3.39.91 \(\int \frac {1}{50} (-500+100 x+e^{\frac {1}{2} (-96-7 x)} (4 x-7 x^2)+e^{\frac {1}{4} (-96-7 x)} (100-215 x+35 x^2)) \, dx\)

Optimal. Leaf size=19 \[ \left (-5+x-\frac {1}{5} e^{-24-\frac {7 x}{4}} x\right )^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 57, normalized size of antiderivative = 3.00, number of steps used = 18, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} -\frac {2}{5} e^{\frac {1}{4} (-7 x-96)} x^2+\frac {1}{25} e^{\frac {1}{2} (-7 x-96)} x^2+x^2+2 e^{\frac {1}{4} (-7 x-96)} x-10 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-500 + 100*x + E^((-96 - 7*x)/2)*(4*x - 7*x^2) + E^((-96 - 7*x)/4)*(100 - 215*x + 35*x^2))/50,x]

[Out]

-10*x + 2*E^((-96 - 7*x)/4)*x + x^2 - (2*E^((-96 - 7*x)/4)*x^2)/5 + (E^((-96 - 7*x)/2)*x^2)/25

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}{50} \int \left (-500+100 x+e^{\frac {1}{2} (-96-7 x)} \left (4 x-7 x^2\right )+e^{\frac {1}{4} (-96-7 x)} \left (100-215 x+35 x^2\right )\right ) \, dx\\ &=-10 x+x^2+\frac {1}{50} \int e^{\frac {1}{2} (-96-7 x)} \left (4 x-7 x^2\right ) \, dx+\frac {1}{50} \int e^{\frac {1}{4} (-96-7 x)} \left (100-215 x+35 x^2\right ) \, dx\\ &=-10 x+x^2+\frac {1}{50} \int e^{\frac {1}{2} (-96-7 x)} (4-7 x) x \, dx+\frac {1}{50} \int \left (100 e^{\frac {1}{4} (-96-7 x)}-215 e^{\frac {1}{4} (-96-7 x)} x+35 e^{\frac {1}{4} (-96-7 x)} x^2\right ) \, dx\\ &=-10 x+x^2+\frac {1}{50} \int \left (4 e^{\frac {1}{2} (-96-7 x)} x-7 e^{\frac {1}{2} (-96-7 x)} x^2\right ) \, dx+\frac {7}{10} \int e^{\frac {1}{4} (-96-7 x)} x^2 \, dx+2 \int e^{\frac {1}{4} (-96-7 x)} \, dx-\frac {43}{10} \int e^{\frac {1}{4} (-96-7 x)} x \, dx\\ &=-\frac {8}{7} e^{\frac {1}{4} (-96-7 x)}-10 x+\frac {86}{35} e^{\frac {1}{4} (-96-7 x)} x+x^2-\frac {2}{5} e^{\frac {1}{4} (-96-7 x)} x^2+\frac {2}{25} \int e^{\frac {1}{2} (-96-7 x)} x \, dx-\frac {7}{50} \int e^{\frac {1}{2} (-96-7 x)} x^2 \, dx+\frac {4}{5} \int e^{\frac {1}{4} (-96-7 x)} x \, dx-\frac {86}{35} \int e^{\frac {1}{4} (-96-7 x)} \, dx\\ &=\frac {64}{245} e^{\frac {1}{4} (-96-7 x)}-10 x+2 e^{\frac {1}{4} (-96-7 x)} x-\frac {4}{175} e^{\frac {1}{2} (-96-7 x)} x+x^2-\frac {2}{5} e^{\frac {1}{4} (-96-7 x)} x^2+\frac {1}{25} e^{\frac {1}{2} (-96-7 x)} x^2+\frac {4}{175} \int e^{\frac {1}{2} (-96-7 x)} \, dx-\frac {2}{25} \int e^{\frac {1}{2} (-96-7 x)} x \, dx+\frac {16}{35} \int e^{\frac {1}{4} (-96-7 x)} \, dx\\ &=-\frac {8 e^{\frac {1}{2} (-96-7 x)}}{1225}-10 x+2 e^{\frac {1}{4} (-96-7 x)} x+x^2-\frac {2}{5} e^{\frac {1}{4} (-96-7 x)} x^2+\frac {1}{25} e^{\frac {1}{2} (-96-7 x)} x^2-\frac {4}{175} \int e^{\frac {1}{2} (-96-7 x)} \, dx\\ &=-10 x+2 e^{\frac {1}{4} (-96-7 x)} x+x^2-\frac {2}{5} e^{\frac {1}{4} (-96-7 x)} x^2+\frac {1}{25} e^{\frac {1}{2} (-96-7 x)} x^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 49, normalized size = 2.58 \begin {gather*} -10 x+x^2+\frac {1}{25} e^{-48-\frac {7 x}{2}} x^2+\frac {1}{10} e^{-7 x/4} \left (\frac {20 x}{e^{24}}-\frac {4 x^2}{e^{24}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-500 + 100*x + E^((-96 - 7*x)/2)*(4*x - 7*x^2) + E^((-96 - 7*x)/4)*(100 - 215*x + 35*x^2))/50,x]

