3.46.54 \(\int \frac {-30 x-6 x^3+(-10+2 e^5-30 x-2 x^3) \log (\frac {1}{5} (-15+3 e^5-45 x-3 x^3))}{e^{10}+e^5 (-5-15 x-x^3)} \, dx\)

Optimal. Leaf size=27 \[ \frac {2 x \log \left (9 \left (-x+\frac {1}{15} \left (-5+e^5-x^3\right )\right )\right )}{e^5} \]

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Rubi [A]  time = 8.23, antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 23, number of rules used = 10, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6688, 12, 6742, 2079, 800, 634, 618, 204, 628, 2523} \begin {gather*} \frac {2 x \log \left (-\frac {3}{5} \left (x^3+15 x-e^5+5\right )\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30*x - 6*x^3 + (-10 + 2*E^5 - 30*x - 2*x^3)*Log[(-15 + 3*E^5 - 45*x - 3*x^3)/5])/(E^10 + E^5*(-5 - 15*x
- x^3)),x]

[Out]

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

Rule 12

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 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 \frac {2 \left (\frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3}+\log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )\right )}{e^5} \, dx\\ &=\frac {2 \int \left (\frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3}+\log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )\right ) \, dx}{e^5}\\ &=\frac {2 \int \log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right ) \, dx}{e^5}+\frac {6 \int \frac {x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}\\ &=\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}-\frac {2 \int \frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}+\frac {6 \int \left (1+\frac {-5+e^5-10 x}{5-e^5+15 x+x^3}\right ) \, dx}{e^5}\\ &=\frac {6 x}{e^5}+\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}+\frac {6 \int \frac {-5+e^5-10 x}{5-e^5+15 x+x^3} \, dx}{e^5}-\frac {6 \int \frac {x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}\\ &=\frac {6 x}{e^5}+\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}+\frac {6 \int \frac {-5+e^5-10 x}{\left (5 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-\sqrt [3]{\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )}+x\right ) \left (5+25 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}+\left (\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}-\left (5 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-\sqrt [3]{\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )}\right ) x+x^2\right )} \, dx}{e^5}-\frac {6 \int \left (1+\frac {-5+e^5-10 x}{5-e^5+15 x+x^3}\right ) \, dx}{e^5}\\ &=\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}-\frac {6 \int \frac {-5+e^5-10 x}{5-e^5+15 x+x^3} \, dx}{e^5}+\frac {6 \int \left (\frac {4 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} \left (5-e^5-50 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}+5\ 2^{2/3} \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )}{3 \left (-10+50 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}+\sqrt [3]{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )^{2/3}\right ) \left (-10 \sqrt [3]{2}+\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}-2 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} x\right )}+\frac {4 \left (e^5 \left (5 \sqrt [3]{2}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )-5 \left (5 \sqrt [3]{2}+\sqrt [3]{2} \sqrt {525-10 e^5+e^{10}}+\frac {50\ 2^{2/3}}{\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}+10 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )-\left (e^5 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}+5 \left (10 \sqrt [3]{2}-\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )\right ) x\right )}{3 \left (10-50 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}-\sqrt [3]{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )^{2/3}\right ) \left (5 \sqrt [3]{2}-\sqrt [3]{2} e^5-\sqrt [3]{2} \sqrt {525-10 e^5+e^{10}}-\frac {50\ 2^{2/3}}{\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-10 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}+\left (10 \sqrt [3]{2}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right ) x-2 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} x^2\right )}\right ) \, dx}{e^5}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 x \log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*x - 6*x^3 + (-10 + 2*E^5 - 30*x - 2*x^3)*Log[(-15 + 3*E^5 - 45*x - 3*x^3)/5])/(E^10 + E^5*(-5 -
15*x - x^3)),x]

[Out]

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

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fricas [A]  time = 0.49, size = 20, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (-5\right )} \log \left (-\frac {3}{5} \, x^{3} - 9 \, x + \frac {3}{5} \, e^{5} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)-2*x^3-30*x-10)*log(3/5*exp(5)-3/5*x^3-9*x-3)-6*x^3-30*x)/(exp(5)^2+(-x^3-15*x-5)*exp(5)),
x, algorithm="fricas")

