3.26.57 \(\int \frac {1-3 x+x \log ^2(3)+e^x (25-50 x+25 x \log ^2(3))+(-2-50 e^x) \log (x+25 e^x x)}{x^3+25 e^x x^3} \, dx\)

Optimal. Leaf size=27 \[ -\frac {\log ^2(3)}{x}+\frac {3 x+\log \left (x+25 e^x x\right )}{x^2} \]

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Rubi [A]  time = 0.54, antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 10, number of rules used = 5, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {6, 6742, 14, 37, 2551} \begin {gather*} \frac {(x+1)^2}{2 x^2}-\frac {\left (1-x \left (2-\log ^2(3)\right )\right )^2}{2 x^2}+\frac {\log \left (25 e^x x+x\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 3*x + x*Log[3]^2 + E^x*(25 - 50*x + 25*x*Log[3]^2) + (-2 - 50*E^x)*Log[x + 25*E^x*x])/(x^3 + 25*E^x*x
^3),x]

[Out]

(1 + x)^2/(2*x^2) - (1 - x*(2 - Log[3]^2))^2/(2*x^2) + Log[x + 25*E^x*x]/x^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

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 {1+x \left (-3+\log ^2(3)\right )+e^x \left (25-50 x+25 x \log ^2(3)\right )+\left (-2-50 e^x\right ) \log \left (x+25 e^x x\right )}{x^3+25 e^x x^3} \, dx\\ &=\int \left (-\frac {1}{\left (1+25 e^x\right ) x^2}+\frac {1-2 x \left (1-\frac {\log ^2(3)}{2}\right )-2 \log \left (x+25 e^x x\right )}{x^3}\right ) \, dx\\ &=-\int \frac {1}{\left (1+25 e^x\right ) x^2} \, dx+\int \frac {1-2 x \left (1-\frac {\log ^2(3)}{2}\right )-2 \log \left (x+25 e^x x\right )}{x^3} \, dx\\ &=-\int \frac {1}{\left (1+25 e^x\right ) x^2} \, dx+\int \left (\frac {1-x \left (2-\log ^2(3)\right )}{x^3}-\frac {2 \log \left (x+25 e^x x\right )}{x^3}\right ) \, dx\\ &=-\left (2 \int \frac {\log \left (x+25 e^x x\right )}{x^3} \, dx\right )-\int \frac {1}{\left (1+25 e^x\right ) x^2} \, dx+\int \frac {1-x \left (2-\log ^2(3)\right )}{x^3} \, dx\\ &=-\frac {\left (1-x \left (2-\log ^2(3)\right )\right )^2}{2 x^2}+\frac {\log \left (x+25 e^x x\right )}{x^2}-\int \frac {1}{\left (1+25 e^x\right ) x^2} \, dx-\int \frac {1+25 e^x (1+x)}{\left (1+25 e^x\right ) x^3} \, dx\\ &=-\frac {\left (1-x \left (2-\log ^2(3)\right )\right )^2}{2 x^2}+\frac {\log \left (x+25 e^x x\right )}{x^2}-\int \frac {1}{\left (1+25 e^x\right ) x^2} \, dx-\int \left (-\frac {1}{\left (1+25 e^x\right ) x^2}+\frac {1+x}{x^3}\right ) \, dx\\ &=-\frac {\left (1-x \left (2-\log ^2(3)\right )\right )^2}{2 x^2}+\frac {\log \left (x+25 e^x x\right )}{x^2}-\int \frac {1+x}{x^3} \, dx\\ &=\frac {(1+x)^2}{2 x^2}-\frac {\left (1-x \left (2-\log ^2(3)\right )\right )^2}{2 x^2}+\frac {\log \left (x+25 e^x x\right )}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 28, normalized size = 1.04 \begin {gather*} \frac {3}{x}-\frac {\log ^2(3)}{x}+\frac {\log \left (x+25 e^x x\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 3*x + x*Log[3]^2 + E^x*(25 - 50*x + 25*x*Log[3]^2) + (-2 - 50*E^x)*Log[x + 25*E^x*x])/(x^3 + 25
*E^x*x^3),x]