[Out]

-10*x + x^2 + (E^(-48 - (7*x)/2)*x^2)/25 + ((20*x)/E^24 - (4*x^2)/E^24)/(10*E^((7*x)/4))

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fricas [A]  time = 0.60, size = 33, normalized size = 1.74 \begin {gather*} \frac {1}{25} \, x^{2} e^{\left (-\frac {7}{2} \, x - 48\right )} + x^{2} - \frac {2}{5} \, {\left (x^{2} - 5 \, x\right )} e^{\left (-\frac {7}{4} \, x - 24\right )} - 10 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*(-7*x^2+4*x)*exp(-7/4*x-24)^2+1/50*(35*x^2-215*x+100)*exp(-7/4*x-24)+2*x-10,x, algorithm="frica
s")

[Out]

1/25*x^2*e^(-7/2*x - 48) + x^2 - 2/5*(x^2 - 5*x)*e^(-7/4*x - 24) - 10*x

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giac [A]  time = 0.23, size = 33, normalized size = 1.74 \begin {gather*} \frac {1}{25} \, x^{2} e^{\left (-\frac {7}{2} \, x - 48\right )} + x^{2} - \frac {2}{5} \, {\left (x^{2} - 5 \, x\right )} e^{\left (-\frac {7}{4} \, x - 24\right )} - 10 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*(-7*x^2+4*x)*exp(-7/4*x-24)^2+1/50*(35*x^2-215*x+100)*exp(-7/4*x-24)+2*x-10,x, algorithm="giac"
)

[Out]

1/25*x^2*e^(-7/2*x - 48) + x^2 - 2/5*(x^2 - 5*x)*e^(-7/4*x - 24) - 10*x

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maple [B]  time = 0.04, size = 36, normalized size = 1.89