[Out]

2*x*e^(-5)*log(-3/5*x^3 - 9*x + 3/5*e^5 - 3)

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giac [A]  time = 0.16, size = 20, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (-5\right )} \log \left (-\frac {3}{5} \, x^{3} - 9 \, x + \frac {3}{5} \, e^{5} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)-2*x^3-30*x-10)*log(3/5*exp(5)-3/5*x^3-9*x-3)-6*x^3-30*x)/(exp(5)^2+(-x^3-15*x-5)*exp(5)),
x, algorithm="giac")

[Out]

2*x*e^(-5)*log(-3/5*x^3 - 9*x + 3/5*e^5 - 3)

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maple [A]  time = 0.07, size = 21, normalized size = 0.78

method	result	size
risch	\(2 x \ln \left (\frac {3 \,{\mathrm e}^{5}}{5}-\frac {3 x^{3}}{5}-9 x -3\right ) {\mathrm e}^{-5}\)	\(21\)
norman	\(2 x \ln \left (\frac {3 \,{\mathrm e}^{5}}{5}-\frac {3 x^{3}}{5}-9 x -3\right ) {\mathrm e}^{-5}\)	\(23\)
default	\(2 \,{\mathrm e}^{-5} \ln \relax (3) x +2 \,{\mathrm e}^{-5} x \ln \left (-x^{3}+{\mathrm e}^{5}-15 x -5\right )+2 \,{\mathrm e}^{-5} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-{\mathrm e}^{5}+15 \textit {\_Z} +5\right )}{\sum }\frac {\left (10 \textit {\_R} +5-{\mathrm e}^{5}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}+5}\right )-2 x \ln \relax (5) {\mathrm e}^{-5}+2 \,{\mathrm e}^{-5} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-{\mathrm e}^{5}+15 \textit {\_Z} +5\right )}{\sum }\frac {\left (-10 \textit {\_R} -5+{\mathrm e}^{5}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}+5}\right )\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(5)-2*x^3-30*x-10)*ln(3/5*exp(5)-3/5*x^3-9*x-3)-6*x^3-30*x)/(exp(5)^2+(-x^3-15*x-5)*exp(5)),x,metho
d=_RETURNVERBOSE)

[Out]

2*x*ln(3/5*exp(5)-3/5*x^3-9*x-3)*exp(-5)

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maxima [C]  time = 0.48, size = 32, normalized size = 1.19 \begin {gather*} 2 \, {\left ({\left (i \, \pi - \log \relax (5) + \log \relax (3)\right )} x + x \log \left (x^{3} + 15 \, x - e^{5} + 5\right )\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)-2*x^3-30*x-10)*log(3/5*exp(5)-3/5*x^3-9*x-3)-6*x^3-30*x)/(exp(5)^2+(-x^3-15*x-5)*exp(5)),
x, algorithm="maxima")

[Out]

2*((I*pi - log(5) + log(3))*x + x*log(x^3 + 15*x - e^5 + 5))*e^(-5)

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mupad [B]  time = 3.55, size = 21, normalized size = 0.78 \begin {gather*} 2\,x\,{\mathrm {e}}^{-5}\,\left (\ln \left (-x^3-15\,x+{\mathrm {e}}^5-5\right )+\ln \left (\frac {3}{5}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x + log((3*exp(5))/5 - 9*x - (3*x^3)/5 - 3)*(30*x - 2*exp(5) + 2*x^3 + 10) + 6*x^3)/(exp(10) - exp(5)
*(15*x + x^3 + 5)),x)

[Out]

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

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sympy [A]  time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} \frac {2 x \log {\left (- \frac {3 x^{3}}{5} - 9 x - 3 + \frac {3 e^{5}}{5} \right )}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(5)-2*x**3-30*x-10)*ln(3/5*exp(5)-3/5*x**3-9*x-3)-6*x**3-30*x)/(exp(5)**2+(-x**3-15*x-5)*exp(
5)),x)

[Out]

2*x*exp(-5)*log(-3*x**3/5 - 9*x - 3 + 3*exp(5)/5)

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