3.74.97 \(\int \frac {5 e^4 x-15 x^3+e^{-5+x} (45+15 e^4 x-15 x^3)+(15+5 e^4 x-5 x^3) \log (-3-e^4 x+x^3)}{3+e^4 x-x^3} \, dx\)

Optimal. Leaf size=26 \[ 5 \left (3 e^{-5+x}+x \log \left (-3+x \left (-e^4+x^2\right )\right )\right ) \]

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Rubi [A]  time = 10.55, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 24, number of rules used = 10, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {6742, 2194, 6688, 2079, 800, 634, 618, 206, 628, 2523} \begin {gather*} 5 x \log \left (x^3-e^4 x-3\right )+15 e^{x-5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^4*x - 15*x^3 + E^(-5 + x)*(45 + 15*E^4*x - 15*x^3) + (15 + 5*E^4*x - 5*x^3)*Log[-3 - E^4*x + x^3])/(3
+ E^4*x - x^3),x]

[Out]

15*E^(-5 + x) + 5*x*Log[-3 - E^4*x + x^3]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2079

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9*a*d^2 + S
qrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Dist[1/d^(2*p), Int[(e + f*x)^m*Simp[(18^(1/3)*b*d)/(3*r) - r/18^(1/
3) + d*x, x]^p*Simp[(b*d)/3 + (12^(1/3)*b^2*d^2)/(3*r^2) + r^2/(3*12^(1/3)) - d*((2^(1/3)*b*d)/(3^(1/3)*r) - r
/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0
]

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 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (15 e^{-5+x}+\frac {5 \left (-e^4 x+3 x^3-3 \log \left (-3-e^4 x+x^3\right )-e^4 x \log \left (-3-e^4 x+x^3\right )+x^3 \log \left (-3-e^4 x+x^3\right )\right )}{-3-e^4 x+x^3}\right ) \, dx\\ &=5 \int \frac {-e^4 x+3 x^3-3 \log \left (-3-e^4 x+x^3\right )-e^4 x \log \left (-3-e^4 x+x^3\right )+x^3 \log \left (-3-e^4 x+x^3\right )}{-3-e^4 x+x^3} \, dx+15 \int e^{-5+x} \, dx\\ &=15 e^{-5+x}+5 \int \left (\frac {x \left (e^4-3 x^2\right )}{3+e^4 x-x^3}+\log \left (-3-e^4 x+x^3\right )\right ) \, dx\\ &=15 e^{-5+x}+5 \int \frac {x \left (e^4-3 x^2\right )}{3+e^4 x-x^3} \, dx+5 \int \log \left (-3-e^4 x+x^3\right ) \, dx\\ &=15 e^{-5+x}+5 x \log \left (-3-e^4 x+x^3\right )-5 \int \frac {x \left (-e^4+3 x^2\right )}{-3-e^4 x+x^3} \, dx+5 \int \left (3-\frac {9+2 e^4 x}{3+e^4 x-x^3}\right ) \, dx\\ &=15 e^{-5+x}+15 x+5 x \log \left (-3-e^4 x+x^3\right )-5 \int \frac {9+2 e^4 x}{3+e^4 x-x^3} \, dx-5 \int \left (3+\frac {9+2 e^4 x}{-3-e^4 x+x^3}\right ) \, dx\\ &=15 e^{-5+x}+5 x \log \left (-3-e^4 x+x^3\right )-5 \int \frac {9+2 e^4 x}{\left (\frac {1}{3} \left (\frac {3^{2/3} e^4}{\sqrt [3]{\frac {1}{2} \left (27+i \sqrt {-729+12 e^{12}}\right )}}+\sqrt [3]{\frac {3}{2} \left (27+i \sqrt {-729+12 e^{12}}\right )}\right )-x\right ) \left (\frac {1}{18} \left (-6 e^4+\frac {6 \sqrt [3]{3} e^8}{\left (\frac {1}{2} \left (27+i \sqrt {-729+12 e^{12}}\right )\right )^{2/3}}+\sqrt [3]{2} \left (3 \left (27+i \sqrt {-729+12 e^{12}}\right )\right )^{2/3}\right )+\frac {1}{3} \left (\frac {3^{2/3} e^4}{\sqrt [3]{\frac {1}{2} \left (27+i \sqrt {-729+12 e^{12}}\right )}}+\sqrt [3]{\frac {3}{2} \left (27+i \sqrt {-729+12 e^{12}}\right )}\right ) x+x^2\right )} \, dx-5 \int \frac {9+2 e^4 x}{-3-e^4 x+x^3} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 24, normalized size = 0.92 \begin {gather*} 5 \left (3 e^{-5+x}+x \log \left (-3-e^4 x+x^3\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^4*x - 15*x^3 + E^(-5 + x)*(45 + 15*E^4*x - 15*x^3) + (15 + 5*E^4*x - 5*x^3)*Log[-3 - E^4*x + x^
3])/(3 + E^4*x - x^3),x]

