3.51.97 \(\int \frac {-48 x-48 x^2-192 x^3-152 x^4-62 x^5-4 x^6+6 x^7+(-16-16 x+4 x^2+4 x^3-x^4) \log (5)}{16 x+16 x^2-4 x^3-4 x^4+x^5} \, dx\)

Optimal. Leaf size=34 \[ \left (1+x^2\right ) \left (x-\frac {(4+x)^2}{2+\frac {4}{x}-x}\right )-\log (5) \log (x) \]

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Rubi [A]  time = 0.17, antiderivative size = 39, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 4, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {2074, 638, 618, 206} \begin {gather*} 2 x^3+10 x^2-\frac {40 (11 x+13)}{-x^2+2 x+4}+42 x-\log (5) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48*x - 48*x^2 - 192*x^3 - 152*x^4 - 62*x^5 - 4*x^6 + 6*x^7 + (-16 - 16*x + 4*x^2 + 4*x^3 - x^4)*Log[5])/
(16*x + 16*x^2 - 4*x^3 - 4*x^4 + x^5),x]

[Out]

42*x + 10*x^2 + 2*x^3 - (40*(13 + 11*x))/(4 + 2*x - x^2) - Log[5]*Log[x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (42+20 x+6 x^2-\frac {80 (31+24 x)}{\left (-4-2 x+x^2\right )^2}-\frac {440}{-4-2 x+x^2}-\frac {\log (5)}{x}\right ) \, dx\\ &=42 x+10 x^2+2 x^3-\log (5) \log (x)-80 \int \frac {31+24 x}{\left (-4-2 x+x^2\right )^2} \, dx-440 \int \frac {1}{-4-2 x+x^2} \, dx\\ &=42 x+10 x^2+2 x^3-\frac {40 (13+11 x)}{4+2 x-x^2}-\log (5) \log (x)+440 \int \frac {1}{-4-2 x+x^2} \, dx+880 \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,-2+2 x\right )\\ &=42 x+10 x^2+2 x^3-\frac {40 (13+11 x)}{4+2 x-x^2}-88 \sqrt {5} \tanh ^{-1}\left (\frac {1-x}{\sqrt {5}}\right )-\log (5) \log (x)-880 \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,-2+2 x\right )\\ &=42 x+10 x^2+2 x^3-\frac {40 (13+11 x)}{4+2 x-x^2}-\log (5) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 37, normalized size = 1.09 \begin {gather*} 42 x+10 x^2+2 x^3+\frac {40 (13+11 x)}{-4-2 x+x^2}-\log (5) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48*x - 48*x^2 - 192*x^3 - 152*x^4 - 62*x^5 - 4*x^6 + 6*x^7 + (-16 - 16*x + 4*x^2 + 4*x^3 - x^4)*Lo
g[5])/(16*x + 16*x^2 - 4*x^3 - 4*x^4 + x^5),x]

[Out]

42*x + 10*x^2 + 2*x^3 + (40*(13 + 11*x))/(-4 - 2*x + x^2) - Log[5]*Log[x]

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fricas [A]  time = 0.84, size = 50, normalized size = 1.47 \begin {gather*} \frac {2 \, x^{5} + 6 \, x^{4} + 14 \, x^{3} - {\left (x^{2} - 2 \, x - 4\right )} \log \relax (5) \log \relax (x) - 124 \, x^{2} + 272 \, x + 520}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+4*x^3+4*x^2-16*x-16)*log(5)+6*x^7-4*x^6-62*x^5-152*x^4-192*x^3-48*x^2-48*x)/(x^5-4*x^4-4*x^3+
16*x^2+16*x),x, algorithm="fricas")

[Out]

(2*x^5 + 6*x^4 + 14*x^3 - (x^2 - 2*x - 4)*log(5)*log(x) - 124*x^2 + 272*x + 520)/(x^2 - 2*x - 4)

