3.84.37 \(\int \frac {e^3 (14-2 x)+14 x-2 x^2+(-7 x+2 x^2+2 x^3) \log (\frac {-7+2 x+2 x^2}{20 x^2})}{-7 x+2 x^2+2 x^3} \, dx\)

Optimal. Leaf size=22 \[ \left (e^3+x\right ) \log \left (\frac {-\frac {7}{2}+x+x^2}{10 x^2}\right ) \]

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Rubi [B]  time = 0.33, antiderivative size = 134, normalized size of antiderivative = 6.09, number of steps used = 13, number of rules used = 7, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1594, 6688, 1628, 632, 31, 2523, 12} \begin {gather*} x \log \left (-\frac {-2 x^2-2 x+7}{20 x^2}\right )-2 e^3 \log (x)-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (2 x-\sqrt {15}+1\right )+\frac {1}{2} \left (1-\sqrt {15}\right ) \log \left (2 x-\sqrt {15}+1\right )-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (2 x+\sqrt {15}+1\right )+\frac {1}{2} \left (1+\sqrt {15}\right ) \log \left (2 x+\sqrt {15}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(14 - 2*x) + 14*x - 2*x^2 + (-7*x + 2*x^2 + 2*x^3)*Log[(-7 + 2*x + 2*x^2)/(20*x^2)])/(-7*x + 2*x^2 +
2*x^3),x]

[Out]

-2*E^3*Log[x] + ((1 - Sqrt[15])*Log[1 - Sqrt[15] + 2*x])/2 - ((1 - Sqrt[15] - 2*E^3)*Log[1 - Sqrt[15] + 2*x])/
2 + ((1 + Sqrt[15])*Log[1 + Sqrt[15] + 2*x])/2 - ((1 + Sqrt[15] - 2*E^3)*Log[1 + Sqrt[15] + 2*x])/2 + x*Log[-1
/20*(7 - 2*x - 2*x^2)/x^2]

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 (14-2 x)+14 x-2 x^2+\left (-7 x+2 x^2+2 x^3\right ) \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )}{x \left (-7+2 x+2 x^2\right )} \, dx\\ &=\int \left (-\frac {2 (-7+x) \left (e^3+x\right )}{x \left (-7+2 x+2 x^2\right )}+\log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right )\right ) \, dx\\ &=-\left (2 \int \frac {(-7+x) \left (e^3+x\right )}{x \left (-7+2 x+2 x^2\right )} \, dx\right )+\int \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right ) \, dx\\ &=x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-2 \int \left (\frac {e^3}{x}+\frac {7+e^3-\left (1-2 e^3\right ) x}{7-2 x-2 x^2}\right ) \, dx-\int \frac {2 (-7+x)}{7-2 x-2 x^2} \, dx\\ &=-2 e^3 \log (x)+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-2 \int \frac {-7+x}{7-2 x-2 x^2} \, dx-2 \int \frac {7+e^3-\left (1-2 e^3\right ) x}{7-2 x-2 x^2} \, dx\\ &=-2 e^3 \log (x)+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )-\left (1-\sqrt {15}\right ) \int \frac {1}{-1+\sqrt {15}-2 x} \, dx-\left (1+\sqrt {15}\right ) \int \frac {1}{-1-\sqrt {15}-2 x} \, dx-\left (-1-\sqrt {15}+2 e^3\right ) \int \frac {1}{-1-\sqrt {15}-2 x} \, dx-\left (-1+\sqrt {15}+2 e^3\right ) \int \frac {1}{-1+\sqrt {15}-2 x} \, dx\\ &=-2 e^3 \log (x)+\frac {1}{2} \left (1-\sqrt {15}\right ) \log \left (1-\sqrt {15}+2 x\right )-\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (1-\sqrt {15}+2 x\right )+\frac {1}{2} \left (1+\sqrt {15}\right ) \log \left (1+\sqrt {15}+2 x\right )-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (1+\sqrt {15}+2 x\right )+x \log \left (-\frac {7-2 x-2 x^2}{20 x^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 135, normalized size = 6.14 \begin {gather*} -\frac {1}{2} \left (1-\sqrt {15}-2 e^3\right ) \log \left (-1+\sqrt {15}-2 x\right )-2 e^3 \log (x)-\frac {1}{2} \left (1+\sqrt {15}-2 e^3\right ) \log \left (1+\sqrt {15}+2 x\right )-2 \left (-\frac {1}{4} \left (1-\sqrt {15}\right ) \log \left (1-\sqrt {15}+2 x\right )-\frac {1}{4} \left (1+\sqrt {15}\right ) \log \left (1+\sqrt {15}+2 x\right )\right )+x \log \left (\frac {-7+2 x+2 x^2}{20 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(14 - 2*x) + 14*x - 2*x^2 + (-7*x + 2*x^2 + 2*x^3)*Log[(-7 + 2*x + 2*x^2)/(20*x^2)])/(-7*x + 2*
x^2 + 2*x^3),x]

[Out]

-1/2*((1 - Sqrt[15] - 2*E^3)*Log[-1 + Sqrt[15] - 2*x]) - 2*E^3*Log[x] - ((1 + Sqrt[15] - 2*E^3)*Log[1 + Sqrt[1
5] + 2*x])/2 - 2*(-1/4*((1 - Sqrt[15])*Log[1 - Sqrt[15] + 2*x]) - ((1 + Sqrt[15])*Log[1 + Sqrt[15] + 2*x])/4)
+ x*Log[(-7 + 2*x + 2*x^2)/(20*x^2)]

