3.30.43 \(\int \frac {25 x^2+(15+3 x^2) \log (5)}{-100 x^2+25 x^3+(-15 x-15 x^2+3 x^3) \log (5)} \, dx\)

Optimal. Leaf size=24 \[ -3+\log \left (-4+x-\frac {5+x}{x+\frac {25 x}{3 \log (5)}}\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2074, 628} \begin {gather*} \log \left (-\left (x^2 (25+\log (125))\right )+5 x (20+\log (125))+15 \log (5)\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*x^2 + (15 + 3*x^2)*Log[5])/(-100*x^2 + 25*x^3 + (-15*x - 15*x^2 + 3*x^3)*Log[5]),x]

[Out]

-Log[x] + Log[15*Log[5] + 5*x*(20 + Log[125]) - x^2*(25 + Log[125])]

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 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 (-\frac {1}{x}+\frac {5 (20+\log (125))-2 x (25+\log (125))}{15 \log (5)+5 x (20+\log (125))-x^2 (25+\log (125))}\right ) \, dx\\ &=-\log (x)+\int \frac {5 (20+\log (125))-2 x (25+\log (125))}{15 \log (5)+x^2 (-25-\log (125))+5 x (20+\log (125))} \, dx\\ &=-\log (x)+\log \left (15 \log (5)+5 x (20+\log (125))-x^2 (25+\log (125))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.25 \begin {gather*} -\log (x)+\log \left (-100 x+25 x^2-15 \log (5)-5 x \log (125)+x^2 \log (125)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^2 + (15 + 3*x^2)*Log[5])/(-100*x^2 + 25*x^3 + (-15*x - 15*x^2 + 3*x^3)*Log[5]),x]

[Out]

-Log[x] + Log[-100*x + 25*x^2 - 15*Log[5] - 5*x*Log[125] + x^2*Log[125]]

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fricas [A]  time = 0.44, size = 27, normalized size = 1.12 \begin {gather*} \log \left (25 \, x^{2} + 3 \, {\left (x^{2} - 5 \, x - 5\right )} \log \relax (5) - 100 \, x\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+15)*log(5)+25*x^2)/((3*x^3-15*x^2-15*x)*log(5)+25*x^3-100*x^2),x, algorithm="fricas")

[Out]

log(25*x^2 + 3*(x^2 - 5*x - 5)*log(5) - 100*x) - log(x)

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giac [A]  time = 0.25, size = 33, normalized size = 1.38 \begin {gather*} \log \left ({\left | 3 \, x^{2} \log \relax (5) + 25 \, x^{2} - 15 \, x \log \relax (5) - 100 \, x - 15 \, \log \relax (5) \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+15)*log(5)+25*x^2)/((3*x^3-15*x^2-15*x)*log(5)+25*x^3-100*x^2),x, algorithm="giac")

[Out]

log(abs(3*x^2*log(5) + 25*x^2 - 15*x*log(5) - 100*x - 15*log(5))) - log(abs(x))

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maple [A]  time = 0.07, size = 30, normalized size = 1.25

method	result	size
risch	\(-\ln \relax (x )+\ln \left (\left (-3 \ln \relax (5)-25\right ) x^{2}+\left (15 \ln \relax (5)+100\right ) x +15 \ln \relax (5)\right )\)	\(30\)
default	\(-\ln \relax (x )+\ln \left (3 x^{2} \ln \relax (5)-15 x \ln \relax (5)+25 x^{2}-15 \ln \relax (5)-100 x \right )\)	\(32\)
norman	\(-\ln \relax (x )+\ln \left (3 x^{2} \ln \relax (5)-15 x \ln \relax (5)+25 x^{2}-15 \ln \relax (5)-100 x \right )\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+15)*ln(5)+25*x^2)/((3*x^3-15*x^2-15*x)*ln(5)+25*x^3-100*x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln((-3*ln(5)-25)*x^2+(15*ln(5)+100)*x+15*ln(5))

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maxima [A]  time = 0.44, size = 30, normalized size = 1.25 \begin {gather*} \log \left (x^{2} {\left (3 \, \log \relax (5) + 25\right )} - 5 \, x {\left (3 \, \log \relax (5) + 20\right )} - 15 \, \log \relax (5)\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+15)*log(5)+25*x^2)/((3*x^3-15*x^2-15*x)*log(5)+25*x^3-100*x^2),x, algorithm="maxima")

[Out]

log(x^2*(3*log(5) + 25) - 5*x*(3*log(5) + 20) - 15*log(5)) - log(x)

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mupad [B]  time = 2.00, size = 31, normalized size = 1.29 \begin {gather*} \ln \left (200\,x+30\,\ln \relax (5)+30\,x\,\ln \relax (5)-6\,x^2\,\ln \relax (5)-50\,x^2\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(3*x^2 + 15) + 25*x^2)/(log(5)*(15*x + 15*x^2 - 3*x^3) + 100*x^2 - 25*x^3),x)

[Out]

log(200*x + 30*log(5) + 30*x*log(5) - 6*x^2*log(5) - 50*x^2) - log(x)

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sympy [A]  time = 0.64, size = 36, normalized size = 1.50 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{2} + \frac {x \left (-100 - 15 \log {\relax (5 )}\right )}{3 \log {\relax (5 )} + 25} - \frac {15 \log {\relax (5 )}}{3 \log {\relax (5 )} + 25} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+15)*ln(5)+25*x**2)/((3*x**3-15*x**2-15*x)*ln(5)+25*x**3-100*x**2),x)

[Out]

-log(x) + log(x**2 + x*(-100 - 15*log(5))/(3*log(5) + 25) - 15*log(5)/(3*log(5) + 25))

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