3.30.87 \(\int \frac {-25-2 x^3+(25-95 x-x^3) \log (5)+(-25+95 x+x^3) \log (\frac {-25+95 x+x^3}{5 x})}{-75 x^2+285 x^3+3 x^5} \, dx\)

Optimal. Leaf size=31 \[ \frac {x+\log (5)-\log \left (20+\frac {x^2}{5}-\frac {5+x}{x}\right )}{3 x} \]

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Rubi [A]  time = 9.28, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 24, number of rules used = 10, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1594, 6742, 2100, 2079, 800, 634, 618, 204, 628, 2525} \begin {gather*} -\frac {\log \left (-\frac {-x^3-95 x+25}{5 x}\right )}{3 x}+\frac {1}{3 x}-\frac {1-\log (5)}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 2*x^3 + (25 - 95*x - x^3)*Log[5] + (-25 + 95*x + x^3)*Log[(-25 + 95*x + x^3)/(5*x)])/(-75*x^2 + 285
*x^3 + 3*x^5),x]

[Out]

1/(3*x) - (1 - Log[5])/(3*x) - Log[-1/5*(25 - 95*x - x^3)/x]/(3*x)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), 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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 2079

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> With[{r = Rt[-9*a*d^2 + S
qrt[3]*d*Sqrt[4*b^3*d + 27*a^2*d^2], 3]}, Dist[1/d^(2*p), Int[(e + f*x)^m*Simp[(18^(1/3)*b*d)/(3*r) - r/18^(1/
3) + d*x, x]^p*Simp[(b*d)/3 + (12^(1/3)*b^2*d^2)/(3*r^2) + r^2/(3*12^(1/3)) - d*((2^(1/3)*b*d)/(3^(1/3)*r) - r
/18^(1/3))*x + d^2*x^2, x]^p, x], x]] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[4*b^3 + 27*a^2*d, 0] && ILtQ[p, 0
]