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fricas [A]  time = 0.51, size = 21, normalized size = 0.95 \begin {gather*} {\left (x + e^{3}\right )} \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2-7*x)*log(1/20*(2*x^2+2*x-7)/x^2)+(-2*x+14)*exp(3)-2*x^2+14*x)/(2*x^3+2*x^2-7*x),x, alg
orithm="fricas")

[Out]

(x + e^3)*log(1/20*(2*x^2 + 2*x - 7)/x^2)

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giac [A]  time = 0.22, size = 39, normalized size = 1.77 \begin {gather*} e^{3} \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, e^{3} \log \relax (x) + x \log \left (\frac {2 \, x^{2} + 2 \, x - 7}{20 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2-7*x)*log(1/20*(2*x^2+2*x-7)/x^2)+(-2*x+14)*exp(3)-2*x^2+14*x)/(2*x^3+2*x^2-7*x),x, alg
orithm="giac")

[Out]

e^3*log(2*x^2 + 2*x - 7) - 2*e^3*log(x) + x*log(1/20*(2*x^2 + 2*x - 7)/x^2)

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maple [B]  time = 0.13, size = 38, normalized size = 1.73

method	result	size
derivativedivides	\(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )-x \ln \left (20\right )+\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right ) {\mathrm e}^{3}\)	\(38\)
default	\(x \ln \left (-\frac {7}{x^{2}}+\frac {2}{x}+2\right )-x \ln \left (20\right )+\ln \left (\frac {7}{x^{2}}-\frac {2}{x}-2\right ) {\mathrm e}^{3}\)	\(38\)
norman	\({\mathrm e}^{3} \ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right )+\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x\)	\(39\)
risch	\(\ln \left (\frac {2 x^{2}+2 x -7}{20 x^{2}}\right ) x -2 \,{\mathrm e}^{3} \ln \left (-x \right )+{\mathrm e}^{3} \ln \left (2 x^{2}+2 x -7\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+2*x^2-7*x)*ln(1/20*(2*x^2+2*x-7)/x^2)+(-2*x+14)*exp(3)-2*x^2+14*x)/(2*x^3+2*x^2-7*x),x,method=_RET
URNVERBOSE)

[Out]

x*ln(-7/x^2+2/x+2)-x*ln(20)+ln(7/x^2-2/x-2)*exp(3)

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maxima [B]  time = 0.54, size = 106, normalized size = 4.82 \begin {gather*} -\frac {1}{15} \, \sqrt {15} e^{3} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) - x {\left (\log \relax (5) + 2 \, \log \relax (2)\right )} + \frac {1}{15} \, {\left (\sqrt {15} \log \left (\frac {2 \, x - \sqrt {15} + 1}{2 \, x + \sqrt {15} + 1}\right ) + 15 \, \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 30 \, \log \relax (x)\right )} e^{3} + x \log \left (2 \, x^{2} + 2 \, x - 7\right ) - 2 \, x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+2*x^2-7*x)*log(1/20*(2*x^2+2*x-7)/x^2)+(-2*x+14)*exp(3)-2*x^2+14*x)/(2*x^3+2*x^2-7*x),x, alg
orithm="maxima")

[Out]

-1/15*sqrt(15)*e^3*log((2*x - sqrt(15) + 1)/(2*x + sqrt(15) + 1)) - x*(log(5) + 2*log(2)) + 1/15*(sqrt(15)*log
((2*x - sqrt(15) + 1)/(2*x + sqrt(15) + 1)) + 15*log(2*x^2 + 2*x - 7) - 30*log(x))*e^3 + x*log(2*x^2 + 2*x - 7
) - 2*x*log(x)

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mupad [B]  time = 5.33, size = 41, normalized size = 1.86 \begin {gather*} x\,\ln \left (\frac {1}{x^2}\right )-x\,\ln \left (20\right )-2\,{\mathrm {e}}^3\,\ln \relax (x)+x\,\ln \left (2\,x^2+2\,x-7\right )+\ln \left (x^2+x-\frac {7}{2}\right )\,{\mathrm {e}}^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((14*x + log((x/10 + x^2/10 - 7/20)/x^2)*(2*x^2 - 7*x + 2*x^3) - 2*x^2 - exp(3)*(2*x - 14))/(2*x^2 - 7*x +
2*x^3),x)

[Out]

x*log(1/x^2) - x*log(20) - 2*exp(3)*log(x) + x*log(2*x + 2*x^2 - 7) + log(x + x^2 - 7/2)*exp(3)

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sympy [B]  time = 0.31, size = 41, normalized size = 1.86 \begin {gather*} x \log {\left (\frac {\frac {x^{2}}{10} + \frac {x}{10} - \frac {7}{20}}{x^{2}} \right )} - 2 e^{3} \log {\relax (x )} + e^{3} \log {\left (2 x^{2} + 2 x - 7 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+2*x**2-7*x)*ln(1/20*(2*x**2+2*x-7)/x**2)+(-2*x+14)*exp(3)-2*x**2+14*x)/(2*x**3+2*x**2-7*x),
x)

[Out]

x*log((x**2/10 + x/10 - 7/20)/x**2) - 2*exp(3)*log(x) + exp(3)*log(2*x**2 + 2*x - 7)

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