3.10 \(\int \frac {\log (c x^n) (a x^m+b \log ^2(c x^n))^2}{x} \, dx\)

Optimal. Leaf size=125 \[ \frac {a^2 x^{2 m} \log \left (c x^n\right )}{2 m}-\frac {a^2 n x^{2 m}}{4 m^2}+\frac {12 a b n^2 x^m \log \left (c x^n\right )}{m^3}-\frac {6 a b n x^m \log ^2\left (c x^n\right )}{m^2}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}-\frac {12 a b n^3 x^m}{m^4}+\frac {b^2 \log ^6\left (c x^n\right )}{6 n} \]

[Out]

-12*a*b*n^3*x^m/m^4-1/4*a^2*n*x^(2*m)/m^2+12*a*b*n^2*x^m*ln(c*x^n)/m^3+1/2*a^2*x^(2*m)*ln(c*x^n)/m-6*a*b*n*x^m
*ln(c*x^n)^2/m^2+2*a*b*x^m*ln(c*x^n)^3/m+1/6*b^2*ln(c*x^n)^6/n

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Rubi [A]  time = 0.17, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2539, 2304, 2305, 2302, 30} \[ \frac {a^2 x^{2 m} \log \left (c x^n\right )}{2 m}-\frac {a^2 n x^{2 m}}{4 m^2}+\frac {12 a b n^2 x^m \log \left (c x^n\right )}{m^3}-\frac {6 a b n x^m \log ^2\left (c x^n\right )}{m^2}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}-\frac {12 a b n^3 x^m}{m^4}+\frac {b^2 \log ^6\left (c x^n\right )}{6 n} \]

Antiderivative was successfully verified.

[In]

Int[(Log[c*x^n]*(a*x^m + b*Log[c*x^n]^2)^2)/x,x]

[Out]

(-12*a*b*n^3*x^m)/m^4 - (a^2*n*x^(2*m))/(4*m^2) + (12*a*b*n^2*x^m*Log[c*x^n])/m^3 + (a^2*x^(2*m)*Log[c*x^n])/(
2*m) - (6*a*b*n*x^m*Log[c*x^n]^2)/m^2 + (2*a*b*x^m*Log[c*x^n]^3)/m + (b^2*Log[c*x^n]^6)/(6*n)

Rule 30

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

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2539

Int[(Log[(c_.)*(x_)^(n_.)]^(r_.)*(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_)^(m_.))^(p_.))/(x_), x_Symbol]
:> Int[ExpandIntegrand[Log[c*x^n]^r/x, (a*x^m + b*Log[c*x^n]^q)^p, x], x] /; FreeQ[{a, b, c, m, n, p, q, r}, x
] && EqQ[r, q - 1] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\log \left (c x^n\right ) \left (a x^m+b \log ^2\left (c x^n\right )\right )^2}{x} \, dx &=\int \left (a^2 x^{-1+2 m} \log \left (c x^n\right )+2 a b x^{-1+m} \log ^3\left (c x^n\right )+\frac {b^2 \log ^5\left (c x^n\right )}{x}\right ) \, dx\\ &=a^2 \int x^{-1+2 m} \log \left (c x^n\right ) \, dx+(2 a b) \int x^{-1+m} \log ^3\left (c x^n\right ) \, dx+b^2 \int \frac {\log ^5\left (c x^n\right )}{x} \, dx\\ &=-\frac {a^2 n x^{2 m}}{4 m^2}+\frac {a^2 x^{2 m} \log \left (c x^n\right )}{2 m}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}+\frac {b^2 \operatorname {Subst}\left (\int x^5 \, dx,x,\log \left (c x^n\right )\right )}{n}-\frac {(6 a b n) \int x^{-1+m} \log ^2\left (c x^n\right ) \, dx}{m}\\ &=-\frac {a^2 n x^{2 m}}{4 m^2}+\frac {a^2 x^{2 m} \log \left (c x^n\right )}{2 m}-\frac {6 a b n x^m \log ^2\left (c x^n\right )}{m^2}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}+\frac {b^2 \log ^6\left (c x^n\right )}{6 n}+\frac {\left (12 a b n^2\right ) \int x^{-1+m} \log \left (c x^n\right ) \, dx}{m^2}\\ &=-\frac {12 a b n^3 x^m}{m^4}-\frac {a^2 n x^{2 m}}{4 m^2}+\frac {12 a b n^2 x^m \log \left (c x^n\right )}{m^3}+\frac {a^2 x^{2 m} \log \left (c x^n\right )}{2 m}-\frac {6 a b n x^m \log ^2\left (c x^n\right )}{m^2}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}+\frac {b^2 \log ^6\left (c x^n\right )}{6 n}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 115, normalized size = 0.92 \[ -\frac {6 a b n x^m \log ^2\left (c x^n\right )}{m^2}+\frac {a x^m \log \left (c x^n\right ) \left (a m^2 x^m+24 b n^2\right )}{2 m^3}+\frac {2 a b x^m \log ^3\left (c x^n\right )}{m}-\frac {a n x^m \left (a m^2 x^m+48 b n^2\right )}{4 m^4}+\frac {b^2 \log ^6\left (c x^n\right )}{6 n} \]

