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

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

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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 {gather*} e^{\frac {45}{x^2 \log \left (\frac {x+19}{3}\right )}+(x+5)^2} \end {gather*}

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

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

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]))

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

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)))

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

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)

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

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

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))

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maxima [A]  time = 0.68, size = 25, normalized size = 1.00 \begin {gather*} e^{\left (x^{2} + 10 \, x - \frac {45}{x^{2} {\left (\log \relax (3) - \log \left (x + 19\right )\right )}} + 25\right )} \end {gather*}

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)

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

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)))

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

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)))

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