3.85.51 $\int \frac{e^4(-15x-6x^2)+e^4(6+3x)\log (e^{2x}(2+x))}{2x^2+x^3}dx$

Optimal. Leaf size=20 $$1-\frac{3e^4\log (e^{2x}(2+x))}{x}$$

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 4, integrand size = 45, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.089, Rules used = {1593, 6742, 72, 2551} $$\begin{array}{c}-\frac{3e^4\log (e^{2x}(x+2))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*E^4*Log[E^(2*x)*(2 + x)])/x

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^4(-15x-6x^2)+e^4(6+3x)\log (e^{2x}(2+x))}{x^2(2+x)}dx\\ & =\int (-\frac{3e^4(5+2x)}{x(2+x)}+\frac{3e^4\log (e^{2x}(2+x))}{x^2})dx\\ & =-(\left(3e^4\right)\int \frac{5+2x}{x(2+x)}dx)+\left(3e^4\right)\int \frac{\log (e^{2x}(2+x))}{x^2}dx\\ & =-\frac{3e^4\log (e^{2x}(2+x))}{x}+\left(3e^4\right)\int \frac{5+2x}{x(2+x)}dx-\left(3e^4\right)\int (\frac{5}{2x}-\frac{1}{2(2+x)})dx\\ & =-\frac{15}{2}{e}^4\log (x)+\frac{3}{2}{e}^4\log (2+x)-\frac{3e^4\log (e^{2x}(2+x))}{x}+\left(3e^4\right)\int (\frac{5}{2x}-\frac{1}{2(2+x)})dx\\ & =-\frac{3e^4\log (e^{2x}(2+x))}{x}\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 18, normalized size = 0.90 $$\begin{array}{c}-\frac{3e^4\log (e^{2x}(2+x))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*E^4*Log[E^(2*x)*(2 + x)])/x

________________________________________________________________________________________

fricas [A]  time = 0.89, size = 16, normalized size = 0.80 $$\begin{array}{c}-\frac{3e^4\log \left((x+2)e^{\left(2x\right)}\right)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="fricas")

[Out]

-3*e^4*log((x + 2)*e^(2*x))/x

________________________________________________________________________________________

giac [A]  time = 0.12, size = 11, normalized size = 0.55 $$\begin{array}{c}-\frac{3e^4\log (x+2)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="giac")

[Out]

-3*e^4*log(x + 2)/x

________________________________________________________________________________________

maple [A]  time = 0.43, size = 19, normalized size = 0.95

method	result	size
norman	$-\frac{3{\mathrm{e}}^4\ln \left({\mathrm{e}}^{2x}(2+x)\right)}{x}$	$19$
default	$-\frac{3{\mathrm{e}}^4\ln \left({\mathrm{e}}^{2x}(2+x)\right)}{x}+6{\mathrm{e}}^4\ln (2x+4)+\frac{15{\mathrm{e}}^4\ln (-2x)}{2}-\frac{15{\mathrm{e}}^4\ln (-2x-4)}{2}-\frac{15{\mathrm{e}}^4\ln (x)}{2}+\frac{3{\mathrm{e}}^4\ln (2+x)}{2}$	$62$
risch	$\frac{3{\mathrm{e}}^4\ln \left({\mathrm{e}}^{-2x}\right)}{x}-\frac{3{\mathrm{e}}^4(-i\pi \mathrm{csgn}\left(i(2+x)\right)\mathrm{csgn}\left(i{\mathrm{e}}^{2x}\right)\mathrm{csgn}\left(i{\mathrm{e}}^{2x}(2+x)\right)+i\pi \mathrm{csgn}\left(i(2+x)\right)\mathrm{csgn}{\left(i{\mathrm{e}}^{2x}(2+x)\right)}^2+i\pi \mathrm{csgn}\left(i{\mathrm{e}}^{2x}\right)\mathrm{csgn}{\left(i{\mathrm{e}}^{2x}(2+x)\right)}^2-i\pi \mathrm{csgn}{\left(i{\mathrm{e}}^{2x}(2+x)\right)}^3+2\ln (2+x))}{2x}$	$124$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6+3*x)*exp(4)*ln((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.40, size = 46, normalized size = 2.30 $$\begin{array}{c}\frac{15}{2}(\log (x+2)-\log (x))e^4-6e^4\log (x+2)+\frac{15}{2}{e}^4\log (x)-\frac{3(x{e}^4+2e^4)\log (x+2)}{2x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*log((2+x)/exp(-2*x))+(-6*x^2-15*x)*exp(4))/(x^3+2*x^2),x, algorithm="maxima")

[Out]

15/2*(log(x + 2) - log(x))*e^4 - 6*e^4*log(x + 2) + 15/2*e^4*log(x) - 3/2*(x*e^4 + 2*e^4)*log(x + 2)/x

________________________________________________________________________________________

mupad [B]  time = 5.34, size = 11, normalized size = 0.55 $$\begin{array}{c}-\frac{3\ln (x+2){\mathrm{e}}^4}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*log(x + 2)*exp(4))/x

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 17, normalized size = 0.85 $$\begin{array}{c}-\frac{3e^4\log \left((x+2)e^{2x}\right)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6+3*x)*exp(4)*ln((2+x)/exp(-2*x))+(-6*x**2-15*x)*exp(4))/(x**3+2*x**2),x)

[Out]

-3*exp(4)*log((x + 2)*exp(2*x))/x

________________________________________________________________________________________