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giac [A]  time = 0.18, size = 38, normalized size = 1.12 \begin {gather*} 2 \, x^{3} + 10 \, x^{2} - \log \relax (5) \log \left ({\left | x \right |}\right ) + 42 \, x + \frac {40 \, {\left (11 \, x + 13\right )}}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+4*x^3+4*x^2-16*x-16)*log(5)+6*x^7-4*x^6-62*x^5-152*x^4-192*x^3-48*x^2-48*x)/(x^5-4*x^4-4*x^3+
16*x^2+16*x),x, algorithm="giac")

[Out]

2*x^3 + 10*x^2 - log(5)*log(abs(x)) + 42*x + 40*(11*x + 13)/(x^2 - 2*x - 4)

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maple [A]  time = 0.08, size = 37, normalized size = 1.09

method	result	size
risch	\(2 x^{3}+10 x^{2}+42 x +\frac {440 x +520}{x^{2}-2 x -4}-\ln \relax (5) \ln \relax (x )\)	\(37\)
default	\(2 x^{3}+10 x^{2}+42 x -\ln \relax (5) \ln \relax (x )-\frac {40 \left (-11 x -13\right )}{x^{2}-2 x -4}\)	\(38\)
norman	\(\frac {2 x^{5}+6 x^{4}+14 x^{3}+24 x +24}{x^{2}-2 x -4}-\ln \relax (5) \ln \relax (x )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4+4*x^3+4*x^2-16*x-16)*ln(5)+6*x^7-4*x^6-62*x^5-152*x^4-192*x^3-48*x^2-48*x)/(x^5-4*x^4-4*x^3+16*x^2+
16*x),x,method=_RETURNVERBOSE)

[Out]

2*x^3+10*x^2+42*x+(440*x+520)/(x^2-2*x-4)-ln(5)*ln(x)

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maxima [A]  time = 0.34, size = 37, normalized size = 1.09 \begin {gather*} 2 \, x^{3} + 10 \, x^{2} - \log \relax (5) \log \relax (x) + 42 \, x + \frac {40 \, {\left (11 \, x + 13\right )}}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4+4*x^3+4*x^2-16*x-16)*log(5)+6*x^7-4*x^6-62*x^5-152*x^4-192*x^3-48*x^2-48*x)/(x^5-4*x^4-4*x^3+
16*x^2+16*x),x, algorithm="maxima")

[Out]

2*x^3 + 10*x^2 - log(5)*log(x) + 42*x + 40*(11*x + 13)/(x^2 - 2*x - 4)

