3.30.44 \(\int \frac {-256+1250 e^{2 x} x^3+1250 x^4+e^x (-1250 x^3-1250 x^4)}{125 x^3} \, dx\) [2944]

Optimal. Leaf size=19 \[ 5 \left (\left (e^x-x\right )^2+\frac {128}{625 x^2}\right ) \]

[Out]

128/125/x^2+5*(exp(x)-x)^2

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Rubi [A]
time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.74, number of steps used = 8, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2225, 2207} \begin {gather*} 5 x^2+\frac {128}{125 x^2}+10 e^x+5 e^{2 x}-10 e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 + 1250*E^(2*x)*x^3 + 1250*x^4 + E^x*(-1250*x^3 - 1250*x^4))/(125*x^3),x]

[Out]

10*E^x + 5*E^(2*x) + 128/(125*x^2) + 5*x^2 - 10*E^x*(1 + x)

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{125} \int \frac {-256+1250 e^{2 x} x^3+1250 x^4+e^x \left (-1250 x^3-1250 x^4\right )}{x^3} \, dx\\ &=\frac {1}{125} \int \left (1250 e^{2 x}-1250 e^x (1+x)+\frac {2 \left (-128+625 x^4\right )}{x^3}\right ) \, dx\\ &=\frac {2}{125} \int \frac {-128+625 x^4}{x^3} \, dx+10 \int e^{2 x} \, dx-10 \int e^x (1+x) \, dx\\ &=5 e^{2 x}-10 e^x (1+x)+\frac {2}{125} \int \left (-\frac {128}{x^3}+625 x\right ) \, dx+10 \int e^x \, dx\\ &=10 e^x+5 e^{2 x}+\frac {128}{125 x^2}+5 x^2-10 e^x (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 26, normalized size = 1.37 \begin {gather*} 5 e^{2 x}+\frac {128}{125 x^2}-10 e^x x+5 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 + 1250*E^(2*x)*x^3 + 1250*x^4 + E^x*(-1250*x^3 - 1250*x^4))/(125*x^3),x]

[Out]

5*E^(2*x) + 128/(125*x^2) - 10*E^x*x + 5*x^2

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Maple [A]
time = 0.07, size = 23, normalized size = 1.21

method	result	size
default	\(5 x^{2}+\frac {128}{125 x^{2}}+5 \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x} x\)	\(23\)
risch	\(5 x^{2}+\frac {128}{125 x^{2}}+5 \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x} x\)	\(23\)
norman	\(\frac {\frac {128}{125}+5 x^{4}-10 \,{\mathrm e}^{x} x^{3}+5 \,{\mathrm e}^{2 x} x^{2}}{x^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/125*(1250*exp(x)^2*x^3+(-1250*x^4-1250*x^3)*exp(x)+1250*x^4-256)/x^3,x,method=_RETURNVERBOSE)

[Out]

5*x^2+128/125/x^2+5*exp(x)^2-10*exp(x)*x

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Maxima [A]
time = 0.28, size = 28, normalized size = 1.47 \begin {gather*} 5 \, x^{2} - 10 \, {\left (x - 1\right )} e^{x} + \frac {128}{125 \, x^{2}} + 5 \, e^{\left (2 \, x\right )} - 10 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(1250*exp(x)^2*x^3+(-1250*x^4-1250*x^3)*exp(x)+1250*x^4-256)/x^3,x, algorithm="maxima")

[Out]

5*x^2 - 10*(x - 1)*e^x + 128/125/x^2 + 5*e^(2*x) - 10*e^x

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Fricas [A]
time = 0.37, size = 28, normalized size = 1.47 \begin {gather*} \frac {625 \, x^{4} - 1250 \, x^{3} e^{x} + 625 \, x^{2} e^{\left (2 \, x\right )} + 128}{125 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(1250*exp(x)^2*x^3+(-1250*x^4-1250*x^3)*exp(x)+1250*x^4-256)/x^3,x, algorithm="fricas")

[Out]

1/125*(625*x^4 - 1250*x^3*e^x + 625*x^2*e^(2*x) + 128)/x^2

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Sympy [A]
time = 0.05, size = 24, normalized size = 1.26 \begin {gather*} 5 x^{2} - 10 x e^{x} + 5 e^{2 x} + \frac {128}{125 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(1250*exp(x)**2*x**3+(-1250*x**4-1250*x**3)*exp(x)+1250*x**4-256)/x**3,x)

[Out]

5*x**2 - 10*x*exp(x) + 5*exp(2*x) + 128/(125*x**2)

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Giac [A]
time = 0.38, size = 28, normalized size = 1.47 \begin {gather*} \frac {625 \, x^{4} - 1250 \, x^{3} e^{x} + 625 \, x^{2} e^{\left (2 \, x\right )} + 128}{125 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(1250*exp(x)^2*x^3+(-1250*x^4-1250*x^3)*exp(x)+1250*x^4-256)/x^3,x, algorithm="giac")

[Out]

1/125*(625*x^4 - 1250*x^3*e^x + 625*x^2*e^(2*x) + 128)/x^2

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Mupad [B]
time = 1.82, size = 22, normalized size = 1.16 \begin {gather*} 5\,{\mathrm {e}}^{2\,x}-10\,x\,{\mathrm {e}}^x+\frac {128}{125\,x^2}+5\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(1250*x^3 + 1250*x^4))/125 - 10*x^3*exp(2*x) - 10*x^4 + 256/125)/x^3,x)

[Out]

5*exp(2*x) - 10*x*exp(x) + 128/(125*x^2) + 5*x^2

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