3.37.30 \(\int \frac {-15 e^{\frac {5}{4 x^2}}+4 x^3-4 e^{2 x} x^3+4 x^4-12 x^5-8 x^6+e^x (12 x^4+4 x^5)}{2 x^3} \, dx\)

Optimal. Leaf size=28 \[ -5+3 e^{\frac {5}{4 x^2}}-\left (-1-e^x+x+x^2\right )^2 \]

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Rubi [A]  time = 0.13, antiderivative size = 54, normalized size of antiderivative = 1.93, number of steps used = 15, number of rules used = 6, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 14, 2194, 2196, 2176, 2209} \begin {gather*} -x^4-2 x^3+2 e^x x^2+x^2+3 e^{\frac {5}{4 x^2}}+2 e^x x+2 x-2 e^x-e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*E^(5/(4*x^2)) + 4*x^3 - 4*E^(2*x)*x^3 + 4*x^4 - 12*x^5 - 8*x^6 + E^x*(12*x^4 + 4*x^5))/(2*x^3),x]

[Out]

3*E^(5/(4*x^2)) - 2*E^x - E^(2*x) + 2*x + 2*E^x*x + x^2 + 2*E^x*x^2 - 2*x^3 - x^4

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 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

Rule 2209

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-15 e^{\frac {5}{4 x^2}}+4 x^3-4 e^{2 x} x^3+4 x^4-12 x^5-8 x^6+e^x \left (12 x^4+4 x^5\right )}{x^3} \, dx\\ &=\frac {1}{2} \int \left (-4 e^{2 x}+4 e^x x (3+x)-\frac {15 e^{\frac {5}{4 x^2}}-4 x^3-4 x^4+12 x^5+8 x^6}{x^3}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {15 e^{\frac {5}{4 x^2}}-4 x^3-4 x^4+12 x^5+8 x^6}{x^3} \, dx\right )-2 \int e^{2 x} \, dx+2 \int e^x x (3+x) \, dx\\ &=-e^{2 x}-\frac {1}{2} \int \left (\frac {15 e^{\frac {5}{4 x^2}}}{x^3}+4 \left (-1-x+3 x^2+2 x^3\right )\right ) \, dx+2 \int \left (3 e^x x+e^x x^2\right ) \, dx\\ &=-e^{2 x}+2 \int e^x x^2 \, dx-2 \int \left (-1-x+3 x^2+2 x^3\right ) \, dx+6 \int e^x x \, dx-\frac {15}{2} \int \frac {e^{\frac {5}{4 x^2}}}{x^3} \, dx\\ &=3 e^{\frac {5}{4 x^2}}-e^{2 x}+2 x+6 e^x x+x^2+2 e^x x^2-2 x^3-x^4-4 \int e^x x \, dx-6 \int e^x \, dx\\ &=3 e^{\frac {5}{4 x^2}}-6 e^x-e^{2 x}+2 x+2 e^x x+x^2+2 e^x x^2-2 x^3-x^4+4 \int e^x \, dx\\ &=3 e^{\frac {5}{4 x^2}}-2 e^x-e^{2 x}+2 x+2 e^x x+x^2+2 e^x x^2-2 x^3-x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 46, normalized size = 1.64 \begin {gather*} 3 e^{\frac {5}{4 x^2}}-e^{2 x}+2 x+x^2-2 x^3-x^4+2 e^x \left (-1+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^(5/(4*x^2)) + 4*x^3 - 4*E^(2*x)*x^3 + 4*x^4 - 12*x^5 - 8*x^6 + E^x*(12*x^4 + 4*x^5))/(2*x^3),
x]

[Out]

3*E^(5/(4*x^2)) - E^(2*x) + 2*x + x^2 - 2*x^3 - x^4 + 2*E^x*(-1 + x + x^2)

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fricas [A]  time = 0.52, size = 41, normalized size = 1.46 \begin {gather*} -x^{4} - 2 \, x^{3} + x^{2} + 2 \, {\left (x^{2} + x - 1\right )} e^{x} + 2 \, x - e^{\left (2 \, x\right )} + 3 \, e^{\left (\frac {5}{4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x)^2*x^3+(4*x^5+12*x^4)*exp(x)-15*exp(5/4/x^2)-8*x^6-12*x^5+4*x^4+4*x^3)/x^3,x, algorith
m="fricas")

