\(\int (d x)^m (a+b \log (c x^n))^2 \, dx\) [151]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 18, antiderivative size = 81 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {2 b^2 n^2 (d x)^{1+m}}{d (1+m)^3}-\frac {2 b n (d x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d (1+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2342, 2341} \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {(d x)^{m+1} \left (a+b \log \left (c x^n\right )\right )^2}{d (m+1)}-\frac {2 b n (d x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{d (m+1)^2}+\frac {2 b^2 n^2 (d x)^{m+1}}{d (m+1)^3} \]

[In]

Int[(d*x)^m*(a + b*Log[c*x^n])^2,x]

[Out]

(2*b^2*n^2*(d*x)^(1 + m))/(d*(1 + m)^3) - (2*b*n*(d*x)^(1 + m)*(a + b*Log[c*x^n]))/(d*(1 + m)^2) + ((d*x)^(1 +
 m)*(a + b*Log[c*x^n])^2)/(d*(1 + m))

Rule 2341

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 2342

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(d x)^{1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d (1+m)}-\frac {(2 b n) \int (d x)^m \left (a+b \log \left (c x^n\right )\right ) \, dx}{1+m} \\ & = \frac {2 b^2 n^2 (d x)^{1+m}}{d (1+m)^3}-\frac {2 b n (d x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \left (a+b \log \left (c x^n\right )\right )^2}{d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {x (d x)^m \left (a^2 (1+m)^2-2 a b (1+m) n+2 b^2 n^2+2 b (1+m) (a+a m-b n) \log \left (c x^n\right )+b^2 (1+m)^2 \log ^2\left (c x^n\right )\right )}{(1+m)^3} \]

[In]

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

[Out]

(x*(d*x)^m*(a^2*(1 + m)^2 - 2*a*b*(1 + m)*n + 2*b^2*n^2 + 2*b*(1 + m)*(a + a*m - b*n)*Log[c*x^n] + b^2*(1 + m)
^2*Log[c*x^n]^2))/(1 + m)^3

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(242\) vs. \(2(81)=162\).

Time = 0.22 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.00

method result size
parallelrisch \(-\frac {-x \left (d x \right )^{m} a^{2} m^{2}-2 x \left (d x \right )^{m} b^{2} n^{2}-2 x \left (d x \right )^{m} a^{2} m -x \left (d x \right )^{m} \ln \left (c \,x^{n}\right )^{2} b^{2}-2 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right ) a b \,m^{2}+2 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right ) b^{2} m n -4 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right ) a b m +2 x \left (d x \right )^{m} a b m n -x \left (d x \right )^{m} a^{2}-x \left (d x \right )^{m} \ln \left (c \,x^{n}\right )^{2} b^{2} m^{2}-2 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right )^{2} b^{2} m +2 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right ) b^{2} n -2 x \left (d x \right )^{m} \ln \left (c \,x^{n}\right ) a b +2 x \left (d x \right )^{m} a b n}{m^{3}+3 m^{2}+3 m +1}\) \(243\)
risch \(\text {Expression too large to display}\) \(2027\)

[In]

int((d*x)^m*(a+b*ln(c*x^n))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (81) = 162\).

Time = 0.28 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.57 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {{\left ({\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} n^{2} x \log \left (x\right )^{2} + {\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} x \log \left (c\right )^{2} + 2 \, {\left (a b m^{2} + 2 \, a b m + a b - {\left (b^{2} m + b^{2}\right )} n\right )} x \log \left (c\right ) + {\left (a^{2} m^{2} + 2 \, b^{2} n^{2} + 2 \, a^{2} m + a^{2} - 2 \, {\left (a b m + a b\right )} n\right )} x + 2 \, {\left ({\left (b^{2} m^{2} + 2 \, b^{2} m + b^{2}\right )} n x \log \left (c\right ) - {\left ({\left (b^{2} m + b^{2}\right )} n^{2} - {\left (a b m^{2} + 2 \, a b m + a b\right )} n\right )} x\right )} \log \left (x\right )\right )} e^{\left (m \log \left (d\right ) + m \log \left (x\right )\right )}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} \]

[In]

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

[Out]

