3.24.1 \(\int \frac {1}{25} e^{\frac {1}{25} (-150+125 x+64 x^2)} (125+575 x+515 x^2-128 x^3) \, dx\)

Optimal. Leaf size=23 \[ e^{-6+x+4 \left (x+\frac {16 x^2}{25}\right )} (5-x) x \]

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Rubi [A]  time = 0.36, antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 28, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {12, 6742, 2235, 2234, 2204, 2244, 2240, 2241} \begin {gather*} 5 e^{\frac {64 x^2}{25}+5 x-6} x-e^{\frac {64 x^2}{25}+5 x-6} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-150 + 125*x + 64*x^2)/25)*(125 + 575*x + 515*x^2 - 128*x^3))/25,x]

[Out]

5*E^(-6 + 5*x + (64*x^2)/25)*x - E^(-6 + 5*x + (64*x^2)/25)*x^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 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2235

Int[(F_)^(v_), x_Symbol] :> Int[F^ExpandToSum[v, x], x] /; FreeQ[F, x] && QuadraticQ[v, x] &&  !QuadraticMatch
Q[v, x]

Rule 2240

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} \left (125+575 x+515 x^2-128 x^3\right ) \, dx\\ &=\frac {1}{25} \int \left (125 e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )}+575 e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x+515 e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x^2-128 e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x^3\right ) \, dx\\ &=5 \int e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} \, dx-\frac {128}{25} \int e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x^3 \, dx+\frac {103}{5} \int e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x^2 \, dx+23 \int e^{\frac {1}{25} \left (-150+125 x+64 x^2\right )} x \, dx\\ &=5 \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx-\frac {128}{25} \int e^{-6+5 x+\frac {64 x^2}{25}} x^3 \, dx+\frac {103}{5} \int e^{-6+5 x+\frac {64 x^2}{25}} x^2 \, dx+23 \int e^{-6+5 x+\frac {64 x^2}{25}} x \, dx\\ &=\frac {575}{128} e^{-6+5 x+\frac {64 x^2}{25}}+\frac {515}{128} e^{-6+5 x+\frac {64 x^2}{25}} x-e^{-6+5 x+\frac {64 x^2}{25}} x^2+2 \int e^{-6+5 x+\frac {64 x^2}{25}} x \, dx-\frac {515}{128} \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx+5 \int e^{-6+5 x+\frac {64 x^2}{25}} x^2 \, dx-\frac {2575}{128} \int e^{-6+5 x+\frac {64 x^2}{25}} x \, dx-\frac {2875}{128} \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx+\frac {5 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{e^{2161/256}}\\ &=\frac {15625 e^{-6+5 x+\frac {64 x^2}{25}}}{16384}+5 e^{-6+5 x+\frac {64 x^2}{25}} x-e^{-6+5 x+\frac {64 x^2}{25}} x^2+\frac {25 \sqrt {\pi } \text {erfi}\left (\frac {1}{80} (125+128 x)\right )}{16 e^{2161/256}}-\frac {125}{128} \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx-\frac {125}{64} \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx-\frac {625}{128} \int e^{-6+5 x+\frac {64 x^2}{25}} x \, dx+\frac {321875 \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx}{16384}-\frac {515 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{128 e^{2161/256}}-\frac {2875 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{128 e^{2161/256}}\\ &=5 e^{-6+5 x+\frac {64 x^2}{25}} x-e^{-6+5 x+\frac {64 x^2}{25}} x^2-\frac {6875 \sqrt {\pi } \text {erfi}\left (\frac {1}{80} (125+128 x)\right )}{1024 e^{2161/256}}+\frac {78125 \int e^{-6+5 x+\frac {64 x^2}{25}} \, dx}{16384}-\frac {125 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{128 e^{2161/256}}-\frac {125 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{64 e^{2161/256}}+\frac {321875 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{16384 e^{2161/256}}\\ &=5 e^{-6+5 x+\frac {64 x^2}{25}} x-e^{-6+5 x+\frac {64 x^2}{25}} x^2-\frac {390625 \sqrt {\pi } \text {erfi}\left (\frac {1}{80} (125+128 x)\right )}{262144 e^{2161/256}}+\frac {78125 \int e^{\frac {25}{256} \left (5+\frac {128 x}{25}\right )^2} \, dx}{16384 e^{2161/256}}\\ &=5 e^{-6+5 x+\frac {64 x^2}{25}} x-e^{-6+5 x+\frac {64 x^2}{25}} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 20, normalized size = 0.87 \begin {gather*} -e^{-6+5 x+\frac {64 x^2}{25}} (-5+x) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-150 + 125*x + 64*x^2)/25)*(125 + 575*x + 515*x^2 - 128*x^3))/25,x]

[Out]

-(E^(-6 + 5*x + (64*x^2)/25)*(-5 + x)*x)

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fricas [A]  time = 1.04, size = 20, normalized size = 0.87 \begin {gather*} -{\left (x^{2} - 5 \, x\right )} e^{\left (\frac {64}{25} \, x^{2} + 5 \, x - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-128*x^3+515*x^2+575*x+125)*exp(64/25*x^2+5*x-6),x, algorithm="fricas")

[Out]

-(x^2 - 5*x)*e^(64/25*x^2 + 5*x - 6)

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giac [A]  time = 0.21, size = 25, normalized size = 1.09 \begin {gather*} -\frac {1}{16384} \, {\left ({\left (128 \, x + 125\right )}^{2} - 113920 \, x - 15625\right )} e^{\left (\frac {64}{25} \, x^{2} + 5 \, x - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-128*x^3+515*x^2+575*x+125)*exp(64/25*x^2+5*x-6),x, algorithm="giac")