Antiderivative was successfully verified.

[In]

Integrate[(Log[c*x^n]*(a*x^m + b*Log[c*x^n]^2)^2)/x,x]

[Out]

-1/4*(a*n*x^m*(48*b*n^2 + a*m^2*x^m))/m^4 + (a*x^m*(24*b*n^2 + a*m^2*x^m)*Log[c*x^n])/(2*m^3) - (6*a*b*n*x^m*L
og[c*x^n]^2)/m^2 + (2*a*b*x^m*Log[c*x^n]^3)/m + (b^2*Log[c*x^n]^6)/(6*n)

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fricas [B]  time = 0.44, size = 267, normalized size = 2.14 \[ \frac {2 \, b^{2} m^{4} n^{5} \log \relax (x)^{6} + 12 \, b^{2} m^{4} n^{4} \log \relax (c) \log \relax (x)^{5} + 30 \, b^{2} m^{4} n^{3} \log \relax (c)^{2} \log \relax (x)^{4} + 40 \, b^{2} m^{4} n^{2} \log \relax (c)^{3} \log \relax (x)^{3} + 30 \, b^{2} m^{4} n \log \relax (c)^{4} \log \relax (x)^{2} + 12 \, b^{2} m^{4} \log \relax (c)^{5} \log \relax (x) + 3 \, {\left (2 \, a^{2} m^{3} n \log \relax (x) + 2 \, a^{2} m^{3} \log \relax (c) - a^{2} m^{2} n\right )} x^{2 \, m} + 24 \, {\left (a b m^{3} n^{3} \log \relax (x)^{3} + a b m^{3} \log \relax (c)^{3} - 3 \, a b m^{2} n \log \relax (c)^{2} + 6 \, a b m n^{2} \log \relax (c) - 6 \, a b n^{3} + 3 \, {\left (a b m^{3} n^{2} \log \relax (c) - a b m^{2} n^{3}\right )} \log \relax (x)^{2} + 3 \, {\left (a b m^{3} n \log \relax (c)^{2} - 2 \, a b m^{2} n^{2} \log \relax (c) + 2 \, a b m n^{3}\right )} \log \relax (x)\right )} x^{m}}{12 \, m^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*x^n)*(a*x^m+b*log(c*x^n)^2)^2/x,x, algorithm="fricas")

[Out]

1/12*(2*b^2*m^4*n^5*log(x)^6 + 12*b^2*m^4*n^4*log(c)*log(x)^5 + 30*b^2*m^4*n^3*log(c)^2*log(x)^4 + 40*b^2*m^4*
n^2*log(c)^3*log(x)^3 + 30*b^2*m^4*n*log(c)^4*log(x)^2 + 12*b^2*m^4*log(c)^5*log(x) + 3*(2*a^2*m^3*n*log(x) +
2*a^2*m^3*log(c) - a^2*m^2*n)*x^(2*m) + 24*(a*b*m^3*n^3*log(x)^3 + a*b*m^3*log(c)^3 - 3*a*b*m^2*n*log(c)^2 + 6
*a*b*m*n^2*log(c) - 6*a*b*n^3 + 3*(a*b*m^3*n^2*log(c) - a*b*m^2*n^3)*log(x)^2 + 3*(a*b*m^3*n*log(c)^2 - 2*a*b*
m^2*n^2*log(c) + 2*a*b*m*n^3)*log(x))*x^m)/m^4

