3.67.3 $\int \frac{e^{\frac{45+(25x^2+10x^3+x^4)\log (\frac{19+x}{3})}{x^2\log (\frac{19+x}{3})}}(-45x+(-1710-90x)\log (\frac{19+x}{3})+(190x^3+48x^4+2x^5){\log }^2(\frac{19+x}{3}))}{(19x^3+x^4){\log }^2(\frac{19+x}{3})}dx$

Optimal. Leaf size=25 $$e^{(5+x{)}^2+\frac{45}{x^2\log (4+\frac{7+x}{3})}}$$

________________________________________________________________________________________

Rubi [A]  time = 4.81, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 4, integrand size = 108, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.037, Rules used = {1593, 6741, 6688, 6706} $$\begin{array}{c}{e}^{\frac{45}{x^2\log \left(\frac{x+19}{3}\right)}+(x+5{)}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^((45 + (25*x^2 + 10*x^3 + x^4)*Log[(19 + x)/3])/(x^2*Log[(19 + x)/3]))*(-45*x + (-1710 - 90*x)*Log[(19
+ x)/3] + (190*x^3 + 48*x^4 + 2*x^5)*Log[(19 + x)/3]^2))/((19*x^3 + x^4)*Log[(19 + x)/3]^2),x]

[Out]

E^((5 + x)^2 + 45/(x^2*Log[(19 + 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 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6741

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{\exp \left(\frac{45+(25x^2+10x^3+x^4)\log \left(\frac{19+x}{3}\right)}{x^2\log \left(\frac{19+x}{3}\right)}\right)(-45x+(-1710-90x)\log \left(\frac{19+x}{3}\right)+(190x^3+48x^4+2x^5){\log }^2\left(\frac{19+x}{3}\right))}{x^3(19+x){\log }^2\left(\frac{19+x}{3}\right)}dx\\ & =\int \frac{\exp \left(\frac{45+(25x^2+10x^3+x^4)\log \left(\frac{19+x}{3}\right)}{x^2\log (\frac{19}{3}+\frac{x}{3})}\right)(-45x+(-1710-90x)\log \left(\frac{19+x}{3}\right)+(190x^3+48x^4+2x^5){\log }^2\left(\frac{19+x}{3}\right))}{x^3(19+x){\log }^2(\frac{19}{3}+\frac{x}{3})}dx\\ & =\int \frac{e^{(5+x{)}^2+\frac{45}{x^2\log \left(\frac{19+x}{3}\right)}}(-45x-90(19+x)\log \left(\frac{19+x}{3}\right)+2x^3(95+24x+x^2){\log }^2\left(\frac{19+x}{3}\right))}{x^3(19+x){\log }^2(\frac{19}{3}+\frac{x}{3})}dx\\ & =e^{(5+x{)}^2+\frac{45}{x^2\log \left(\frac{19+x}{3}\right)}}\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 25, normalized size = 1.00 $$\begin{array}{c}{e}^{25+10x+x^2+\frac{45}{x^2\log \left(\frac{19+x}{3}\right)}}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^((45 + (25*x^2 + 10*x^3 + x^4)*Log[(19 + x)/3])/(x^2*Log[(19 + x)/3]))*(-45*x + (-1710 - 90*x)*Lo
g[(19 + x)/3] + (190*x^3 + 48*x^4 + 2*x^5)*Log[(19 + x)/3]^2))/((19*x^3 + x^4)*Log[(19 + x)/3]^2),x]

[Out]

E^(25 + 10*x + x^2 + 45/(x^2*Log[(19 + x)/3]))

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 36, normalized size = 1.44 $$\begin{array}{c}{e}^{\left(\frac{(x^4+10x^3+25x^2)\log (\frac{1}{3}x+\frac{19}{3})+45}{x^2\log (\frac{1}{3}x+\frac{19}{3})}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="fricas")

[Out]

e^(((x^4 + 10*x^3 + 25*x^2)*log(1/3*x + 19/3) + 45)/(x^2*log(1/3*x + 19/3)))