[Out]

-x^4 - 2*x^3 + x^2 + 2*(x^2 + x - 1)*e^x + 2*x - e^(2*x) + 3*e^(5/4/x^2)

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giac [A]  time = 0.14, size = 47, normalized size = 1.68 \begin {gather*} -x^{4} - 2 \, x^{3} + 2 \, x^{2} e^{x} + x^{2} + 2 \, x e^{x} + 2 \, x - e^{\left (2 \, x\right )} - 2 \, e^{x} + 3 \, e^{\left (\frac {5}{4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x)^2*x^3+(4*x^5+12*x^4)*exp(x)-15*exp(5/4/x^2)-8*x^6-12*x^5+4*x^4+4*x^3)/x^3,x, algorith
m="giac")

[Out]

-x^4 - 2*x^3 + 2*x^2*e^x + x^2 + 2*x*e^x + 2*x - e^(2*x) - 2*e^x + 3*e^(5/4/x^2)

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maple [A]  time = 0.03, size = 46, normalized size = 1.64

method	result	size
risch	\(-x^{4}-2 x^{3}+x^{2}-{\mathrm e}^{2 x}+2 x +\frac {\left (4 x^{2}+4 x -4\right ) {\mathrm e}^{x}}{2}+3 \,{\mathrm e}^{\frac {5}{4 x^{2}}}\)	\(46\)
default	\(-x^{4}-2 x^{3}+x^{2}+2 x -{\mathrm e}^{2 x}+3 \,{\mathrm e}^{\frac {5}{4 x^{2}}}+2 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} x^{2}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-4*exp(x)^2*x^3+(4*x^5+12*x^4)*exp(x)-15*exp(5/4/x^2)-8*x^6-12*x^5+4*x^4+4*x^3)/x^3,x,method=_RETURNV
ERBOSE)

[Out]

-x^4-2*x^3+x^2-exp(2*x)+2*x+1/2*(4*x^2+4*x-4)*exp(x)+3*exp(5/4/x^2)

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maxima [B]  time = 0.86, size = 50, normalized size = 1.79 \begin {gather*} -x^{4} - 2 \, x^{3} + x^{2} + 2 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 6 \, {\left (x - 1\right )} e^{x} + 2 \, x - e^{\left (2 \, x\right )} + 3 \, e^{\left (\frac {5}{4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x)^2*x^3+(4*x^5+12*x^4)*exp(x)-15*exp(5/4/x^2)-8*x^6-12*x^5+4*x^4+4*x^3)/x^3,x, algorith
m="maxima")

[Out]

-x^4 - 2*x^3 + x^2 + 2*(x^2 - 2*x + 2)*e^x + 6*(x - 1)*e^x + 2*x - e^(2*x) + 3*e^(5/4/x^2)

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mupad [B]  time = 2.39, size = 41, normalized size = 1.46 \begin {gather*} 2\,x-{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{\frac {5}{4\,x^2}}+x^2-2\,x^3-x^4+2\,{\mathrm {e}}^x\,\left (x^2+x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((15*exp(5/(4*x^2)))/2 - (exp(x)*(12*x^4 + 4*x^5))/2 + 2*x^3*exp(2*x) - 2*x^3 - 2*x^4 + 6*x^5 + 4*x^6)/x^
3,x)

[Out]

2*x - exp(2*x) + 3*exp(5/(4*x^2)) + x^2 - 2*x^3 - x^4 + 2*exp(x)*(x + x^2 - 1)

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sympy [A]  time = 0.38, size = 42, normalized size = 1.50 \begin {gather*} - x^{4} - 2 x^{3} + x^{2} + 2 x + \left (2 x^{2} + 2 x - 2\right ) e^{x} + 3 e^{\frac {5}{4 x^{2}}} - e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-4*exp(x)**2*x**3+(4*x**5+12*x**4)*exp(x)-15*exp(5/4/x**2)-8*x**6-12*x**5+4*x**4+4*x**3)/x**3,x
)

[Out]

-x**4 - 2*x**3 + x**2 + 2*x + (2*x**2 + 2*x - 2)*exp(x) + 3*exp(5/(4*x**2)) - exp(2*x)

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