3.71.25 \(\int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+(500 x+300 x^2+60 x^3+4 x^4) \log (\frac {25}{x^3})}{4 x} \, dx\)

Optimal. Leaf size=20 \[ x+x^3+\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 14, 32, 2313, 43} \begin {gather*} x^3+\frac {1}{4} (x+5)^4 \log \left (\frac {25}{x^3}\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1875 - 1496*x - 450*x^2 - 48*x^3 - 3*x^4 + (500*x + 300*x^2 + 60*x^3 + 4*x^4)*Log[25/x^3])/(4*x),x]

[Out]

x + x^3 + ((5 + x)^4*Log[25/x^3])/4

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4+\left (500 x+300 x^2+60 x^3+4 x^4\right ) \log \left (\frac {25}{x^3}\right )}{x} \, dx\\ &=\frac {1}{4} \int \left (\frac {-1875-1496 x-450 x^2-48 x^3-3 x^4}{x}+4 (5+x)^3 \log \left (\frac {25}{x^3}\right )\right ) \, dx\\ &=\frac {1}{4} \int \frac {-1875-1496 x-450 x^2-48 x^3-3 x^4}{x} \, dx+\int (5+x)^3 \log \left (\frac {25}{x^3}\right ) \, dx\\ &=\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right )+\frac {1}{4} \int \left (-1496-\frac {1875}{x}-450 x-48 x^2-3 x^3\right ) \, dx+3 \int \frac {(5+x)^4}{4 x} \, dx\\ &=-374 x-\frac {225 x^2}{4}-4 x^3-\frac {3 x^4}{16}+\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right )-\frac {1875 \log (x)}{4}+\frac {3}{4} \int \frac {(5+x)^4}{x} \, dx\\ &=-374 x-\frac {225 x^2}{4}-4 x^3-\frac {3 x^4}{16}+\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right )-\frac {1875 \log (x)}{4}+\frac {3}{4} \int \left (500+\frac {625}{x}+150 x+20 x^2+x^3\right ) \, dx\\ &=x+x^3+\frac {1}{4} (5+x)^4 \log \left (\frac {25}{x^3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.01, size = 57, normalized size = 2.85 \begin {gather*} x+x^3+125 x \log \left (\frac {25}{x^3}\right )+\frac {75}{2} x^2 \log \left (\frac {25}{x^3}\right )+5 x^3 \log \left (\frac {25}{x^3}\right )+\frac {1}{4} x^4 \log \left (\frac {25}{x^3}\right )-\frac {1875 \log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1875 - 1496*x - 450*x^2 - 48*x^3 - 3*x^4 + (500*x + 300*x^2 + 60*x^3 + 4*x^4)*Log[25/x^3])/(4*x),x
]

[Out]

x + x^3 + 125*x*Log[25/x^3] + (75*x^2*Log[25/x^3])/2 + 5*x^3*Log[25/x^3] + (x^4*Log[25/x^3])/4 - (1875*Log[x])
/4

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fricas [A]  time = 0.95, size = 31, normalized size = 1.55 \begin {gather*} x^{3} + \frac {1}{4} \, {\left (x^{4} + 20 \, x^{3} + 150 \, x^{2} + 500 \, x + 625\right )} \log \left (\frac {25}{x^{3}}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x^2-1496*x-1875)/x,x, algorithm="fric
as")

[Out]

x^3 + 1/4*(x^4 + 20*x^3 + 150*x^2 + 500*x + 625)*log(25/x^3) + x

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giac [A]  time = 0.23, size = 34, normalized size = 1.70 \begin {gather*} x^{3} + \frac {1}{4} \, {\left (x^{4} + 20 \, x^{3} + 150 \, x^{2} + 500 \, x\right )} \log \left (\frac {25}{x^{3}}\right ) + x - \frac {1875}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x^2-1496*x-1875)/x,x, algorithm="giac
")

[Out]

x^3 + 1/4*(x^4 + 20*x^3 + 150*x^2 + 500*x)*log(25/x^3) + x - 1875/4*log(x)

