3.17.23 $\int \frac{10-2e^x+(-5+e^x(1+x))\log (4x^2)}{{\log }^2(4x^2)}dx$

Optimal. Leaf size=15 $$\frac{(-5+e^x)x}{\log \left(4x^2\right)}$$

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Rubi [A]  time = 0.20, antiderivative size = 25, normalized size of antiderivative = 1.67, number of steps used = 10, number of rules used = 6, integrand size = 32, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.188, Rules used = {6742, 2360, 2297, 2300, 2178, 2288} $$\begin{array}{c}\frac{e^{x}x}{\log \left(4x^2\right)}-\frac{5x}{\log \left(4x^2\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(10 - 2*E^x + (-5 + E^x*(1 + x))*Log[4*x^2])/Log[4*x^2]^2,x]

[Out]

(-5*x)/Log[4*x^2] + (E^x*x)/Log[4*x^2]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{5(-2+\log \left(4x^2\right))}{{\log }^2\left(4x^2\right)}+\frac{e^x(-2+\log \left(4x^2\right)+x\log \left(4x^2\right))}{{\log }^2\left(4x^2\right)})dx\\ & =-(5\int \frac{-2+\log \left(4x^2\right)}{{\log }^2\left(4x^2\right)}dx)+\int \frac{e^x(-2+\log \left(4x^2\right)+x\log \left(4x^2\right))}{{\log }^2\left(4x^2\right)}dx\\ & =\frac{e^{x}x}{\log \left(4x^2\right)}-5\int (-\frac{2}{{\log }^2\left(4x^2\right)}+\frac{1}{\log \left(4x^2\right)})dx\\ & =\frac{e^{x}x}{\log \left(4x^2\right)}-5\int \frac{1}{\log \left(4x^2\right)}dx+10\int \frac{1}{{\log }^2\left(4x^2\right)}dx\\ & =-\frac{5x}{\log \left(4x^2\right)}+\frac{e^{x}x}{\log \left(4x^2\right)}+5\int \frac{1}{\log \left(4x^2\right)}dx-\frac{(5x)\mathrm{Subst}(\int \frac{e^{x/2}}{x}dx,x,\log \left(4x^2\right))}{4\sqrt{x^2}}\\ & =-\frac{5x\text{Ei}(\frac{1}{2}\log \left(4x^2\right))}{4\sqrt{x^2}}-\frac{5x}{\log \left(4x^2\right)}+\frac{e^{x}x}{\log \left(4x^2\right)}+\frac{(5x)\mathrm{Subst}(\int \frac{e^{x/2}}{x}dx,x,\log \left(4x^2\right))}{4\sqrt{x^2}}\\ & =-\frac{5x}{\log \left(4x^2\right)}+\frac{e^{x}x}{\log \left(4x^2\right)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.10, size = 15, normalized size = 1.00 $$\begin{array}{c}\frac{(-5+e^x)x}{\log \left(4x^2\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(10 - 2*E^x + (-5 + E^x*(1 + x))*Log[4*x^2])/Log[4*x^2]^2,x]

[Out]

((-5 + E^x)*x)/Log[4*x^2]

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fricas [A]  time = 0.75, size = 17, normalized size = 1.13 $$\begin{array}{c}\frac{x{e}^x-5x}{\log \left(4x^2\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="fricas")

[Out]

(x*e^x - 5*x)/log(4*x^2)

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giac [A]  time = 0.27, size = 17, normalized size = 1.13 $$\begin{array}{c}\frac{x{e}^x-5x}{\log \left(4x^2\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="giac")

[Out]

(x*e^x - 5*x)/log(4*x^2)

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maple [A]  time = 0.38, size = 22, normalized size = 1.47

method	result	size
norman	$\frac{{\mathrm{e}}^{x}x-5x}{\ln \left(4x^2\right)}$	$22$
default	$-\frac{5x}{\ln \left(4x^2\right)}+\frac{x{\mathrm{e}}^x}{\ln \left(4x^2\right)}$	$29$
risch	$\frac{2ix({\mathrm{e}}^x-5)}{\pi \mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)-2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2+\pi \mathrm{csgn}{\left(i{x}^2\right)}^3+4i\ln (2)+4i\ln (x)}$	$66$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(1/2*x)^2-5)*ln(4*x^2)-2*exp(1/2*x)^2+10)/ln(4*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

(x*exp(1/2*x)^2-5*x)/ln(4*x^2)

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maxima [A]  time = 0.55, size = 17, normalized size = 1.13 $$\begin{array}{c}\frac{x{e}^x-5x}{2(\log (2)+\log (x))}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)^2-5)*log(4*x^2)-2*exp(1/2*x)^2+10)/log(4*x^2)^2,x, algorithm="maxima")

[Out]

1/2*(x*e^x - 5*x)/(log(2) + log(x))

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mupad [B]  time = 1.08, size = 14, normalized size = 0.93 $$\begin{array}{c}\frac{x({\mathrm{e}}^x-5)}{\ln \left(4x^2\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*x^2)*(exp(x)*(x + 1) - 5) - 2*exp(x) + 10)/log(4*x^2)^2,x)

[Out]

(x*(exp(x) - 5))/log(4*x^2)

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sympy [A]  time = 0.28, size = 20, normalized size = 1.33 $$\begin{array}{c}\frac{x{e}^x}{\log \left(4x^2\right)}-\frac{5x}{\log \left(4x^2\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(1/2*x)**2-5)*ln(4*x**2)-2*exp(1/2*x)**2+10)/ln(4*x**2)**2,x)

[Out]

x*exp(x)/log(4*x**2) - 5*x/log(4*x**2)

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