[Out]

3/x - Log[3]^2/x + Log[x + 25*E^x*x]/x^2

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fricas [A]  time = 0.79, size = 25, normalized size = 0.93 \begin {gather*} -\frac {x \log \relax (3)^{2} - 3 \, x - \log \left (25 \, x e^{x} + x\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*exp(x)-2)*log(25*exp(x)*x+x)+(25*x*log(3)^2-50*x+25)*exp(x)+x*log(3)^2-3*x+1)/(25*exp(x)*x^3+x
^3),x, algorithm="fricas")

[Out]

-(x*log(3)^2 - 3*x - log(25*x*e^x + x))/x^2

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giac [A]  time = 0.34, size = 25, normalized size = 0.93 \begin {gather*} -\frac {x \log \relax (3)^{2} - 3 \, x - \log \left (25 \, x e^{x} + x\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*exp(x)-2)*log(25*exp(x)*x+x)+(25*x*log(3)^2-50*x+25)*exp(x)+x*log(3)^2-3*x+1)/(25*exp(x)*x^3+x
^3),x, algorithm="giac")

[Out]

-(x*log(3)^2 - 3*x - log(25*x*e^x + x))/x^2

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maple [A]  time = 0.14, size = 24, normalized size = 0.89

method	result	size
norman	\(\frac {\left (-\ln \relax (3)^{2}+3\right ) x +\ln \left (25 \,{\mathrm e}^{x} x +x \right )}{x^{2}}\)	\(24\)
risch	\(\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{25}\right )}{x^{2}}+\frac {-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right )^{2}-i \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{x}+\frac {1}{25}\right )\right )^{3}-2 x \ln \relax (3)^{2}+6 x +2 \ln \relax (x )}{2 x^{2}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-50*exp(x)-2)*ln(25*exp(x)*x+x)+(25*x*ln(3)^2-50*x+25)*exp(x)+x*ln(3)^2-3*x+1)/(25*exp(x)*x^3+x^3),x,met
hod=_RETURNVERBOSE)

[Out]

((-ln(3)^2+3)*x+ln(25*exp(x)*x+x))/x^2

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maxima [A]  time = 0.91, size = 27, normalized size = 1.00 \begin {gather*} -\frac {{\left (\log \relax (3)^{2} - 3\right )} x - \log \relax (x) - \log \left (25 \, e^{x} + 1\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*exp(x)-2)*log(25*exp(x)*x+x)+(25*x*log(3)^2-50*x+25)*exp(x)+x*log(3)^2-3*x+1)/(25*exp(x)*x^3+x
^3),x, algorithm="maxima")

[Out]

-((log(3)^2 - 3)*x - log(x) - log(25*e^x + 1))/x^2

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mupad [B]  time = 1.56, size = 22, normalized size = 0.81 \begin {gather*} \frac {\ln \left (x+25\,x\,{\mathrm {e}}^x\right )-x\,\left ({\ln \relax (3)}^2-3\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(25*x*log(3)^2 - 50*x + 25) - 3*x + x*log(3)^2 - log(x + 25*x*exp(x))*(50*exp(x) + 2) + 1)/(25*x^3
*exp(x) + x^3),x)

[Out]

(log(x + 25*x*exp(x)) - x*(log(3)^2 - 3))/x^2

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sympy [A]  time = 0.22, size = 20, normalized size = 0.74 \begin {gather*} - \frac {-3 + \log {\relax (3 )}^{2}}{x} + \frac {\log {\left (25 x e^{x} + x \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*exp(x)-2)*ln(25*exp(x)*x+x)+(25*x*ln(3)**2-50*x+25)*exp(x)+x*ln(3)**2-3*x+1)/(25*exp(x)*x**3+x
**3),x)

[Out]

-(-3 + log(3)**2)/x + log(25*x*exp(x) + x)/x**2

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