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giac [B]  time = 0.19, size = 286, normalized size = 2.29 \[ \frac {1}{6} \, b^{2} n^{5} \log \relax (x)^{6} + b^{2} n^{4} \log \relax (c) \log \relax (x)^{5} + \frac {5}{2} \, b^{2} n^{3} \log \relax (c)^{2} \log \relax (x)^{4} + \frac {10}{3} \, b^{2} n^{2} \log \relax (c)^{3} \log \relax (x)^{3} + \frac {5}{2} \, b^{2} n \log \relax (c)^{4} \log \relax (x)^{2} + b^{2} \log \relax (c)^{5} \log \relax (x) + \frac {2 \, a b n^{3} x^{m} \log \relax (x)^{3}}{m} + \frac {6 \, a b n^{2} x^{m} \log \relax (c) \log \relax (x)^{2}}{m} + \frac {6 \, a b n x^{m} \log \relax (c)^{2} \log \relax (x)}{m} - \frac {6 \, a b n^{3} x^{m} \log \relax (x)^{2}}{m^{2}} + \frac {2 \, a b x^{m} \log \relax (c)^{3}}{m} - \frac {12 \, a b n^{2} x^{m} \log \relax (c) \log \relax (x)}{m^{2}} - \frac {6 \, a b n x^{m} \log \relax (c)^{2}}{m^{2}} + \frac {a^{2} n x^{2 \, m} \log \relax (x)}{2 \, m} + \frac {12 \, a b n^{3} x^{m} \log \relax (x)}{m^{3}} + \frac {a^{2} x^{2 \, m} \log \relax (c)}{2 \, m} + \frac {12 \, a b n^{2} x^{m} \log \relax (c)}{m^{3}} - \frac {a^{2} n x^{2 \, m}}{4 \, m^{2}} - \frac {12 \, a b n^{3} x^{m}}{m^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*x^n)*(a*x^m+b*log(c*x^n)^2)^2/x,x, algorithm="giac")

[Out]

1/6*b^2*n^5*log(x)^6 + b^2*n^4*log(c)*log(x)^5 + 5/2*b^2*n^3*log(c)^2*log(x)^4 + 10/3*b^2*n^2*log(c)^3*log(x)^
3 + 5/2*b^2*n*log(c)^4*log(x)^2 + b^2*log(c)^5*log(x) + 2*a*b*n^3*x^m*log(x)^3/m + 6*a*b*n^2*x^m*log(c)*log(x)
^2/m + 6*a*b*n*x^m*log(c)^2*log(x)/m - 6*a*b*n^3*x^m*log(x)^2/m^2 + 2*a*b*x^m*log(c)^3/m - 12*a*b*n^2*x^m*log(
c)*log(x)/m^2 - 6*a*b*n*x^m*log(c)^2/m^2 + 1/2*a^2*n*x^(2*m)*log(x)/m + 12*a*b*n^3*x^m*log(x)/m^3 + 1/2*a^2*x^
(2*m)*log(c)/m + 12*a*b*n^2*x^m*log(c)/m^3 - 1/4*a^2*n*x^(2*m)/m^2 - 12*a*b*n^3*x^m/m^4

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maple [C]  time = 3.16, size = 14983, normalized size = 119.86 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*x^n)*(a*x^m+b*ln(c*x^n)^2)^2/x,x)

[Out]

