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

Optimal. Leaf size=19 \[ 11+e^x+\log \left (\left (\left (16+\frac {5}{x}\right )^2+x\right )^2\right ) \]

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Rubi [A]  time = 0.46, antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6741, 6742, 2194, 1587} \begin {gather*} 2 \log \left (x^3+256 x^2+160 x+25\right )+e^x-4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 1587

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]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

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 6741

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

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 {-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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.26 \begin {gather*} e^x-4 \log (x)+2 \log \left (25+160 x+256 x^2+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.62, size = 23, normalized size = 1.21 \begin {gather*} e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.32, size = 23, normalized size = 1.21 \begin {gather*} e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 24, normalized size = 1.26

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 23, normalized size = 1.21 \begin {gather*} e^{x} + 2 \, \log \left (x^{3} + 256 \, x^{2} + 160 \, x + 25\right ) - 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.35, size = 23, normalized size = 1.21 \begin {gather*} 2\,\ln \left (x^3+256\,x^2+160\,x+25\right )+{\mathrm {e}}^x-4\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.16, size = 24, normalized size = 1.26 \begin {gather*} e^{x} - 4 \log {\relax (x )} + 2 \log {\left (x^{3} + 256 x^{2} + 160 x + 25 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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