[Out]

5*(3*E^(-5 + x) + x*Log[-3 - E^4*x + x^3])

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fricas [A]  time = 0.85, size = 21, normalized size = 0.81 \begin {gather*} 5 \, x \log \left (x^{3} - x e^{4} - 3\right ) + 15 \, e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)-5*x^3+15)*log(-x*exp(4)+x^3-3)+(15*x*exp(4)-15*x^3+45)*exp(x-5)+5*x*exp(4)-15*x^3)/(x*e
xp(4)-x^3+3),x, algorithm="fricas")

[Out]

5*x*log(x^3 - x*e^4 - 3) + 15*e^(x - 5)

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giac [A]  time = 0.17, size = 24, normalized size = 0.92 \begin {gather*} 5 \, {\left (x e^{5} \log \left (x^{3} - x e^{4} - 3\right ) + 3 \, e^{x}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)-5*x^3+15)*log(-x*exp(4)+x^3-3)+(15*x*exp(4)-15*x^3+45)*exp(x-5)+5*x*exp(4)-15*x^3)/(x*e
xp(4)-x^3+3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 22, normalized size = 0.85

method	result	size
default	\(15 \,{\mathrm e}^{x -5}+5 x \ln \left (-x \,{\mathrm e}^{4}+x^{3}-3\right )\)	\(22\)
norman	\(15 \,{\mathrm e}^{x -5}+5 x \ln \left (-x \,{\mathrm e}^{4}+x^{3}-3\right )\)	\(22\)
risch	\(15 \,{\mathrm e}^{x -5}+5 x \ln \left (-x \,{\mathrm e}^{4}+x^{3}-3\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x*exp(4)-5*x^3+15)*ln(-x*exp(4)+x^3-3)+(15*x*exp(4)-15*x^3+45)*exp(x-5)+5*x*exp(4)-15*x^3)/(x*exp(4)-x
^3+3),x,method=_RETURNVERBOSE)

[Out]

15*exp(x-5)+5*x*ln(-x*exp(4)+x^3-3)

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maxima [A]  time = 0.37, size = 24, normalized size = 0.92 \begin {gather*} 5 \, {\left (x e^{5} \log \left (x^{3} - x e^{4} - 3\right ) + 3 \, e^{x}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)-5*x^3+15)*log(-x*exp(4)+x^3-3)+(15*x*exp(4)-15*x^3+45)*exp(x-5)+5*x*exp(4)-15*x^3)/(x*e
xp(4)-x^3+3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.15, size = 21, normalized size = 0.81 \begin {gather*} 15\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x+5\,x\,\ln \left (x^3-{\mathrm {e}}^4\,x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^3 - x*exp(4) - 3)*(5*x*exp(4) - 5*x^3 + 15) + 5*x*exp(4) + exp(x - 5)*(15*x*exp(4) - 15*x^3 + 45) -
15*x^3)/(x*exp(4) - x^3 + 3),x)

[Out]

15*exp(-5)*exp(x) + 5*x*log(x^3 - x*exp(4) - 3)

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sympy [A]  time = 0.70, size = 20, normalized size = 0.77 \begin {gather*} 5 x \log {\left (x^{3} - x e^{4} - 3 \right )} + 15 e^{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x*exp(4)-5*x**3+15)*ln(-x*exp(4)+x**3-3)+(15*x*exp(4)-15*x**3+45)*exp(x-5)+5*x*exp(4)-15*x**3)/(
x*exp(4)-x**3+3),x)

[Out]

5*x*log(x**3 - x*exp(4) - 3) + 15*exp(x - 5)

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