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maple [A]  time = 0.09, size = 35, normalized size = 1.75

method	result	size
risch	\(\frac {\left (x^{4}+20 x^{3}+150 x^{2}+500 x \right ) \ln \left (\frac {25}{x^{3}}\right )}{4}+x^{3}+x -\frac {1875 \ln \relax (x )}{4}\)	\(35\)
norman	\(x +x^{3}+125 \ln \left (\frac {25}{x^{3}}\right ) x +\frac {75 \ln \left (\frac {25}{x^{3}}\right ) x^{2}}{2}+5 \ln \left (\frac {25}{x^{3}}\right ) x^{3}+\frac {\ln \left (\frac {25}{x^{3}}\right ) x^{4}}{4}-\frac {1875 \ln \relax (x )}{4}\)	\(52\)
derivativedivides	\(\frac {1875 \ln \left (\frac {1}{x}\right )}{4}+x^{3}+x +75 x^{2} \ln \relax (5)+10 x^{3} \ln \relax (5)+\frac {x^{4} \ln \relax (5)}{2}+250 x \ln \relax (5)+\frac {75 x^{2} \ln \left (\frac {1}{x^{3}}\right )}{2}+5 x^{3} \ln \left (\frac {1}{x^{3}}\right )+\frac {x^{4} \ln \left (\frac {1}{x^{3}}\right )}{4}+125 x \ln \left (\frac {1}{x^{3}}\right )\)	\(72\)
default	\(\frac {1875 \ln \left (\frac {1}{x}\right )}{4}+x^{3}+x +75 x^{2} \ln \relax (5)+10 x^{3} \ln \relax (5)+\frac {x^{4} \ln \relax (5)}{2}+250 x \ln \relax (5)+\frac {75 x^{2} \ln \left (\frac {1}{x^{3}}\right )}{2}+5 x^{3} \ln \left (\frac {1}{x^{3}}\right )+\frac {x^{4} \ln \left (\frac {1}{x^{3}}\right )}{4}+125 x \ln \left (\frac {1}{x^{3}}\right )\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((4*x^4+60*x^3+300*x^2+500*x)*ln(25/x^3)-3*x^4-48*x^3-450*x^2-1496*x-1875)/x,x,method=_RETURNVERBOSE)

[Out]

1/4*(x^4+20*x^3+150*x^2+500*x)*ln(25/x^3)+x^3+x-1875/4*ln(x)

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maxima [B]  time = 0.37, size = 51, normalized size = 2.55 \begin {gather*} \frac {1}{4} \, x^{4} \log \left (\frac {25}{x^{3}}\right ) + 5 \, x^{3} \log \left (\frac {25}{x^{3}}\right ) + x^{3} + \frac {75}{2} \, x^{2} \log \left (\frac {25}{x^{3}}\right ) + 125 \, x \log \left (\frac {25}{x^{3}}\right ) + x - \frac {1875}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x^4+60*x^3+300*x^2+500*x)*log(25/x^3)-3*x^4-48*x^3-450*x^2-1496*x-1875)/x,x, algorithm="maxi
ma")

[Out]

1/4*x^4*log(25/x^3) + 5*x^3*log(25/x^3) + x^3 + 75/2*x^2*log(25/x^3) + 125*x*log(25/x^3) + x - 1875/4*log(x)

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mupad [B]  time = 4.17, size = 55, normalized size = 2.75 \begin {gather*} \frac {625\,\ln \left (\frac {1}{x^3}\right )}{4}+x\,\left (125\,\ln \left (\frac {25}{x^3}\right )+1\right )+x^3\,\left (5\,\ln \left (\frac {25}{x^3}\right )+1\right )+\frac {75\,x^2\,\ln \left (\frac {25}{x^3}\right )}{2}+\frac {x^4\,\ln \left (\frac {25}{x^3}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(374*x + (225*x^2)/2 + 12*x^3 + (3*x^4)/4 - (log(25/x^3)*(500*x + 300*x^2 + 60*x^3 + 4*x^4))/4 + 1875/4)/
x,x)

[Out]

(625*log(1/x^3))/4 + x*(125*log(25/x^3) + 1) + x^3*(5*log(25/x^3) + 1) + (75*x^2*log(25/x^3))/2 + (x^4*log(25/
x^3))/4

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sympy [B]  time = 0.16, size = 37, normalized size = 1.85 \begin {gather*} x^{3} + x + \left (\frac {x^{4}}{4} + 5 x^{3} + \frac {75 x^{2}}{2} + 125 x\right ) \log {\left (\frac {25}{x^{3}} \right )} - \frac {1875 \log {\relax (x )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((4*x**4+60*x**3+300*x**2+500*x)*ln(25/x**3)-3*x**4-48*x**3-450*x**2-1496*x-1875)/x,x)

[Out]

x**3 + x + (x**4/4 + 5*x**3 + 75*x**2/2 + 125*x)*log(25/x**3) - 1875*log(x)/4

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