Rule 2100

Int[(Pm_)/(Qn_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*Log[Qn])/(n*Coe
ff[Qn, x, n]), x] + Dist[1/(n*Coeff[Qn, x, n]), Int[ExpandToSum[n*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*D[Qn, x
], x]/Qn, x], x] /; EqQ[m, n - 1]] /; PolyQ[Pm, x] && PolyQ[Qn, x]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && 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 {-25-2 x^3+\left (25-95 x-x^3\right ) \log (5)+\left (-25+95 x+x^3\right ) \log \left (\frac {-25+95 x+x^3}{5 x}\right )}{x^2 \left (-75+285 x+3 x^3\right )} \, dx\\ &=\int \left (\frac {25 (1-\log (5))+95 x \log (5)+x^3 (2+\log (5))}{3 x^2 \left (25-95 x-x^3\right )}+\frac {\log \left (\frac {-25+95 x+x^3}{5 x}\right )}{3 x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {25 (1-\log (5))+95 x \log (5)+x^3 (2+\log (5))}{x^2 \left (25-95 x-x^3\right )} \, dx+\frac {1}{3} \int \frac {\log \left (\frac {-25+95 x+x^3}{5 x}\right )}{x^2} \, dx\\ &=-\frac {\log \left (-\frac {25-95 x-x^3}{5 x}\right )}{3 x}+\frac {1}{3} \int \frac {-25-2 x^3}{x^2 \left (25-95 x-x^3\right )} \, dx+\frac {1}{3} \int \left (\frac {19}{5 x}+\frac {-1805-15 x-19 x^2}{5 \left (-25+95 x+x^3\right )}+\frac {1-\log (5)}{x^2}\right ) \, dx\\ &=-\frac {1-\log (5)}{3 x}+\frac {19 \log (x)}{15}-\frac {\log \left (-\frac {25-95 x-x^3}{5 x}\right )}{3 x}+\frac {1}{15} \int \frac {-1805-15 x-19 x^2}{-25+95 x+x^3} \, dx+\frac {1}{3} \int \left (-\frac {1}{x^2}-\frac {19}{5 x}+\frac {1805+15 x+19 x^2}{5 \left (-25+95 x+x^3\right )}\right ) \, dx\\ &=\frac {1}{3 x}-\frac {1-\log (5)}{3 x}-\frac {\log \left (-\frac {25-95 x-x^3}{5 x}\right )}{3 x}-\frac {19}{45} \log \left (-25+95 x+x^3\right )+\frac {1}{45} \int \frac {-3610-45 x}{-25+95 x+x^3} \, dx+\frac {1}{15} \int \frac {1805+15 x+19 x^2}{-25+95 x+x^3} \, dx\\ &=\frac {1}{3 x}-\frac {1-\log (5)}{3 x}-\frac {\log \left (-\frac {25-95 x-x^3}{5 x}\right )}{3 x}+\frac {1}{45} \int \frac {-3610-45 x}{\left (\frac {\sqrt [3]{5} \left (38 \sqrt [3]{\frac {15}{45+\sqrt {413565}}}-\sqrt [3]{2 \left (45+\sqrt {413565}\right )}\right )}{6^{2/3}}+x\right ) \left (\frac {1}{18} \left (570+10830 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}+\sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}\right )-\frac {\left (19\ 5^{2/3} \sqrt [3]{\frac {6}{45+\sqrt {413565}}}-\sqrt [3]{\frac {5}{2} \left (45+\sqrt {413565}\right )}\right ) x}{3^{2/3}}+x^2\right )} \, dx+\frac {1}{45} \int \frac {3610+45 x}{-25+95 x+x^3} \, dx\\ &=\frac {1}{3 x}-\frac {1-\log (5)}{3 x}-\frac {\log \left (-\frac {25-95 x-x^3}{5 x}\right )}{3 x}+\frac {1}{45} \int \frac {3610+45 x}{\left (\frac {\sqrt [3]{5} \left (38 \sqrt [3]{\frac {15}{45+\sqrt {413565}}}-\sqrt [3]{2 \left (45+\sqrt {413565}\right )}\right )}{6^{2/3}}+x\right ) \left (\frac {1}{18} \left (570+10830 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}+\sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}\right )-\frac {\left (19\ 5^{2/3} \sqrt [3]{\frac {6}{45+\sqrt {413565}}}-\sqrt [3]{\frac {5}{2} \left (45+\sqrt {413565}\right )}\right ) x}{3^{2/3}}+x^2\right )} \, dx+\frac {1}{45} \int \left (\frac {90 \left (-1444+114\ 15^{2/3} \sqrt [3]{\frac {2}{45+\sqrt {413565}}}-3\ 2^{2/3} \sqrt [3]{15 \left (45+\sqrt {413565}\right )}\right )}{\left (-570+10830 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}+\sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}\right ) \left (38\ 15^{2/3} \sqrt [3]{\frac {2}{45+\sqrt {413565}}}-2^{2/3} \sqrt [3]{15 \left (45+\sqrt {413565}\right )}+6 x\right )}+\frac {90 \left (1710+54872\ 15^{2/3} \sqrt [3]{\frac {2}{45+\sqrt {413565}}}+32490 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}-1444\ 2^{2/3} \sqrt [3]{15 \left (45+\sqrt {413565}\right )}+3 \sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}-3 \left (1444-2^{2/3} \left (\frac {57\ 30^{2/3}}{\sqrt [3]{45+\sqrt {413565}}}-3 \sqrt [3]{15 \left (45+\sqrt {413565}\right )}\right )\right ) x\right )}{\left (570-10830 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}-\sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}\right ) \left (570+10830 \sqrt [3]{15} \left (\frac {2}{45+\sqrt {413565}}\right )^{2/3}+\sqrt [3]{2} \left (15 \left (45+\sqrt {413565}\right )\right )^{2/3}-3\ 2^{2/3} \left (\frac {19\ 30^{2/3}}{\sqrt [3]{45+\sqrt {413565}}}-\sqrt [3]{15 \left (45+\sqrt {413565}\right )}\right ) x+18 x^2\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 28, normalized size = 0.90 \begin {gather*} \frac {1}{3} \left (\frac {2 \log (5)}{x}-\frac {\log \left (95-\frac {25}{x}+x^2\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 2*x^3 + (25 - 95*x - x^3)*Log[5] + (-25 + 95*x + x^3)*Log[(-25 + 95*x + x^3)/(5*x)])/(-75*x^2
+ 285*x^3 + 3*x^5),x]

