3.41.4 $\int \frac{e^{-\frac{4+x+x^4}{x^3}}(3x^4+(36+42x+6x^2+3x^3-3x^5)\log (2+2x))}{x^3+x^4}dx$

Optimal. Leaf size=23 $$3e^{-x-\frac{4+x}{x^3}}x\log (2+2x)$$

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Rubi [B]  time = 1.75, antiderivative size = 51, normalized size of antiderivative = 2.22, number of steps used = 9, number of rules used = 5, integrand size = 56, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.089, Rules used = {1593, 6741, 6742, 2288, 2554} $$\begin{array}{c}-\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(-x^4+2x+12)\log (2x+2)}{(-\frac{12}{x^4}-\frac{2}{x^3}+1)x^3}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(3*x^4 + (36 + 42*x + 6*x^2 + 3*x^3 - 3*x^5)*Log[2 + 2*x])/(E^((4 + x + x^4)/x^3)*(x^3 + x^4)),x]

[Out]

(-3*E^(-4/x^3 - x^(-2) - x)*(12 + 2*x - x^4)*Log[2 + 2*x])/((1 - 12/x^4 - 2/x^3)*x^3)

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6741

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{-\frac{4+x+x^4}{x^3}}(3x^4+(36+42x+6x^2+3x^3-3x^5)\log (2+2x))}{x^3(1+x)}dx\\ & =\int \frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(3x^4+(36+42x+6x^2+3x^3-3x^5)\log (2+2x))}{x^3(1+x)}dx\\ & =\int (\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}x}{1+x}+\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(12+2x+x^3-x^4)\log (2+2x)}{x^3})dx\\ & =3\int \frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}x}{1+x}dx+3\int \frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(12+2x+x^3-x^4)\log (2+2x)}{x^3}dx\\ & =-\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(12+2x-x^4)\log (2+2x)}{(1-\frac{12}{x^4}-\frac{2}{x^3})x^3}+3\int (e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}+\frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}}{-1-x})dx-3\int \frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}x}{1+x}dx\\ & =-\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(12+2x-x^4)\log (2+2x)}{(1-\frac{12}{x^4}-\frac{2}{x^3})x^3}+3\int e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}dx-3\int (e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}+\frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}}{-1-x})dx+3\int \frac{e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}}{-1-x}dx\\ & =-\frac{3e^{-\frac{4}{x^3}-\frac{1}{x^2}-x}(12+2x-x^4)\log (2+2x)}{(1-\frac{12}{x^4}-\frac{2}{x^3})x^3}\end{array}\end{array}$$

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Mathematica [A]  time = 0.27, size = 22, normalized size = 0.96 $$\begin{array}{c}3e^{-\frac{4+x+x^4}{x^3}}x\log (2(1+x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(3*x^4 + (36 + 42*x + 6*x^2 + 3*x^3 - 3*x^5)*Log[2 + 2*x])/(E^((4 + x + x^4)/x^3)*(x^3 + x^4)),x]

[Out]

(3*x*Log[2*(1 + x)])/E^((4 + x + x^4)/x^3)

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fricas [A]  time = 0.57, size = 21, normalized size = 0.91 $$\begin{array}{c}3x{e}^{(-\frac{x^4+x+4}{x^3})}\log (2x+2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+3*x^3+6*x^2+42*x+36)*log(2*x+2)+3*x^4)/(x^4+x^3)/exp((x^4+x+4)/x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 21, normalized size = 0.91 $$\begin{array}{c}3x{e}^{(-\frac{x^4+x+4}{x^3})}\log (2x+2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+3*x^3+6*x^2+42*x+36)*log(2*x+2)+3*x^4)/(x^4+x^3)/exp((x^4+x+4)/x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 22, normalized size = 0.96

method	result	size
risch	$3x{\mathrm{e}}^{-\frac{x^4+x+4}{x^3}}\ln (2x+2)$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^5+3*x^3+6*x^2+42*x+36)*ln(2*x+2)+3*x^4)/(x^4+x^3)/exp((x^4+x+4)/x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.53, size = 28, normalized size = 1.22 $$\begin{array}{c}3(x\log (2)+x\log (x+1))e^{(-x-\frac{1}{x^2}-\frac{4}{x^3})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^5+3*x^3+6*x^2+42*x+36)*log(2*x+2)+3*x^4)/(x^4+x^3)/exp((x^4+x+4)/x^3),x, algorithm="maxima")

[Out]

3*(x*log(2) + x*log(x + 1))*e^(-x - 1/x^2 - 4/x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 $$\begin{array}{c}\int \frac{{\mathrm{e}}^{-\frac{x^4+x+4}{x^3}}(3x^4+\ln (2x+2)(-3x^5+3x^3+6x^2+42x+36))}{x^4+x^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(x + x^4 + 4)/x^3)*(3*x^4 + log(2*x + 2)*(42*x + 6*x^2 + 3*x^3 - 3*x^5 + 36)))/(x^3 + x^4),x)

[Out]

int((exp(-(x + x^4 + 4)/x^3)*(3*x^4 + log(2*x + 2)*(42*x + 6*x^2 + 3*x^3 - 3*x^5 + 36)))/(x^3 + x^4), x)

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sympy [A]  time = 0.48, size = 20, normalized size = 0.87 $$\begin{array}{c}3x{e}^{-\frac{x^4+x+4}{x^3}}\log (2x+2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**5+3*x**3+6*x**2+42*x+36)*ln(2*x+2)+3*x**4)/(x**4+x**3)/exp((x**4+x+4)/x**3),x)

[Out]

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

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