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {7 x}{2}-48} x^{2}}{25}+\frac {\left (-20 x^{2}+100 x \right ) {\mathrm e}^{-\frac {7 x}{4}-24}}{50}+x^{2}-10 x\)	\(36\)
norman	\(x^{2}-10 x +2 \,{\mathrm e}^{-\frac {7 x}{4}-24} x -\frac {2 \,{\mathrm e}^{-\frac {7 x}{4}-24} x^{2}}{5}+\frac {{\mathrm e}^{-\frac {7 x}{2}-48} x^{2}}{25}\)	\(41\)
default	\(x^{2}-10 x +\frac {9216 \,{\mathrm e}^{-\frac {7 x}{2}-48}}{1225}+\frac {768 \,{\mathrm e}^{-\frac {7 x}{2}-48} \left (-\frac {7 x}{4}-24\right )}{1225}+\frac {16 \,{\mathrm e}^{-\frac {7 x}{2}-48} \left (-\frac {7 x}{4}-24\right )^{2}}{1225}-\frac {1816 \,{\mathrm e}^{-\frac {7 x}{4}-24} \left (-\frac {7 x}{4}-24\right )}{245}-\frac {25152 \,{\mathrm e}^{-\frac {7 x}{4}-24}}{245}-\frac {32 \,{\mathrm e}^{-\frac {7 x}{4}-24} \left (-\frac {7 x}{4}-24\right )^{2}}{245}\)	\(86\)
derivativedivides	\(-\frac {262 x}{7}-\frac {25152}{49}+\frac {16 \left (-\frac {7 x}{4}-24\right )^{2}}{49}+\frac {9216 \,{\mathrm e}^{-\frac {7 x}{2}-48}}{1225}+\frac {768 \,{\mathrm e}^{-\frac {7 x}{2}-48} \left (-\frac {7 x}{4}-24\right )}{1225}+\frac {16 \,{\mathrm e}^{-\frac {7 x}{2}-48} \left (-\frac {7 x}{4}-24\right )^{2}}{1225}-\frac {1816 \,{\mathrm e}^{-\frac {7 x}{4}-24} \left (-\frac {7 x}{4}-24\right )}{245}-\frac {25152 \,{\mathrm e}^{-\frac {7 x}{4}-24}}{245}-\frac {32 \,{\mathrm e}^{-\frac {7 x}{4}-24} \left (-\frac {7 x}{4}-24\right )^{2}}{245}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/50*(-7*x^2+4*x)*exp(-7/4*x-24)^2+1/50*(35*x^2-215*x+100)*exp(-7/4*x-24)+2*x-10,x,method=_RETURNVERBOSE)

[Out]

1/25*exp(-7/2*x-48)*x^2+1/50*(-20*x^2+100*x)*exp(-7/4*x-24)+x^2-10*x

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maxima [A]  time = 0.35, size = 33, normalized size = 1.74 \begin {gather*} \frac {1}{25} \, x^{2} e^{\left (-\frac {7}{2} \, x - 48\right )} + x^{2} - \frac {2}{5} \, {\left (x^{2} - 5 \, x\right )} e^{\left (-\frac {7}{4} \, x - 24\right )} - 10 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*(-7*x^2+4*x)*exp(-7/4*x-24)^2+1/50*(35*x^2-215*x+100)*exp(-7/4*x-24)+2*x-10,x, algorithm="maxim
a")

[Out]

1/25*x^2*e^(-7/2*x - 48) + x^2 - 2/5*(x^2 - 5*x)*e^(-7/4*x - 24) - 10*x

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mupad [B]  time = 0.08, size = 38, normalized size = 2.00 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-\frac {7\,x}{2}-48}\,\left (5\,{\mathrm {e}}^{\frac {7\,x}{4}+24}-1\right )\,\left (x+50\,{\mathrm {e}}^{\frac {7\,x}{4}+24}-5\,x\,{\mathrm {e}}^{\frac {7\,x}{4}+24}\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + (exp(- (7*x)/2 - 48)*(4*x - 7*x^2))/50 + (exp(- (7*x)/4 - 24)*(35*x^2 - 215*x + 100))/50 - 10,x)

[Out]

-(x*exp(- (7*x)/2 - 48)*(5*exp((7*x)/4 + 24) - 1)*(x + 50*exp((7*x)/4 + 24) - 5*x*exp((7*x)/4 + 24)))/25

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sympy [B]  time = 0.14, size = 41, normalized size = 2.16 \begin {gather*} \frac {x^{2} e^{- \frac {7 x}{2} - 48}}{25} + x^{2} - 10 x + \frac {\left (- 50 x^{2} + 250 x\right ) e^{- \frac {7 x}{4} - 24}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/50*(-7*x**2+4*x)*exp(-7/4*x-24)**2+1/50*(35*x**2-215*x+100)*exp(-7/4*x-24)+2*x-10,x)

[Out]

x**2*exp(-7*x/2 - 48)/25 + x**2 - 10*x + (-50*x**2 + 250*x)*exp(-7*x/4 - 24)/125

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