3.90.19 \(\int \frac {180 e^{\frac {18}{x^2}}+3 x^3+2 e^{2 x} x^3}{5 x^3} \, dx\)

Optimal. Leaf size=25 \[ -e^{\frac {18}{x^2}}+\frac {1}{5} \left (-1+e^{2 x}-2 x\right )+x \]

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Rubi [A]  time = 0.04, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2194, 2209} \begin {gather*} -e^{\frac {18}{x^2}}+\frac {3 x}{5}+\frac {e^{2 x}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(180*E^(18/x^2) + 3*x^3 + 2*E^(2*x)*x^3)/(5*x^3),x]

[Out]

-E^(18/x^2) + E^(2*x)/5 + (3*x)/5

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 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 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}{5} \int \frac {180 e^{\frac {18}{x^2}}+3 x^3+2 e^{2 x} x^3}{x^3} \, dx\\ &=\frac {1}{5} \int \left (2 e^{2 x}+\frac {3 \left (60 e^{\frac {18}{x^2}}+x^3\right )}{x^3}\right ) \, dx\\ &=\frac {2}{5} \int e^{2 x} \, dx+\frac {3}{5} \int \frac {60 e^{\frac {18}{x^2}}+x^3}{x^3} \, dx\\ &=\frac {e^{2 x}}{5}+\frac {3}{5} \int \left (1+\frac {60 e^{\frac {18}{x^2}}}{x^3}\right ) \, dx\\ &=\frac {e^{2 x}}{5}+\frac {3 x}{5}+36 \int \frac {e^{\frac {18}{x^2}}}{x^3} \, dx\\ &=-e^{\frac {18}{x^2}}+\frac {e^{2 x}}{5}+\frac {3 x}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.96 \begin {gather*} -e^{\frac {18}{x^2}}+\frac {e^{2 x}}{5}+\frac {3 x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(180*E^(18/x^2) + 3*x^3 + 2*E^(2*x)*x^3)/(5*x^3),x]

[Out]

-E^(18/x^2) + E^(2*x)/5 + (3*x)/5

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fricas [A]  time = 0.43, size = 18, normalized size = 0.72 \begin {gather*} \frac {3}{5} \, x + \frac {1}{5} \, e^{\left (2 \, x\right )} - e^{\left (\frac {18}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*exp(x)^2*x^3+180*exp(9/x^2)^2+3*x^3)/x^3,x, algorithm="fricas")

[Out]

3/5*x + 1/5*e^(2*x) - e^(18/x^2)

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giac [A]  time = 0.12, size = 18, normalized size = 0.72 \begin {gather*} \frac {3}{5} \, x + \frac {1}{5} \, e^{\left (2 \, x\right )} - e^{\left (\frac {18}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*exp(x)^2*x^3+180*exp(9/x^2)^2+3*x^3)/x^3,x, algorithm="giac")

[Out]

3/5*x + 1/5*e^(2*x) - e^(18/x^2)

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maple [A]  time = 0.02, size = 19, normalized size = 0.76

method	result	size
default	\(\frac {{\mathrm e}^{2 x}}{5}-{\mathrm e}^{\frac {18}{x^{2}}}+\frac {3 x}{5}\)	\(19\)
risch	\(\frac {{\mathrm e}^{2 x}}{5}-{\mathrm e}^{\frac {18}{x^{2}}}+\frac {3 x}{5}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(2*exp(x)^2*x^3+180*exp(9/x^2)^2+3*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

3/5*x+1/5*exp(x)^2-exp(1/x^2)^18

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maxima [A]  time = 0.36, size = 18, normalized size = 0.72 \begin {gather*} \frac {3}{5} \, x + \frac {1}{5} \, e^{\left (2 \, x\right )} - e^{\left (\frac {18}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*exp(x)^2*x^3+180*exp(9/x^2)^2+3*x^3)/x^3,x, algorithm="maxima")

[Out]

3/5*x + 1/5*e^(2*x) - e^(18/x^2)

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mupad [B]  time = 7.08, size = 18, normalized size = 0.72 \begin {gather*} \frac {3\,x}{5}+\frac {{\mathrm {e}}^{2\,x}}{5}-{\mathrm {e}}^{\frac {18}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((36*exp(18/x^2) + (2*x^3*exp(2*x))/5 + (3*x^3)/5)/x^3,x)

[Out]

(3*x)/5 + exp(2*x)/5 - exp(18/x^2)

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sympy [A]  time = 0.29, size = 17, normalized size = 0.68 \begin {gather*} \frac {3 x}{5} - e^{\frac {18}{x^{2}}} + \frac {e^{2 x}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(2*exp(x)**2*x**3+180*exp(9/x**2)**2+3*x**3)/x**3,x)

[Out]

3*x/5 - exp(18/x**2) + exp(2*x)/5

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