\(\int \frac {-100-320 x+2 x^3+e^x (25 x+160 x^2+256 x^3+x^4)}{25 x+160 x^2+256 x^3+x^4} \, dx\) [202]

Optimal result

Integrand size = 51, antiderivative size = 19 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=11+e^x+\log \left (\left (\left (16+\frac {5}{x}\right )^2+x\right )^2\right ) \]

[Out]

exp(x)+11+ln(((5/x+16)^2+x)^2)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6873, 6874, 2225, 1601} \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=2 \log \left (x^3+256 x^2+160 x+25\right )+e^x-4 \log (x) \]

[In]

Int[(-100 - 320*x + 2*x^3 + E^x*(25*x + 160*x^2 + 256*x^3 + x^4))/(25*x + 160*x^2 + 256*x^3 + x^4),x]

[Out]

E^x - 4*Log[x] + 2*Log[25 + 160*x + 256*x^2 + x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2225

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 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{x \left (25+160 x+256 x^2+x^3\right )} \, dx \\ & = \int \left (e^x+\frac {2 \left (-50-160 x+x^3\right )}{x \left (25+160 x+256 x^2+x^3\right )}\right ) \, dx \\ & = 2 \int \frac {-50-160 x+x^3}{x \left (25+160 x+256 x^2+x^3\right )} \, dx+\int e^x \, dx \\ & = e^x+2 \int \left (-\frac {2}{x}+\frac {160+512 x+3 x^2}{25+160 x+256 x^2+x^3}\right ) \, dx \\ & = e^x-4 \log (x)+2 \int \frac {160+512 x+3 x^2}{25+160 x+256 x^2+x^3} \, dx \\ & = e^x-4 \log (x)+2 \log \left (25+160 x+256 x^2+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^x-4 \log (x)+2 \log \left (25+160 x+256 x^2+x^3\right ) \]

[In]

Integrate[(-100 - 320*x + 2*x^3 + E^x*(25*x + 160*x^2 + 256*x^3 + x^4))/(25*x + 160*x^2 + 256*x^3 + x^4),x]

[Out]

E^x - 4*Log[x] + 2*Log[25 + 160*x + 256*x^2 + x^3]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26

method	result	size
norman	\({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\)	\(24\)
risch	\({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\)	\(24\)
parallelrisch	\({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\)	\(24\)
parts	\({\mathrm e}^{x}-4 \ln \left (x \right )+2 \ln \left (x^{3}+256 x^{2}+160 x +25\right )\)	\(24\)
default	\({\mathrm e}^{x}+\left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\left (256 \textit {\_R1}^{2}+160 \textit {\_R1} +25\right ) {\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\left (-\textit {\_R}^{2}-256 \textit {\_R} -160\right ) \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )-4 \ln \left (x \right )-320 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{3 \textit {\_R}^{2}+512 \textit {\_R} +160}\right )-25 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {{\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-160 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R1} \,{\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )-256 \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3}+256 \textit {\_Z}^{2}+160 \textit {\_Z} +25\right )}{\sum }\frac {\textit {\_R1}^{2} {\mathrm e}^{\textit {\_R1}} \operatorname {Ei}_{1}\left (-x +\textit {\_R1} \right )}{3 \textit {\_R1}^{2}+512 \textit {\_R1} +160}\right )\)	\(311\)

[In]

int(((x^4+256*x^3+160*x^2+25*x)*exp(x)+2*x^3-320*x-100)/(x^4+256*x^3+160*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

exp(x)-4*ln(x)+2*ln(x^3+256*x^2+160*x+25)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]

[In]

integrate(((x^4+256*x^3+160*x^2+25*x)*exp(x)+2*x^3-320*x-100)/(x^4+256*x^3+160*x^2+25*x),x, algorithm="fricas"
)

[Out]

e^x + 2*log(x^3 + 256*x^2 + 160*x + 25) - 4*log(x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} - 4 \log {\left (x \right )} + 2 \log {\left (x^{3} + 256 x^{2} + 160 x + 25 \right )} \]

[In]

integrate(((x**4+256*x**3+160*x**2+25*x)*exp(x)+2*x**3-320*x-100)/(x**4+256*x**3+160*x**2+25*x),x)

[Out]

exp(x) - 4*log(x) + 2*log(x**3 + 256*x**2 + 160*x + 25)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]

[In]

integrate(((x^4+256*x^3+160*x^2+25*x)*exp(x)+2*x^3-320*x-100)/(x^4+256*x^3+160*x^2+25*x),x, algorithm="maxima"
)

[Out]

e^x + 2*log(x^3 + 256*x^2 + 160*x + 25) - 4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \left (x\right ) \]

[In]

integrate(((x^4+256*x^3+160*x^2+25*x)*exp(x)+2*x^3-320*x-100)/(x^4+256*x^3+160*x^2+25*x),x, algorithm="giac")

[Out]

e^x + 2*log(x^3 + 256*x^2 + 160*x + 25) - 4*log(x)

Mupad [B] (verification not implemented)

Time = 8.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {-100-320 x+2 x^3+e^x \left (25 x+160 x^2+256 x^3+x^4\right )}{25 x+160 x^2+256 x^3+x^4} \, dx=2\,\ln \left (x^3+256\,x^2+160\,x+25\right )+{\mathrm {e}}^x-4\,\ln \left (x\right ) \]

[In]

int(-(320*x - exp(x)*(25*x + 160*x^2 + 256*x^3 + x^4) - 2*x^3 + 100)/(25*x + 160*x^2 + 256*x^3 + x^4),x)

[Out]

2*log(160*x + 256*x^2 + x^3 + 25) + exp(x) - 4*log(x)