3.53.73 \(\int \frac {-15-5 x^2+e^{4 e^2-2 x} (-10 x-2 x^3)}{5 x+x^3} \, dx\)

Optimal. Leaf size=27 \[ e^{4 e^2-2 x}+\log (5)-\log \left (2 x^3 \left (5+x^2\right )\right ) \]

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Rubi [A]  time = 0.36, antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1593, 6725, 2194, 446, 72} \begin {gather*} -\log \left (x^2+5\right )+e^{4 e^2-2 x}-3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 - 5*x^2 + E^(4*E^2 - 2*x)*(-10*x - 2*x^3))/(5*x + x^3),x]

[Out]

E^(4*E^2 - 2*x) - 3*Log[x] - Log[5 + x^2]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15-5 x^2+e^{4 e^2-2 x} \left (-10 x-2 x^3\right )}{x \left (5+x^2\right )} \, dx\\ &=\int \left (-2 e^{4 e^2-2 x}-\frac {5 \left (3+x^2\right )}{x \left (5+x^2\right )}\right ) \, dx\\ &=-\left (2 \int e^{4 e^2-2 x} \, dx\right )-5 \int \frac {3+x^2}{x \left (5+x^2\right )} \, dx\\ &=e^{4 e^2-2 x}-\frac {5}{2} \operatorname {Subst}\left (\int \frac {3+x}{x (5+x)} \, dx,x,x^2\right )\\ &=e^{4 e^2-2 x}-\frac {5}{2} \operatorname {Subst}\left (\int \left (\frac {3}{5 x}+\frac {2}{5 (5+x)}\right ) \, dx,x,x^2\right )\\ &=e^{4 e^2-2 x}-3 \log (x)-\log \left (5+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.89 \begin {gather*} e^{4 e^2-2 x}-3 \log (x)-\log \left (5+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 - 5*x^2 + E^(4*E^2 - 2*x)*(-10*x - 2*x^3))/(5*x + x^3),x]

[Out]

E^(4*E^2 - 2*x) - 3*Log[x] - Log[5 + x^2]

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fricas [A]  time = 0.48, size = 22, normalized size = 0.81 \begin {gather*} e^{\left (-2 \, x + 4 \, e^{2}\right )} - \log \left (x^{2} + 5\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-10*x)*exp(4*exp(2)-2*x)-5*x^2-15)/(x^3+5*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 22, normalized size = 0.81 \begin {gather*} e^{\left (-2 \, x + 4 \, e^{2}\right )} - \log \left (x^{2} + 5\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-10*x)*exp(4*exp(2)-2*x)-5*x^2-15)/(x^3+5*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 23, normalized size = 0.85

method	result	size
norman	\({\mathrm e}^{4 \,{\mathrm e}^{2}-2 x}-3 \ln \relax (x )-\ln \left (x^{2}+5\right )\)	\(23\)
risch	\({\mathrm e}^{4 \,{\mathrm e}^{2}-2 x}-3 \ln \relax (x )-\ln \left (x^{2}+5\right )\)	\(23\)
derivativedivides	\(-60 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )-5 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )-80 \,{\mathrm e}^{4} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )+40 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R} \ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )+20 \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )+{\mathrm e}^{4 \,{\mathrm e}^{2}-2 x}+4 \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\left (3 \textit {\_R1}^{2} {\mathrm e}^{2}-12 \textit {\_R1} \,{\mathrm e}^{4}+20 \,{\mathrm e}^{2}+16 \,{\mathrm e}^{6}-5 \textit {\_R1} \right ) {\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-80 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-64 \,{\mathrm e}^{6} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )+48 \,{\mathrm e}^{4} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-12 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1}^{2} {\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )\)	\(708\)
default	\(-60 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )-5 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )-80 \,{\mathrm e}^{4} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )+40 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R} \ln \left (4 \,{\mathrm e}^{2}-2 x -\textit {\_R} \right )}{48 \,{\mathrm e}^{4}-24 \textit {\_R} \,{\mathrm e}^{2}+3 \textit {\_R}^{2}+20}\right )+20 \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )+{\mathrm e}^{4 \,{\mathrm e}^{2}-2 x}+4 \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\left (3 \textit {\_R1}^{2} {\mathrm e}^{2}-12 \textit {\_R1} \,{\mathrm e}^{4}+20 \,{\mathrm e}^{2}+16 \,{\mathrm e}^{6}-5 \textit {\_R1} \right ) {\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-80 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-64 \,{\mathrm e}^{6} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )+48 \,{\mathrm e}^{4} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )-12 \,{\mathrm e}^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (\textit {\_Z}^{3}-12 \textit {\_Z}^{2} {\mathrm e}^{2}+\left (48 \,{\mathrm e}^{4}+20\right ) \textit {\_Z} -80 \,{\mathrm e}^{2}-64 \,{\mathrm e}^{6}\right )}{\sum }\frac {\textit {\_R1}^{2} {\mathrm e}^{\textit {\_R1}} \expIntegralEi \left (1, -4 \,{\mathrm e}^{2}+2 x +\textit {\_R1} \right )}{-48 \,{\mathrm e}^{4}+24 \textit {\_R1} \,{\mathrm e}^{2}-3 \textit {\_R1}^{2}-20}\right )\)	\(708\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3-10*x)*exp(4*exp(2)-2*x)-5*x^2-15)/(x^3+5*x),x,method=_RETURNVERBOSE)

[Out]

exp(4*exp(2)-2*x)-3*ln(x)-ln(x^2+5)

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maxima [A]  time = 0.43, size = 22, normalized size = 0.81 \begin {gather*} e^{\left (-2 \, x + 4 \, e^{2}\right )} - \log \left (x^{2} + 5\right ) - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-10*x)*exp(4*exp(2)-2*x)-5*x^2-15)/(x^3+5*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 23, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{4\,{\mathrm {e}}^2}\,{\mathrm {e}}^{-2\,x}-3\,\ln \relax (x)-\ln \left (x^2+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4*exp(2) - 2*x)*(10*x + 2*x^3) + 5*x^2 + 15)/(5*x + x^3),x)

[Out]

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

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sympy [A]  time = 0.14, size = 20, normalized size = 0.74 \begin {gather*} e^{- 2 x + 4 e^{2}} - 3 \log {\relax (x )} - \log {\left (x^{2} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3-10*x)*exp(4*exp(2)-2*x)-5*x**2-15)/(x**3+5*x),x)

[Out]

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

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