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mupad [B]  time = 3.37, size = 325, normalized size = 9.56 \begin {gather*} 42\,x-\ln \relax (5)\,\ln \relax (x)+10\,x^2+2\,x^3-\frac {\frac {16\,\ln \relax (5)}{5}-\frac {4\,\ln \left (625\right )}{5}+x\,\left (\frac {14\,\ln \relax (5)}{5}-\frac {7\,\ln \left (625\right )}{10}+440\right )+520}{-x^2+2\,x+4}-\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\frac {\sqrt {5}\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )\,\left (\frac {64\,\ln \relax (5)}{5}-\frac {6\,\ln \left (625\right )}{5}+x\,\left (\frac {72\,\ln \relax (5)}{5}-\frac {3\,\ln \left (625\right )}{5}\right )-\frac {3\,\sqrt {5}\,\left (32\,x+8\right )\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )}{100}\right )\,3{}\mathrm {i}}{100}+\frac {\sqrt {5}\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )\,\left (\frac {64\,\ln \relax (5)}{5}-\frac {6\,\ln \left (625\right )}{5}+x\,\left (\frac {72\,\ln \relax (5)}{5}-\frac {3\,\ln \left (625\right )}{5}\right )+\frac {3\,\sqrt {5}\,\left (32\,x+8\right )\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )}{100}\right )\,3{}\mathrm {i}}{100}}{2\,x\,\left (\frac {3\,\ln \relax (5)\,\ln \left (625\right )}{25}+\frac {24\,{\ln \relax (5)}^2}{25}-\frac {9\,{\ln \left (625\right )}^2}{100}\right )-\frac {12\,\ln \relax (5)\,\ln \left (625\right )}{5}+\frac {48\,{\ln \relax (5)}^2}{5}-\frac {3\,\sqrt {5}\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )\,\left (\frac {64\,\ln \relax (5)}{5}-\frac {6\,\ln \left (625\right )}{5}+x\,\left (\frac {72\,\ln \relax (5)}{5}-\frac {3\,\ln \left (625\right )}{5}\right )-\frac {3\,\sqrt {5}\,\left (32\,x+8\right )\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )}{100}\right )}{100}+\frac {3\,\sqrt {5}\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )\,\left (\frac {64\,\ln \relax (5)}{5}-\frac {6\,\ln \left (625\right )}{5}+x\,\left (\frac {72\,\ln \relax (5)}{5}-\frac {3\,\ln \left (625\right )}{5}\right )+\frac {3\,\sqrt {5}\,\left (32\,x+8\right )\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )}{100}\right )}{100}}\right )\,\left (4\,\ln \relax (5)-\ln \left (625\right )\right )\,3{}\mathrm {i}}{50} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x + log(5)*(16*x - 4*x^2 - 4*x^3 + x^4 + 16) + 48*x^2 + 192*x^3 + 152*x^4 + 62*x^5 + 4*x^6 - 6*x^7)/(
16*x + 16*x^2 - 4*x^3 - 4*x^4 + x^5),x)

[Out]

42*x - log(5)*log(x) + 10*x^2 + 2*x^3 - ((16*log(5))/5 - (4*log(625))/5 + x*((14*log(5))/5 - (7*log(625))/10 +
440) + 520)/(2*x - x^2 + 4) - (5^(1/2)*atan(((5^(1/2)*(4*log(5) - log(625))*((64*log(5))/5 - (6*log(625))/5 +
x*((72*log(5))/5 - (3*log(625))/5) - (3*5^(1/2)*(32*x + 8)*(4*log(5) - log(625)))/100)*3i)/100 + (5^(1/2)*(4*
log(5) - log(625))*((64*log(5))/5 - (6*log(625))/5 + x*((72*log(5))/5 - (3*log(625))/5) + (3*5^(1/2)*(32*x + 8
)*(4*log(5) - log(625)))/100)*3i)/100)/(2*x*((3*log(5)*log(625))/25 + (24*log(5)^2)/25 - (9*log(625)^2)/100) -
(12*log(5)*log(625))/5 + (48*log(5)^2)/5 - (3*5^(1/2)*(4*log(5) - log(625))*((64*log(5))/5 - (6*log(625))/5 +
x*((72*log(5))/5 - (3*log(625))/5) - (3*5^(1/2)*(32*x + 8)*(4*log(5) - log(625)))/100))/100 + (3*5^(1/2)*(4*l
og(5) - log(625))*((64*log(5))/5 - (6*log(625))/5 + x*((72*log(5))/5 - (3*log(625))/5) + (3*5^(1/2)*(32*x + 8)
*(4*log(5) - log(625)))/100))/100))*(4*log(5) - log(625))*3i)/50

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sympy [A]  time = 0.43, size = 32, normalized size = 0.94 \begin {gather*} 2 x^{3} + 10 x^{2} + 42 x + \frac {440 x + 520}{x^{2} - 2 x - 4} - \log {\relax (5 )} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4+4*x**3+4*x**2-16*x-16)*ln(5)+6*x**7-4*x**6-62*x**5-152*x**4-192*x**3-48*x**2-48*x)/(x**5-4*x
**4-4*x**3+16*x**2+16*x),x)

[Out]

2*x**3 + 10*x**2 + 42*x + (440*x + 520)/(x**2 - 2*x - 4) - log(5)*log(x)

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