3.34.3 \(\int \frac {-1+15 x+(-1+5 x) \log (x-10 x^2+25 x^3)}{(-1+5 x) \log (2)} \, dx\)

Optimal. Leaf size=16 \[ \frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)} \]

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Rubi [A]  time = 0.06, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6742, 43, 2487, 31, 8} \begin {gather*} \frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 15*x + (-1 + 5*x)*Log[x - 10*x^2 + 25*x^3])/((-1 + 5*x)*Log[2]),x]

[Out]

(x*Log[(1 - 5*x)^2*x])/Log[2]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2487

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + (Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p
*(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] - Dist[r*s*(p + q), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1
), x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && NeQ[p + q, 0] && IGtQ[s, 0] &&
LtQ[s, 4]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-1+15 x+(-1+5 x) \log \left (x-10 x^2+25 x^3\right )}{-1+5 x} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {-1+15 x}{-1+5 x}+\log \left (x (-1+5 x)^2\right )\right ) \, dx}{\log (2)}\\ &=\frac {\int \frac {-1+15 x}{-1+5 x} \, dx}{\log (2)}+\frac {\int \log \left (x (-1+5 x)^2\right ) \, dx}{\log (2)}\\ &=\frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)}+\frac {\int \left (3+\frac {2}{-1+5 x}\right ) \, dx}{\log (2)}-\frac {2 \int \frac {1}{-1+5 x} \, dx}{\log (2)}-\frac {3 \int 1 \, dx}{\log (2)}\\ &=\frac {x \log \left ((1-5 x)^2 x\right )}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 16, normalized size = 1.00 \begin {gather*} \frac {x \log \left (x (-1+5 x)^2\right )}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 15*x + (-1 + 5*x)*Log[x - 10*x^2 + 25*x^3])/((-1 + 5*x)*Log[2]),x]

[Out]

(x*Log[x*(-1 + 5*x)^2])/Log[2]

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fricas [A]  time = 0.57, size = 19, normalized size = 1.19 \begin {gather*} \frac {x \log \left (25 \, x^{3} - 10 \, x^{2} + x\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(25*x^3-10*x^2+x)+15*x-1)/(5*x-1)/log(2),x, algorithm="fricas")

[Out]

x*log(25*x^3 - 10*x^2 + x)/log(2)

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giac [A]  time = 0.22, size = 19, normalized size = 1.19 \begin {gather*} \frac {x \log \left (25 \, x^{3} - 10 \, x^{2} + x\right )}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(25*x^3-10*x^2+x)+15*x-1)/(5*x-1)/log(2),x, algorithm="giac")

[Out]

x*log(25*x^3 - 10*x^2 + x)/log(2)

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maple [A]  time = 0.48, size = 20, normalized size = 1.25

method	result	size
default	\(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \relax (2)}\)	\(20\)
norman	\(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \relax (2)}\)	\(20\)
risch	\(\frac {\ln \left (25 x^{3}-10 x^{2}+x \right ) x}{\ln \relax (2)}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x-1)*ln(25*x^3-10*x^2+x)+15*x-1)/(5*x-1)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/ln(2)*ln(25*x^3-10*x^2+x)*x

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maxima [B]  time = 0.43, size = 33, normalized size = 2.06 \begin {gather*} \frac {2 \, {\left (5 \, x - 1\right )} \log \left (5 \, x - 1\right ) + 5 \, x \log \relax (x) + 2 \, \log \left (5 \, x - 1\right )}{5 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*log(25*x^3-10*x^2+x)+15*x-1)/(5*x-1)/log(2),x, algorithm="maxima")

[Out]

1/5*(2*(5*x - 1)*log(5*x - 1) + 5*x*log(x) + 2*log(5*x - 1))/log(2)

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mupad [B]  time = 2.21, size = 16, normalized size = 1.00 \begin {gather*} \frac {x\,\ln \left (x\,{\left (5\,x-1\right )}^2\right )}{\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x + log(x - 10*x^2 + 25*x^3)*(5*x - 1) - 1)/(log(2)*(5*x - 1)),x)

[Out]

(x*log(x*(5*x - 1)^2))/log(2)

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sympy [A]  time = 0.17, size = 17, normalized size = 1.06 \begin {gather*} \frac {x \log {\left (25 x^{3} - 10 x^{2} + x \right )}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x-1)*ln(25*x**3-10*x**2+x)+15*x-1)/(5*x-1)/ln(2),x)

[Out]

x*log(25*x**3 - 10*x**2 + x)/log(2)

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