\(\int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx\) [9884]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=e^{4 x}+4 x \left (\frac {5}{x^2}+x\right ) \]

[Out]

exp(x)^4+4*(5/x^2+x)*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2225} \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=4 x^2+e^{4 x}+\frac {20}{x} \]

[In]

Int[(-20 + 4*E^(4*x)*x^2 + 8*x^3)/x^2,x]

[Out]

E^(4*x) + 20/x + 4*x^2

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 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{align*} \text {integral}& = \int \left (4 e^{4 x}+\frac {4 \left (-5+2 x^3\right )}{x^2}\right ) \, dx \\ & = 4 \int e^{4 x} \, dx+4 \int \frac {-5+2 x^3}{x^2} \, dx \\ & = e^{4 x}+4 \int \left (-\frac {5}{x^2}+2 x\right ) \, dx \\ & = e^{4 x}+\frac {20}{x}+4 x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=4 \left (\frac {e^{4 x}}{4}+\frac {5}{x}+x^2\right ) \]

[In]

Integrate[(-20 + 4*E^(4*x)*x^2 + 8*x^3)/x^2,x]

[Out]

4*(E^(4*x)/4 + 5/x + x^2)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
default	\({\mathrm e}^{4 x}+4 x^{2}+\frac {20}{x}\)	\(16\)
risch	\({\mathrm e}^{4 x}+4 x^{2}+\frac {20}{x}\)	\(16\)
parts	\({\mathrm e}^{4 x}+4 x^{2}+\frac {20}{x}\)	\(16\)
norman	\(\frac {4 x^{3}+x \,{\mathrm e}^{4 x}+20}{x}\)	\(18\)
parallelrisch	\(\frac {4 x^{3}+x \,{\mathrm e}^{4 x}+20}{x}\)	\(18\)

[In]

int((4*x^2*exp(x)^4+8*x^3-20)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(x)^4+4*x^2+20/x

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=\frac {4 \, x^{3} + x e^{\left (4 \, x\right )} + 20}{x} \]

[In]

integrate((4*x^2*exp(x)^4+8*x^3-20)/x^2,x, algorithm="fricas")

[Out]

(4*x^3 + x*e^(4*x) + 20)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=4 x^{2} + e^{4 x} + \frac {20}{x} \]

[In]

integrate((4*x**2*exp(x)**4+8*x**3-20)/x**2,x)

[Out]

4*x**2 + exp(4*x) + 20/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=4 \, x^{2} + \frac {20}{x} + e^{\left (4 \, x\right )} \]

[In]

integrate((4*x^2*exp(x)^4+8*x^3-20)/x^2,x, algorithm="maxima")

[Out]

4*x^2 + 20/x + e^(4*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx=\frac {4 \, x^{3} + x e^{\left (4 \, x\right )} + 20}{x} \]

[In]

integrate((4*x^2*exp(x)^4+8*x^3-20)/x^2,x, algorithm="giac")

[Out]

(4*x^3 + x*e^(4*x) + 20)/x

Mupad [B] (verification not implemented)

Time = 16.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-20+4 e^{4 x} x^2+8 x^3}{x^2} \, dx={\mathrm {e}}^{4\,x}+\frac {20}{x}+4\,x^2 \]

[In]

int((4*x^2*exp(4*x) + 8*x^3 - 20)/x^2,x)

[Out]

exp(4*x) + 20/x + 4*x^2