathrm e}^{5 x -6} x^{5}+25 \,{\mathrm e}^{10 x -12} x^{4}\) \(31\)
derivativedivides \(25 x^{6}+\frac {1296 \,{\mathrm e}^{10 x -12}}{25}+\frac {864 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )}{25}+\frac {216 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{2}}{25}+\frac {24 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{3}}{25}+\frac {{\mathrm e}^{10 x -12} \left (5 x -6\right )^{4}}{25}+\frac {2592 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )}{25}+\frac {15552 \,{\mathrm e}^{5 x -6}}{125}+\frac {864 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{2}}{25}+\frac {144 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{3}}{25}+\frac {12 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{4}}{25}+\frac {2 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{5}}{125}\) \(164\)
default \(25 x^{6}+\frac {1296 \,{\mathrm e}^{10 x -12}}{25}+\frac {864 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )}{25}+\frac {216 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{2}}{25}+\frac {24 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{3}}{25}+\frac {{\mathrm e}^{10 x -12} \left (5 x -6\right )^{4}}{25}+\frac {2592 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )}{25}+\frac {15552 \,{\mathrm e}^{5 x -6}}{125}+\frac {864 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{2}}{25}+\frac {144 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{3}}{25}+\frac {12 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{4}}{25}+\frac {2 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{5}}{125}\) \(164\)
parts \(25 x^{6}+\frac {1296 \,{\mathrm e}^{10 x -12}}{25}+\frac {864 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )}{25}+\frac {216 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{2}}{25}+\frac {24 \,{\mathrm e}^{10 x -12} \left (5 x -6\right )^{3}}{25}+\frac {{\mathrm e}^{10 x -12} \left (5 x -6\right )^{4}}{25}+\frac {2592 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )}{25}+\frac {15552 \,{\mathrm e}^{5 x -6}}{125}+\frac {864 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{2}}{25}+\frac {144 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{3}}{25}+\frac {12 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{4}}{25}+\frac {2 \,{\mathrm e}^{5 x -6} \left (5 x -6\right )^{5}}{125}\) \(164\)

[In]

int((250*x^4+100*x^3)*exp(5*x-6)^2+(250*x^5+250*x^4)*exp(5*x-6)+150*x^5,x,method=_RETURNVERBOSE)

[Out]

25*x^6+50*exp(5*x-6)*x^5+25*exp(5*x-6)^2*x^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25 \, x^{6} + 50 \, x^{5} e^{\left (5 \, x - 6\right )} + 25 \, x^{4} e^{\left (10 \, x - 12\right )} \]

[In]

integrate((250*x^4+100*x^3)*exp(5*x-6)^2+(250*x^5+250*x^4)*exp(5*x-6)+150*x^5,x, algorithm="fricas")

[Out]

25*x^6 + 50*x^5*e^(5*x - 6) + 25*x^4*e^(10*x - 12)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25 x^{6} + 50 x^{5} e^{5 x - 6} + 25 x^{4} e^{10 x - 12} \]

[In]

integrate((250*x**4+100*x**3)*exp(5*x-6)**2+(250*x**5+250*x**4)*exp(5*x-6)+150*x**5,x)

[Out]

25*x**6 + 50*x**5*exp(5*x - 6) + 25*x**4*exp(10*x - 12)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (21) = 42\).

Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.00 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25 \, x^{6} + 25 \, x^{4} e^{\left (10 \, x - 12\right )} + \frac {2}{25} \, {\left (625 \, x^{5} - 625 \, x^{4} + 500 \, x^{3} - 300 \, x^{2} + 120 \, x - 24\right )} e^{\left (5 \, x - 6\right )} + \frac {2}{25} \, {\left (625 \, x^{4} - 500 \, x^{3} + 300 \, x^{2} - 120 \, x + 24\right )} e^{\left (5 \, x - 6\right )} \]

[In]

integrate((250*x^4+100*x^3)*exp(5*x-6)^2+(250*x^5+250*x^4)*exp(5*x-6)+150*x^5,x, algorithm="maxima")

[Out]

25*x^6 + 25*x^4*e^(10*x - 12) + 2/25*(625*x^5 - 625*x^4 + 500*x^3 - 300*x^2 + 120*x - 24)*e^(5*x - 6) + 2/25*(
625*x^4 - 500*x^3 + 300*x^2 - 120*x + 24)*e^(5*x - 6)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25 \, x^{6} + 50 \, x^{5} e^{\left (5 \, x - 6\right )} + 25 \, x^{4} e^{\left (10 \, x - 12\right )} \]

[In]

integrate((250*x^4+100*x^3)*exp(5*x-6)^2+(250*x^5+250*x^4)*exp(5*x-6)+150*x^5,x, algorithm="giac")

[Out]

25*x^6 + 50*x^5*e^(5*x - 6) + 25*x^4*e^(10*x - 12)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \left (150 x^5+e^{-12+10 x} \left (100 x^3+250 x^4\right )+e^{-6+5 x} \left (250 x^4+250 x^5\right )\right ) \, dx=25\,x^4\,{\mathrm {e}}^{-12}\,{\left ({\mathrm {e}}^{5\,x}+x\,{\mathrm {e}}^6\right )}^2 \]

[In]

int(exp(10*x - 12)*(100*x^3 + 250*x^4) + exp(5*x - 6)*(250*x^4 + 250*x^5) + 150*x^5,x)

[Out]

25*x^4*exp(-12)*(exp(5*x) + x*exp(6))^2