3.7.74 \(\int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx\)

Optimal. Leaf size=23 \[ 4-e^{5/x}+16 e^{10+2 x^3}+x \]

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {14, 2209} \begin {gather*} 16 e^{2 x^3+10}+x-e^{5/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^(5/x) + x^2 + 96*E^(10 + 2*x^3)*x^4)/x^2,x]

[Out]

-E^(5/x) + 16*E^(10 + 2*x^3) + 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 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} &=\int \left (96 e^{10+2 x^3} x^2+\frac {5 e^{5/x}+x^2}{x^2}\right ) \, dx\\ &=96 \int e^{10+2 x^3} x^2 \, dx+\int \frac {5 e^{5/x}+x^2}{x^2} \, dx\\ &=16 e^{10+2 x^3}+\int \left (1+\frac {5 e^{5/x}}{x^2}\right ) \, dx\\ &=16 e^{10+2 x^3}+x+5 \int \frac {e^{5/x}}{x^2} \, dx\\ &=-e^{5/x}+16 e^{10+2 x^3}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} -e^{5/x}+16 e^{10+2 x^3}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(5/x) + x^2 + 96*E^(10 + 2*x^3)*x^4)/x^2,x]

[Out]

-E^(5/x) + 16*E^(10 + 2*x^3) + x

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fricas [A]  time = 0.88, size = 22, normalized size = 0.96 \begin {gather*} x + e^{\left (2 \, x^{3} + 4 \, \log \relax (2) + 10\right )} - e^{\frac {5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4*exp(2*log(2)+x^3+5)^2+5*exp(5/x)+x^2)/x^2,x, algorithm="fricas")

[Out]

x + e^(2*x^3 + 4*log(2) + 10) - e^(5/x)

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giac [A]  time = 0.55, size = 20, normalized size = 0.87 \begin {gather*} x + 16 \, e^{\left (2 \, x^{3} + 10\right )} - e^{\frac {5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4*exp(2*log(2)+x^3+5)^2+5*exp(5/x)+x^2)/x^2,x, algorithm="giac")

[Out]

x + 16*e^(2*x^3 + 10) - e^(5/x)

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maple [A]  time = 0.03, size = 21, normalized size = 0.91

method	result	size
risch	\(16 \,{\mathrm e}^{2 x^{3}+10}+x -{\mathrm e}^{\frac {5}{x}}\)	\(21\)
default	\(x -{\mathrm e}^{\frac {5}{x}}+16 \,{\mathrm e}^{2 x^{3}} {\mathrm e}^{10}\)	\(23\)
norman	\(\frac {x^{2}+16 x \,{\mathrm e}^{2 x^{3}+10}-x \,{\mathrm e}^{\frac {5}{x}}}{x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^4*exp(2*ln(2)+x^3+5)^2+5*exp(5/x)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

16*exp(2*x^3+10)+x-exp(5/x)

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maxima [A]  time = 0.40, size = 20, normalized size = 0.87 \begin {gather*} x + 16 \, e^{\left (2 \, x^{3} + 10\right )} - e^{\frac {5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x^4*exp(2*log(2)+x^3+5)^2+5*exp(5/x)+x^2)/x^2,x, algorithm="maxima")

[Out]

x + 16*e^(2*x^3 + 10) - e^(5/x)

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mupad [B]  time = 0.64, size = 20, normalized size = 0.87 \begin {gather*} x-{\mathrm {e}}^{5/x}+16\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(5/x) + x^2 + 6*x^4*exp(4*log(2) + 2*x^3 + 10))/x^2,x)

[Out]

x - exp(5/x) + 16*exp(10)*exp(2*x^3)

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sympy [A]  time = 0.28, size = 15, normalized size = 0.65 \begin {gather*} x - e^{\frac {5}{x}} + 16 e^{2 x^{3} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**4*exp(2*ln(2)+x**3+5)**2+5*exp(5/x)+x**2)/x**2,x)

[Out]

x - exp(5/x) + 16*exp(2*x**3 + 10)

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