((b^2*m^2 + 2*b^2*m + b^2)*n^2*x*log(x)^2 + (b^2*m^2 + 2*b^2*m + b^2)*x*log(c)^2 + 2*(a*b*m^2 + 2*a*b*m + a*b
- (b^2*m + b^2)*n)*x*log(c) + (a^2*m^2 + 2*b^2*n^2 + 2*a^2*m + a^2 - 2*(a*b*m + a*b)*n)*x + 2*((b^2*m^2 + 2*b^
2*m + b^2)*n*x*log(c) - ((b^2*m + b^2)*n^2 - (a*b*m^2 + 2*a*b*m + a*b)*n)*x)*log(x))*e^(m*log(d) + m*log(x))/(
m^3 + 3*m^2 + 3*m + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (73) = 146\).

Time = 6.38 (sec) , antiderivative size = 502, normalized size of antiderivative = 6.20 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} m^{2} x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 a^{2} m x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {a^{2} x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 a b m^{2} x \left (d x\right )^{m} \log {\left (c x^{n} \right )}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 a b m n x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {4 a b m x \left (d x\right )^{m} \log {\left (c x^{n} \right )}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 a b n x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 a b x \left (d x\right )^{m} \log {\left (c x^{n} \right )}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {b^{2} m^{2} x \left (d x\right )^{m} \log {\left (c x^{n} \right )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 b^{2} m n x \left (d x\right )^{m} \log {\left (c x^{n} \right )}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 b^{2} m x \left (d x\right )^{m} \log {\left (c x^{n} \right )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {2 b^{2} n^{2} x \left (d x\right )^{m}}{m^{3} + 3 m^{2} + 3 m + 1} - \frac {2 b^{2} n x \left (d x\right )^{m} \log {\left (c x^{n} \right )}}{m^{3} + 3 m^{2} + 3 m + 1} + \frac {b^{2} x \left (d x\right )^{m} \log {\left (c x^{n} \right )}^{2}}{m^{3} + 3 m^{2} + 3 m + 1} & \text {for}\: m \neq -1 \\\frac {\begin {cases} \frac {a^{2} \log {\left (c x^{n} \right )} + a b \log {\left (c x^{n} \right )}^{2} + \frac {b^{2} \log {\left (c x^{n} \right )}^{3}}{3}}{n} & \text {for}\: n \neq 0 \\\left (a^{2} + 2 a b \log {\left (c \right )} + b^{2} \log {\left (c \right )}^{2}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*m**2*x*(d*x)**m/(m**3 + 3*m**2 + 3*m + 1) + 2*a**2*m*x*(d*x)**m/(m**3 + 3*m**2 + 3*m + 1) + a*
*2*x*(d*x)**m/(m**3 + 3*m**2 + 3*m + 1) + 2*a*b*m**2*x*(d*x)**m*log(c*x**n)/(m**3 + 3*m**2 + 3*m + 1) - 2*a*b*
m*n*x*(d*x)**m/(m**3 + 3*m**2 + 3*m + 1) + 4*a*b*m*x*(d*x)**m*log(c*x**n)/(m**3 + 3*m**2 + 3*m + 1) - 2*a*b*n*
x*(d*x)**m/(m**3 + 3*m**2 + 3*m + 1) + 2*a*b*x*(d*x)**m*log(c*x**n)/(m**3 + 3*m**2 + 3*m + 1) + b**2*m**2*x*(d
*x)**m*log(c*x**n)**2/(m**3 + 3*m**2 + 3*m + 1) - 2*b**2*m*n*x*(d*x)**m*log(c*x**n)/(m**3 + 3*m**2 + 3*m + 1)
+ 2*b**2*m*x*(d*x)**m*log(c*x**n)**2/(m**3 + 3*m**2 + 3*m + 1) + 2*b**2*n**2*x*(d*x)**m/(m**3 + 3*m**2 + 3*m +
 1) - 2*b**2*n*x*(d*x)**m*log(c*x**n)/(m**3 + 3*m**2 + 3*m + 1) + b**2*x*(d*x)**m*log(c*x**n)**2/(m**3 + 3*m**
2 + 3*m + 1), Ne(m, -1)), (Piecewise(((a**2*log(c*x**n) + a*b*log(c*x**n)**2 + b**2*log(c*x**n)**3/3)/n, Ne(n,
 0)), ((a**2 + 2*a*b*log(c) + b**2*log(c)**2)*log(x), True))/d, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.63 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=-\frac {2 \, a b d^{m} n x x^{m}}{{\left (m + 1\right )}^{2}} - 2 \, {\left (\frac {d^{m} n x x^{m} \log \left (c x^{n}\right )}{{\left (m + 1\right )}^{2}} - \frac {d^{m} n^{2} x x^{m}}{{\left (m + 1\right )}^{3}}\right )} b^{2} + \frac {\left (d x\right )^{m + 1} b^{2} \log \left (c x^{n}\right )^{2}}{d {\left (m + 1\right )}} + \frac {2 \, \left (d x\right )^{m + 1} a b \log \left (c x^{n}\right )}{d {\left (m + 1\right )}} + \frac {\left (d x\right )^{m + 1} a^{2}}{d {\left (m + 1\right )}} \]