result too large to display

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maxima [B]  time = 0.66, size = 530, normalized size = 4.24 \[ \frac {1}{10} \, {\left (\frac {2 \, b^{2} \log \left (c x^{n}\right )^{5}}{n} + \frac {20 \, a b x^{m} \log \left (c x^{n}\right )^{2}}{m} - 40 \, a b {\left (\frac {n x^{m} \log \left (c x^{n}\right )}{m^{2}} - \frac {n^{2} x^{m}}{m^{3}}\right )} + \frac {5 \, a^{2} x^{2 \, m}}{m}\right )} \log \left (c x^{n}\right ) + \frac {2 \, b^{2} m^{4} n^{5} \log \relax (x)^{6} - 12 \, b^{2} m^{4} n^{4} \log \relax (c) \log \relax (x)^{5} + 30 \, b^{2} m^{4} n^{3} \log \relax (c)^{2} \log \relax (x)^{4} - 40 \, b^{2} m^{4} n^{2} \log \relax (c)^{3} \log \relax (x)^{3} + 30 \, b^{2} m^{4} n \log \relax (c)^{4} \log \relax (x)^{2} - 12 \, b^{2} m^{4} \log \relax (c)^{5} \log \relax (x) - 12 \, b^{2} m^{4} \log \relax (x) \log \left (x^{n}\right )^{5} - 15 \, a^{2} m^{2} n x^{2 \, m} + 30 \, {\left (b^{2} m^{4} n \log \relax (x)^{2} - 2 \, b^{2} m^{4} \log \relax (c) \log \relax (x)\right )} \log \left (x^{n}\right )^{4} - 120 \, {\left (m^{2} n \log \relax (c)^{2} - 4 \, m n^{2} \log \relax (c) + 6 \, n^{3}\right )} a b x^{m} - 40 \, {\left (b^{2} m^{4} n^{2} \log \relax (x)^{3} - 3 \, b^{2} m^{4} n \log \relax (c) \log \relax (x)^{2} + 3 \, b^{2} m^{4} \log \relax (c)^{2} \log \relax (x)\right )} \log \left (x^{n}\right )^{3} + 30 \, {\left (b^{2} m^{4} n^{3} \log \relax (x)^{4} - 4 \, b^{2} m^{4} n^{2} \log \relax (c) \log \relax (x)^{3} + 6 \, b^{2} m^{4} n \log \relax (c)^{2} \log \relax (x)^{2} - 4 \, b^{2} m^{4} \log \relax (c)^{3} \log \relax (x) - 4 \, a b m^{2} n x^{m}\right )} \log \left (x^{n}\right )^{2} - 12 \, {\left (b^{2} m^{4} n^{4} \log \relax (x)^{5} - 5 \, b^{2} m^{4} n^{3} \log \relax (c) \log \relax (x)^{4} + 10 \, b^{2} m^{4} n^{2} \log \relax (c)^{2} \log \relax (x)^{3} - 10 \, b^{2} m^{4} n \log \relax (c)^{3} \log \relax (x)^{2} + 5 \, b^{2} m^{4} \log \relax (c)^{4} \log \relax (x) + 20 \, {\left (m^{2} n \log \relax (c) - 2 \, m n^{2}\right )} a b x^{m}\right )} \log \left (x^{n}\right )}{60 \, m^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*x^n)*(a*x^m+b*log(c*x^n)^2)^2/x,x, algorithm="maxima")

[Out]

1/10*(2*b^2*log(c*x^n)^5/n + 20*a*b*x^m*log(c*x^n)^2/m - 40*a*b*(n*x^m*log(c*x^n)/m^2 - n^2*x^m/m^3) + 5*a^2*x
^(2*m)/m)*log(c*x^n) + 1/60*(2*b^2*m^4*n^5*log(x)^6 - 12*b^2*m^4*n^4*log(c)*log(x)^5 + 30*b^2*m^4*n^3*log(c)^2
*log(x)^4 - 40*b^2*m^4*n^2*log(c)^3*log(x)^3 + 30*b^2*m^4*n*log(c)^4*log(x)^2 - 12*b^2*m^4*log(c)^5*log(x) - 1
2*b^2*m^4*log(x)*log(x^n)^5 - 15*a^2*m^2*n*x^(2*m) + 30*(b^2*m^4*n*log(x)^2 - 2*b^2*m^4*log(c)*log(x))*log(x^n
)^4 - 120*(m^2*n*log(c)^2 - 4*m*n^2*log(c) + 6*n^3)*a*b*x^m - 40*(b^2*m^4*n^2*log(x)^3 - 3*b^2*m^4*n*log(c)*lo
g(x)^2 + 3*b^2*m^4*log(c)^2*log(x))*log(x^n)^3 + 30*(b^2*m^4*n^3*log(x)^4 - 4*b^2*m^4*n^2*log(c)*log(x)^3 + 6*
b^2*m^4*n*log(c)^2*log(x)^2 - 4*b^2*m^4*log(c)^3*log(x) - 4*a*b*m^2*n*x^m)*log(x^n)^2 - 12*(b^2*m^4*n^4*log(x)
^5 - 5*b^2*m^4*n^3*log(c)*log(x)^4 + 10*b^2*m^4*n^2*log(c)^2*log(x)^3 - 10*b^2*m^4*n*log(c)^3*log(x)^2 + 5*b^2
*m^4*log(c)^4*log(x) + 20*(m^2*n*log(c) - 2*m*n^2)*a*b*x^m)*log(x^n))/m^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,x^n\right )\,{\left (a\,x^m+b\,{\ln \left (c\,x^n\right )}^2\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(c*x^n)*(a*x^m + b*log(c*x^n)^2)^2)/x,x)

