3.33.50 $\int \frac{e^{5+x}(-9+24x-16x^2)+e^5(9-24x+16x^2)+12e^4\log (2)}{e^5(9-24x+16x^2)}dx$

Optimal. Leaf size=22 $$4-e^x+x-\frac{3\log (2)}{e(-3+4x)}$$

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Rubi [A]  time = 0.25, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 5, integrand size = 54, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.093, Rules used = {12, 27, 6742, 2194, 683} $$\begin{array}{c}x-e^x+\frac{\log (4096)}{4e(3-4x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(5 + x)*(-9 + 24*x - 16*x^2) + E^5*(9 - 24*x + 16*x^2) + 12*E^4*Log[2])/(E^5*(9 - 24*x + 16*x^2)),x]

[Out]

-E^x + x + Log[4096]/(4*E*(3 - 4*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{e^{5+x}(-9+24x-16x^2)+e^5(9-24x+16x^2)+12e^4\log (2)}{9-24x+16x^2}dx}{e^5}\\ & =\frac{\int \frac{e^{5+x}(-9+24x-16x^2)+e^5(9-24x+16x^2)+12e^4\log (2)}{(-3+4x{)}^2}dx}{e^5}\\ & =\frac{\int (-e^{5+x}+\frac{e^4(9e-24ex+16e{x}^2+\log (4096))}{(-3+4x{)}^2})dx}{e^5}\\ & =-\frac{\int e^{5+x}dx}{e^5}+\frac{\int \frac{9e-24ex+16e{x}^2+\log (4096)}{(-3+4x{)}^2}dx}{e}\\ & =-e^x+\frac{\int (e+\frac{\log (4096)}{(-3+4x{)}^2})dx}{e}\\ & =-e^x+x+\frac{\log (4096)}{4e(3-4x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.14 $$\begin{array}{c}\frac{-e^{1+x}+ex+\frac{\log (4096)}{12-16x}}{e}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + x)*(-9 + 24*x - 16*x^2) + E^5*(9 - 24*x + 16*x^2) + 12*E^4*Log[2])/(E^5*(9 - 24*x + 16*x^2))
,x]

[Out]

(-E^(1 + x) + E*x + Log[4096]/(12 - 16*x))/E

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fricas [A]  time = 0.56, size = 40, normalized size = 1.82 $$\begin{array}{c}\frac{((4x^2-3x)e^5-(4x-3)e^{(x+5)}-3e^4\log (2))e^{(-5)}}{4x-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="fricas")

[Out]

((4*x^2 - 3*x)*e^5 - (4*x - 3)*e^(x + 5) - 3*e^4*log(2))*e^(-5)/(4*x - 3)

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giac [B]  time = 0.28, size = 42, normalized size = 1.91 $$\begin{array}{c}\frac{(4x^2{e}^5-3x{e}^5-4x{e}^{(x+5)}-3e^4\log (2)+3e^{(x+5)})e^{(-5)}}{4x-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="giac")

[Out]

(4*x^2*e^5 - 3*x*e^5 - 4*x*e^(x + 5) - 3*e^4*log(2) + 3*e^(x + 5))*e^(-5)/(4*x - 3)

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maple [A]  time = 0.55, size = 18, normalized size = 0.82

method	result	size
risch	$x-\frac{3{\mathrm{e}}^{-1}\ln (2)}{4(x-\frac{3}{4})}-{\mathrm{e}}^x$	$18$
norman	$\frac{4x^2-4{\mathrm{e}}^{x}x+3{\mathrm{e}}^x-\frac{3(4{\mathrm{e}}^4\ln (2)+3{\mathrm{e}}^5){\mathrm{e}}^{-5}}{4}}{4x-3}$	$41$
default	${\mathrm{e}}^{-5}(x{\mathrm{e}}^5-\frac{3{\mathrm{e}}^4\ln (2)}{4x-3}-9{\mathrm{e}}^5(-\frac{{\mathrm{e}}^x}{16(x-\frac{3}{4})}-\frac{{\mathrm{e}}^{\frac{3}{4}}\text{expIntegralEi}(1,-x+\frac{3}{4})}{16})+24{\mathrm{e}}^5(-\frac{3{\mathrm{e}}^x}{64(x-\frac{3}{4})}-\frac{7{\mathrm{e}}^{\frac{3}{4}}\text{expIntegralEi}(1,-x+\frac{3}{4})}{64})-16{\mathrm{e}}^5(\frac{{\mathrm{e}}^x}{16}-\frac{9{\mathrm{e}}^x}{256(x-\frac{3}{4})}-\frac{33{\mathrm{e}}^{\frac{3}{4}}\text{expIntegralEi}(1,-x+\frac{3}{4})}{256}))$	$103$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*ln(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x,method=_R
ETURNVERBOSE)

[Out]

x-3/4*exp(-1)*ln(2)/(x-3/4)-exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\frac{1}{4}((4x-\frac{9}{4x-3}+6\log (4x-3))e^5+6(\frac{3}{4x-3}-\log (4x-3))e^5-\frac{32(2x^2{e}^5-3x{e}^5)e^x}{16x^2-24x+9}+\frac{9e^{\frac{23}{4}}{E}_2(-x+\frac{3}{4})}{4x-3}-\frac{12e^4\log (2)}{4x-3}-\frac{9e^5}{4x-3}+288\int \frac{e^{(x+5)}}{64x^3-144x^2+108x-27}dx)e^{(-5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="maxima")

[Out]

1/4*((4*x - 9/(4*x - 3) + 6*log(4*x - 3))*e^5 + 6*(3/(4*x - 3) - log(4*x - 3))*e^5 - 32*(2*x^2*e^5 - 3*x*e^5)*
e^x/(16*x^2 - 24*x + 9) + 9*e^(23/4)*exp_integral_e(2, -x + 3/4)/(4*x - 3) - 12*e^4*log(2)/(4*x - 3) - 9*e^5/(
4*x - 3) + 288*integrate(e^(x + 5)/(64*x^3 - 144*x^2 + 108*x - 27), x))*e^(-5)

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mupad [B]  time = 0.15, size = 24, normalized size = 1.09 $$\begin{array}{c}x-{\mathrm{e}}^x+\frac{3{\mathrm{e}}^4\ln (2)}{3{\mathrm{e}}^5-4x{\mathrm{e}}^5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*(12*exp(4)*log(2) + exp(5)*(16*x^2 - 24*x + 9) - exp(5)*exp(x)*(16*x^2 - 24*x + 9)))/(16*x^2 - 24
*x + 9),x)

[Out]

x - exp(x) + (3*exp(4)*log(2))/(3*exp(5) - 4*x*exp(5))

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sympy [A]  time = 0.25, size = 20, normalized size = 0.91 $$\begin{array}{c}x-e^x-\frac{3\log (2)}{4ex-3e}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**2+24*x-9)*exp(5)*exp(x)+12*exp(4)*ln(2)+(16*x**2-24*x+9)*exp(5))/(16*x**2-24*x+9)/exp(5),x)

[Out]

x - exp(x) - 3*log(2)/(4*E*x - 3*E)

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