3.244 \(\int x^3 \log (\frac {7+5 x}{x^2}) \, dx\)

Optimal. Leaf size=54 \[ \frac {x^4}{16}+\frac {7 x^3}{60}-\frac {49 x^2}{200}+\frac {1}{4} x^4 \log \left (\frac {5 x+7}{x^2}\right )+\frac {343 x}{500}-\frac {2401 \log (5 x+7)}{2500} \]

[Out]

343/500*x-49/200*x^2+7/60*x^3+1/16*x^4-2401/2500*ln(7+5*x)+1/4*x^4*ln((7+5*x)/x^2)

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2495, 30, 43} \[ \frac {x^4}{16}+\frac {7 x^3}{60}-\frac {49 x^2}{200}+\frac {1}{4} x^4 \log \left (\frac {5 x+7}{x^2}\right )+\frac {343 x}{500}-\frac {2401 \log (5 x+7)}{2500} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Log[(7 + 5*x)/x^2],x]

[Out]

(343*x)/500 - (49*x^2)/200 + (7*x^3)/60 + x^4/16 - (2401*Log[7 + 5*x])/2500 + (x^4*Log[(7 + 5*x)/x^2])/4

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^3 \log \left (\frac {7+5 x}{x^2}\right ) \, dx &=\frac {1}{4} x^4 \log \left (\frac {7+5 x}{x^2}\right )+\frac {\int x^3 \, dx}{2}-\frac {5}{4} \int \frac {x^4}{7+5 x} \, dx\\ &=\frac {x^4}{8}+\frac {1}{4} x^4 \log \left (\frac {7+5 x}{x^2}\right )-\frac {5}{4} \int \left (-\frac {343}{625}+\frac {49 x}{125}-\frac {7 x^2}{25}+\frac {x^3}{5}+\frac {2401}{625 (7+5 x)}\right ) \, dx\\ &=\frac {343 x}{500}-\frac {49 x^2}{200}+\frac {7 x^3}{60}+\frac {x^4}{16}-\frac {2401 \log (7+5 x)}{2500}+\frac {1}{4} x^4 \log \left (\frac {7+5 x}{x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 54, normalized size = 1.00 \[ \frac {x^4}{16}+\frac {7 x^3}{60}-\frac {49 x^2}{200}+\frac {1}{4} x^4 \log \left (\frac {5 x+7}{x^2}\right )+\frac {343 x}{500}-\frac {2401 \log (5 x+7)}{2500} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Log[(7 + 5*x)/x^2],x]

[Out]

(343*x)/500 - (49*x^2)/200 + (7*x^3)/60 + x^4/16 - (2401*Log[7 + 5*x])/2500 + (x^4*Log[(7 + 5*x)/x^2])/4

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fricas [A]  time = 0.44, size = 42, normalized size = 0.78 \[ \frac {1}{4} \, x^{4} \log \left (\frac {5 \, x + 7}{x^{2}}\right ) + \frac {1}{16} \, x^{4} + \frac {7}{60} \, x^{3} - \frac {49}{200} \, x^{2} + \frac {343}{500} \, x - \frac {2401}{2500} \, \log \left (5 \, x + 7\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log((7+5*x)/x^2),x, algorithm="fricas")

[Out]

1/4*x^4*log((5*x + 7)/x^2) + 1/16*x^4 + 7/60*x^3 - 49/200*x^2 + 343/500*x - 2401/2500*log(5*x + 7)

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giac [A]  time = 0.17, size = 43, normalized size = 0.80 \[ \frac {1}{4} \, x^{4} \log \left (\frac {5 \, x + 7}{x^{2}}\right ) + \frac {1}{16} \, x^{4} + \frac {7}{60} \, x^{3} - \frac {49}{200} \, x^{2} + \frac {343}{500} \, x - \frac {2401}{2500} \, \log \left ({\left | 5 \, x + 7 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log((7+5*x)/x^2),x, algorithm="giac")

[Out]

1/4*x^4*log((5*x + 7)/x^2) + 1/16*x^4 + 7/60*x^3 - 49/200*x^2 + 343/500*x - 2401/2500*log(abs(5*x + 7))

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maple [A]  time = 0.08, size = 53, normalized size = 0.98 \[ \frac {x^{4} \ln \left (\frac {\frac {7}{x}+5}{x}\right )}{4}+\frac {x^{4}}{16}+\frac {7 x^{3}}{60}-\frac {49 x^{2}}{200}+\frac {343 x}{500}+\frac {2401 \ln \left (\frac {1}{x}\right )}{2500}-\frac {2401 \ln \left (\frac {7}{x}+5\right )}{2500} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*ln((7+5*x)/x^2),x)

[Out]

1/4*x^4*ln((7/x+5)/x)-2401/2500*ln(7/x+5)+1/16*x^4+7/60*x^3-49/200*x^2+343/500*x+2401/2500*ln(1/x)

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maxima [A]  time = 0.68, size = 42, normalized size = 0.78 \[ \frac {1}{4} \, x^{4} \log \left (\frac {5 \, x + 7}{x^{2}}\right ) + \frac {1}{16} \, x^{4} + \frac {7}{60} \, x^{3} - \frac {49}{200} \, x^{2} + \frac {343}{500} \, x - \frac {2401}{2500} \, \log \left (5 \, x + 7\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log((7+5*x)/x^2),x, algorithm="maxima")

[Out]

1/4*x^4*log((5*x + 7)/x^2) + 1/16*x^4 + 7/60*x^3 - 49/200*x^2 + 343/500*x - 2401/2500*log(5*x + 7)

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mupad [B]  time = 0.59, size = 56, normalized size = 1.04 \[ \frac {343\,x}{500}-\frac {2401\,\ln \left (x\,\left (5\,x+7\right )\right )}{3750}-\frac {2401\,\ln \left (\frac {5\,x+7}{x^2}\right )}{7500}+\frac {x^4\,\ln \left (\frac {5\,x+7}{x^2}\right )}{4}-\frac {49\,x^2}{200}+\frac {7\,x^3}{60}+\frac {x^4}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log((5*x + 7)/x^2),x)

[Out]

(343*x)/500 - (2401*log(x*(5*x + 7)))/3750 - (2401*log((5*x + 7)/x^2))/7500 + (x^4*log((5*x + 7)/x^2))/4 - (49
*x^2)/200 + (7*x^3)/60 + x^4/16

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sympy [A]  time = 0.17, size = 48, normalized size = 0.89 \[ \frac {x^{4} \log {\left (\frac {5 x + 7}{x^{2}} \right )}}{4} + \frac {x^{4}}{16} + \frac {7 x^{3}}{60} - \frac {49 x^{2}}{200} + \frac {343 x}{500} - \frac {2401 \log {\left (5 x + 7 \right )}}{2500} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*ln((7+5*x)/x**2),x)

[Out]

x**4*log((5*x + 7)/x**2)/4 + x**4/16 + 7*x**3/60 - 49*x**2/200 + 343*x/500 - 2401*log(5*x + 7)/2500

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