[Out]

int((log(c*x^n)*(a*x^m + b*log(c*x^n)^2)^2)/x, x)

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sympy [A]  time = 37.59, size = 298, normalized size = 2.38 \[ - a^{2} n \left (\begin {cases} \frac {\begin {cases} \frac {x^{2 m}}{2 m} & \text {for}\: m \neq 0 \\\log {\relax (x )} & \text {otherwise} \end {cases}}{2 m} & \text {for}\: m > -\infty \wedge m < \infty \wedge m \neq 0 \\\frac {\log {\relax (x )}^{2}}{2} & \text {otherwise} \end {cases}\right ) + a^{2} \left (\begin {cases} \frac {x^{2 m}}{2 m} & \text {for}\: 2 m - 1 \neq -1 \\\log {\relax (x )} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} + 2 a b \left (\begin {cases} \frac {n^{3} x^{m} \log {\relax (x )}^{3}}{m} + \frac {3 n^{2} x^{m} \log {\relax (c )} \log {\relax (x )}^{2}}{m} + \frac {3 n x^{m} \log {\relax (c )}^{2} \log {\relax (x )}}{m} + \frac {x^{m} \log {\relax (c )}^{3}}{m} - \frac {3 n^{3} x^{m} \log {\relax (x )}^{2}}{m^{2}} - \frac {6 n^{2} x^{m} \log {\relax (c )} \log {\relax (x )}}{m^{2}} - \frac {3 n x^{m} \log {\relax (c )}^{2}}{m^{2}} + \frac {6 n^{3} x^{m} \log {\relax (x )}}{m^{3}} + \frac {6 n^{2} x^{m} \log {\relax (c )}}{m^{3}} - \frac {6 n^{3} x^{m}}{m^{4}} & \text {for}\: m \neq 0 \\\begin {cases} \frac {\log {\left (c x^{n} \right )}^{4}}{4 n} & \text {for}\: \left |{c x^{n}}\right | < 1 \\\frac {\log {\left (\frac {x^{- n}}{c} \right )}^{4}}{4 n} & \text {for}\: \frac {1}{\left |{c x^{n}}\right |} < 1 \\\frac {6 {G_{5, 5}^{5, 0}\left (\begin {matrix} & 1, 1, 1, 1, 1 \\0, 0, 0, 0, 0 & \end {matrix} \middle | {c x^{n}} \right )}}{n} + \frac {6 {G_{5, 5}^{0, 5}\left (\begin {matrix} 1, 1, 1, 1, 1 & \\ & 0, 0, 0, 0, 0 \end {matrix} \middle | {c x^{n}} \right )}}{n} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases}\right ) - b^{2} \left (\begin {cases} - \log {\relax (c )}^{5} \log {\relax (x )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{6}}{6 n} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*x**n)*(a*x**m+b*ln(c*x**n)**2)**2/x,x)

[Out]

-a**2*n*Piecewise((Piecewise((x**(2*m)/(2*m), Ne(m, 0)), (log(x), True))/(2*m), (m > -oo) & (m < oo) & Ne(m, 0
)), (log(x)**2/2, True)) + a**2*Piecewise((x**(2*m)/(2*m), Ne(2*m - 1, -1)), (log(x), True))*log(c*x**n) + 2*a
*b*Piecewise((n**3*x**m*log(x)**3/m + 3*n**2*x**m*log(c)*log(x)**2/m + 3*n*x**m*log(c)**2*log(x)/m + x**m*log(
c)**3/m - 3*n**3*x**m*log(x)**2/m**2 - 6*n**2*x**m*log(c)*log(x)/m**2 - 3*n*x**m*log(c)**2/m**2 + 6*n**3*x**m*
log(x)/m**3 + 6*n**2*x**m*log(c)/m**3 - 6*n**3*x**m/m**4, Ne(m, 0)), (Piecewise((log(c*x**n)**4/(4*n), Abs(c*x
**n) < 1), (log(x**(-n)/c)**4/(4*n), 1/Abs(c*x**n) < 1), (6*meijerg(((), (1, 1, 1, 1, 1)), ((0, 0, 0, 0, 0), (
)), c*x**n)/n + 6*meijerg(((1, 1, 1, 1, 1), ()), ((), (0, 0, 0, 0, 0)), c*x**n)/n, True)), True)) - b**2*Piece
wise((-log(c)**5*log(x), Eq(n, 0)), (-log(c*x**n)**6/(6*n), True))

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