[Out]

-1/16384*((128*x + 125)^2 - 113920*x - 15625)*e^(64/25*x^2 + 5*x - 6)

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maple [A]  time = 0.04, size = 18, normalized size = 0.78

method	result	size
gosper	\(-{\mathrm e}^{\frac {64}{25} x^{2}+5 x -6} \left (x -5\right ) x\)	\(18\)
risch	\(\frac {\left (-25 x^{2}+125 x \right ) {\mathrm e}^{\frac {64}{25} x^{2}+5 x -6}}{25}\)	\(23\)
default	\(5 x \,{\mathrm e}^{\frac {64}{25} x^{2}+5 x -6}-x^{2} {\mathrm e}^{\frac {64}{25} x^{2}+5 x -6}\)	\(32\)
norman	\(5 x \,{\mathrm e}^{\frac {64}{25} x^{2}+5 x -6}-x^{2} {\mathrm e}^{\frac {64}{25} x^{2}+5 x -6}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-128*x^3+515*x^2+575*x+125)*exp(64/25*x^2+5*x-6),x,method=_RETURNVERBOSE)

[Out]

-exp(64/25*x^2+5*x-6)*(x-5)*x

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maxima [C]  time = 0.53, size = 251, normalized size = 10.91 \begin {gather*} -\frac {25}{16} i \, \sqrt {\pi } \operatorname {erf}\left (\frac {8}{5} i \, x + \frac {25}{16} i\right ) e^{\left (-\frac {2161}{256}\right )} - \frac {25}{262144} \, {\left (\frac {19200 \, {\left (128 \, x + 125\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )}{\left (-{\left (128 \, x + 125\right )}^{2}\right )^{\frac {3}{2}}} - \frac {15625 \, \sqrt {\pi } {\left (128 \, x + 125\right )} {\left (\operatorname {erf}\left (\frac {1}{80} \, \sqrt {-{\left (128 \, x + 125\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (128 \, x + 125\right )}^{2}}} + 30000 \, e^{\left (\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )} - 4096 \, \Gamma \left (2, -\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )\right )} e^{\left (-\frac {2161}{256}\right )} - \frac {2575}{262144} \, {\left (\frac {256 \, {\left (128 \, x + 125\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )}{\left (-{\left (128 \, x + 125\right )}^{2}\right )^{\frac {3}{2}}} - \frac {625 \, \sqrt {\pi } {\left (128 \, x + 125\right )} {\left (\operatorname {erf}\left (\frac {1}{80} \, \sqrt {-{\left (128 \, x + 125\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (128 \, x + 125\right )}^{2}}} + 800 \, e^{\left (\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )}\right )} e^{\left (-\frac {2161}{256}\right )} - \frac {575}{2048} \, {\left (\frac {25 \, \sqrt {\pi } {\left (128 \, x + 125\right )} {\left (\operatorname {erf}\left (\frac {1}{80} \, \sqrt {-{\left (128 \, x + 125\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (128 \, x + 125\right )}^{2}}} - 16 \, e^{\left (\frac {1}{6400} \, {\left (128 \, x + 125\right )}^{2}\right )}\right )} e^{\left (-\frac {2161}{256}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-128*x^3+515*x^2+575*x+125)*exp(64/25*x^2+5*x-6),x, algorithm="maxima")

[Out]

-25/16*I*sqrt(pi)*erf(8/5*I*x + 25/16*I)*e^(-2161/256) - 25/262144*(19200*(128*x + 125)^3*gamma(3/2, -1/6400*(
128*x + 125)^2)/(-(128*x + 125)^2)^(3/2) - 15625*sqrt(pi)*(128*x + 125)*(erf(1/80*sqrt(-(128*x + 125)^2)) - 1)
/sqrt(-(128*x + 125)^2) + 30000*e^(1/6400*(128*x + 125)^2) - 4096*gamma(2, -1/6400*(128*x + 125)^2))*e^(-2161/
256) - 2575/262144*(256*(128*x + 125)^3*gamma(3/2, -1/6400*(128*x + 125)^2)/(-(128*x + 125)^2)^(3/2) - 625*sqr
t(pi)*(128*x + 125)*(erf(1/80*sqrt(-(128*x + 125)^2)) - 1)/sqrt(-(128*x + 125)^2) + 800*e^(1/6400*(128*x + 125
)^2))*e^(-2161/256) - 575/2048*(25*sqrt(pi)*(128*x + 125)*(erf(1/80*sqrt(-(128*x + 125)^2)) - 1)/sqrt(-(128*x
+ 125)^2) - 16*e^(1/6400*(128*x + 125)^2))*e^(-2161/256)

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mupad [B]  time = 0.08, size = 17, normalized size = 0.74 \begin {gather*} -x\,{\mathrm {e}}^{\frac {64\,x^2}{25}+5\,x-6}\,\left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5*x + (64*x^2)/25 - 6)*(575*x + 515*x^2 - 128*x^3 + 125))/25,x)

[Out]

-x*exp(5*x + (64*x^2)/25 - 6)*(x - 5)

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sympy [A]  time = 0.18, size = 19, normalized size = 0.83 \begin {gather*} \left (- x^{2} + 5 x\right ) e^{\frac {64 x^{2}}{25} + 5 x - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-128*x**3+515*x**2+575*x+125)*exp(64/25*x**2+5*x-6),x)

[Out]

(-x**2 + 5*x)*exp(64*x**2/25 + 5*x - 6)

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