________________________________________________________________________________________

giac [A]  time = 1.98, size = 22, normalized size = 0.88 $$\begin{array}{c}{e}^{(x^2+10x+\frac{45}{x^2\log (\frac{1}{3}x+\frac{19}{3})}+25)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="giac")

[Out]

e^(x^2 + 10*x + 45/(x^2*log(1/3*x + 19/3)) + 25)

________________________________________________________________________________________

maple [B]  time = 0.44, size = 48, normalized size = 1.92

method	result	size
risch	${\mathrm{e}}^{\frac{\ln (\frac{x}{3}+\frac{19}{3})x^4+10\ln (\frac{x}{3}+\frac{19}{3})x^3+25x^2\ln (\frac{x}{3}+\frac{19}{3})+45}{x^2\ln (\frac{x}{3}+\frac{19}{3})}}$	$48$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5+48*x^4+190*x^3)*ln(1/3*x+19/3)^2+(-90*x-1710)*ln(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2)*ln(1/3
*x+19/3)+45)/x^2/ln(1/3*x+19/3))/(x^4+19*x^3)/ln(1/3*x+19/3)^2,x,method=_RETURNVERBOSE)

[Out]

exp((ln(1/3*x+19/3)*x^4+10*ln(1/3*x+19/3)*x^3+25*x^2*ln(1/3*x+19/3)+45)/x^2/ln(1/3*x+19/3))

________________________________________________________________________________________

maxima [A]  time = 0.68, size = 25, normalized size = 1.00 $$\begin{array}{c}{e}^{(x^2+10x-\frac{45}{x^2(\log (3)-\log (x+19))}+25)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+48*x^4+190*x^3)*log(1/3*x+19/3)^2+(-90*x-1710)*log(1/3*x+19/3)-45*x)*exp(((x^4+10*x^3+25*x^2
)*log(1/3*x+19/3)+45)/x^2/log(1/3*x+19/3))/(x^4+19*x^3)/log(1/3*x+19/3)^2,x, algorithm="maxima")

[Out]

e^(x^2 + 10*x - 45/(x^2*(log(3) - log(x + 19))) + 25)

________________________________________________________________________________________

mupad [B]  time = 4.48, size = 25, normalized size = 1.00 $$\begin{array}{c}{\mathrm{e}}^{10x}{\mathrm{e}}^{x^2}{\mathrm{e}}^{25}{\mathrm{e}}^{\frac{45}{x^2\ln (\frac{x}{3}+\frac{19}{3})}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(x/3 + 19/3)*(25*x^2 + 10*x^3 + x^4) + 45)/(x^2*log(x/3 + 19/3)))*(45*x + log(x/3 + 19/3)*(90*x
+ 1710) - log(x/3 + 19/3)^2*(190*x^3 + 48*x^4 + 2*x^5)))/(log(x/3 + 19/3)^2*(19*x^3 + x^4)),x)

[Out]

exp(10*x)*exp(x^2)*exp(25)*exp(45/(x^2*log(x/3 + 19/3)))

________________________________________________________________________________________

sympy [A]  time = 0.48, size = 36, normalized size = 1.44 $$\begin{array}{c}{e}^{\frac{(x^4+10x^3+25x^2)\log (\frac{x}{3}+\frac{19}{3})+45}{x^2\log (\frac{x}{3}+\frac{19}{3})}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5+48*x**4+190*x**3)*ln(1/3*x+19/3)**2+(-90*x-1710)*ln(1/3*x+19/3)-45*x)*exp(((x**4+10*x**3+25
*x**2)*ln(1/3*x+19/3)+45)/x**2/ln(1/3*x+19/3))/(x**4+19*x**3)/ln(1/3*x+19/3)**2,x)

[Out]

exp(((x**4 + 10*x**3 + 25*x**2)*log(x/3 + 19/3) + 45)/(x**2*log(x/3 + 19/3)))

________________________________________________________________________________________