[Out]

((2*Log[5])/x - Log[95 - 25/x + x^2]/x)/3

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fricas [A]  time = 0.67, size = 24, normalized size = 0.77 \begin {gather*} \frac {\log \relax (5) - \log \left (\frac {x^{3} + 95 \, x - 25}{5 \, x}\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+95*x-25)*log(1/5*(x^3+95*x-25)/x)+(-x^3-95*x+25)*log(5)-2*x^3-25)/(3*x^5+285*x^3-75*x^2),x, al
gorithm="fricas")

[Out]

1/3*(log(5) - log(1/5*(x^3 + 95*x - 25)/x))/x

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giac [A]  time = 0.18, size = 29, normalized size = 0.94 \begin {gather*} \frac {2 \, \log \relax (5)}{3 \, x} - \frac {\log \left (x^{3} + 95 \, x - 25\right )}{3 \, x} + \frac {\log \relax (x)}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+95*x-25)*log(1/5*(x^3+95*x-25)/x)+(-x^3-95*x+25)*log(5)-2*x^3-25)/(3*x^5+285*x^3-75*x^2),x, al
gorithm="giac")

[Out]

2/3*log(5)/x - 1/3*log(x^3 + 95*x - 25)/x + 1/3*log(x)/x

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maple [A]  time = 0.08, size = 26, normalized size = 0.84

method	result	size
norman	\(\frac {\frac {\ln \relax (5)}{3}-\frac {\ln \left (\frac {x^{3}+95 x -25}{5 x}\right )}{3}}{x}\)	\(26\)
default	\(-\frac {\ln \left (\frac {x^{3}+95 x -25}{x}\right )}{3 x}+\frac {2 \ln \relax (5)}{3 x}\)	\(27\)
risch	\(-\frac {\ln \left (\frac {x^{3}+95 x -25}{5 x}\right )}{3 x}+\frac {\ln \relax (5)}{3 x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+95*x-25)*ln(1/5*(x^3+95*x-25)/x)+(-x^3-95*x+25)*ln(5)-2*x^3-25)/(3*x^5+285*x^3-75*x^2),x,method=_RET
URNVERBOSE)

[Out]

(1/3*ln(5)-1/3*ln(1/5*(x^3+95*x-25)/x))/x

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maxima [A]  time = 0.70, size = 23, normalized size = 0.74 \begin {gather*} \frac {2 \, \log \relax (5) - \log \left (x^{3} + 95 \, x - 25\right ) + \log \relax (x)}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+95*x-25)*log(1/5*(x^3+95*x-25)/x)+(-x^3-95*x+25)*log(5)-2*x^3-25)/(3*x^5+285*x^3-75*x^2),x, al
gorithm="maxima")

[Out]

1/3*(2*log(5) - log(x^3 + 95*x - 25) + log(x))/x

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mupad [B]  time = 1.92, size = 25, normalized size = 0.81 \begin {gather*} \frac {2\,\ln \relax (5)-\ln \left (\frac {x^3+95\,x-25}{x}\right )}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3 - log((19*x + x^3/5 - 5)/x)*(95*x + x^3 - 25) + log(5)*(95*x + x^3 - 25) + 25)/(285*x^3 - 75*x^2 +
3*x^5),x)

[Out]

(2*log(5) - log((95*x + x^3 - 25)/x))/(3*x)

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sympy [A]  time = 0.20, size = 22, normalized size = 0.71 \begin {gather*} - \frac {\log {\left (\frac {\frac {x^{3}}{5} + 19 x - 5}{x} \right )}}{3 x} + \frac {\log {\relax (5 )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+95*x-25)*ln(1/5*(x**3+95*x-25)/x)+(-x**3-95*x+25)*ln(5)-2*x**3-25)/(3*x**5+285*x**3-75*x**2),
x)

[Out]

-log((x**3/5 + 19*x - 5)/x)/(3*x) + log(5)/(3*x)

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