\(\int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx\) [674]

Optimal result

Integrand size = 31, antiderivative size = 23 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=4-e^{5/x}+16 e^{10+2 x^3}+x \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {14, 2240} \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=16 e^{2 x^3+10}+x-e^{5/x} \]

[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 2240

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=-e^{5/x}+16 e^{2 \left (5+x^3\right )}+x \]

[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^(2*(5 + x^3)) + x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=x + e^{\left (2 \, x^{3} + 4 \, \log \left (2\right ) + 10\right )} - e^{\frac {5}{x}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=x - e^{\frac {5}{x}} + 16 e^{2 x^{3} + 10} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=x + 16 \, e^{\left (2 \, x^{3} + 10\right )} - e^{\frac {5}{x}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=x + 16 \, e^{\left (2 \, x^{3} + 10\right )} - e^{\frac {5}{x}} \]

[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)

Mupad [B] (verification not implemented)

Time = 7.91 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {5 e^{5/x}+x^2+96 e^{10+2 x^3} x^4}{x^2} \, dx=x-{\mathrm {e}}^{5/x}+16\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{2\,x^3} \]

[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)