[In]

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

[Out]

-2*a*b*d^m*n*x*x^m/(m + 1)^2 - 2*(d^m*n*x*x^m*log(c*x^n)/(m + 1)^2 - d^m*n^2*x*x^m/(m + 1)^3)*b^2 + (d*x)^(m +
 1)*b^2*log(c*x^n)^2/(d*(m + 1)) + 2*(d*x)^(m + 1)*a*b*log(c*x^n)/(d*(m + 1)) + (d*x)^(m + 1)*a^2/(d*(m + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (81) = 162\).

Time = 0.36 (sec) , antiderivative size = 402, normalized size of antiderivative = 4.96 \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {b^{2} d^{m} m^{2} n^{2} x x^{m} \log \left (x\right )^{2}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} + \frac {2 \, b^{2} d^{m} m n^{2} x x^{m} \log \left (x\right )^{2}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} - \frac {2 \, b^{2} d^{m} m n^{2} x x^{m} \log \left (x\right )}{m^{3} + 3 \, m^{2} + 3 \, m + 1} + \frac {2 \, b^{2} d^{m} m n x x^{m} \log \left (c\right ) \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac {b^{2} d^{m} n^{2} x x^{m} \log \left (x\right )^{2}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} + \frac {2 \, a b d^{m} m n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} - \frac {2 \, b^{2} d^{m} n^{2} x x^{m} \log \left (x\right )}{m^{3} + 3 \, m^{2} + 3 \, m + 1} + \frac {2 \, b^{2} d^{m} n x x^{m} \log \left (c\right ) \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac {2 \, b^{2} d^{m} n^{2} x x^{m}}{m^{3} + 3 \, m^{2} + 3 \, m + 1} - \frac {2 \, b^{2} d^{m} n x x^{m} \log \left (c\right )}{m^{2} + 2 \, m + 1} + \frac {2 \, a b d^{m} n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} - \frac {2 \, a b d^{m} n x x^{m}}{m^{2} + 2 \, m + 1} + \frac {\left (d x\right )^{m} b^{2} x \log \left (c\right )^{2}}{m + 1} + \frac {2 \, \left (d x\right )^{m} a b x \log \left (c\right )}{m + 1} + \frac {\left (d x\right )^{m} a^{2} x}{m + 1} \]

[In]

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

[Out]

b^2*d^m*m^2*n^2*x*x^m*log(x)^2/(m^3 + 3*m^2 + 3*m + 1) + 2*b^2*d^m*m*n^2*x*x^m*log(x)^2/(m^3 + 3*m^2 + 3*m + 1
) - 2*b^2*d^m*m*n^2*x*x^m*log(x)/(m^3 + 3*m^2 + 3*m + 1) + 2*b^2*d^m*m*n*x*x^m*log(c)*log(x)/(m^2 + 2*m + 1) +
 b^2*d^m*n^2*x*x^m*log(x)^2/(m^3 + 3*m^2 + 3*m + 1) + 2*a*b*d^m*m*n*x*x^m*log(x)/(m^2 + 2*m + 1) - 2*b^2*d^m*n
^2*x*x^m*log(x)/(m^3 + 3*m^2 + 3*m + 1) + 2*b^2*d^m*n*x*x^m*log(c)*log(x)/(m^2 + 2*m + 1) + 2*b^2*d^m*n^2*x*x^
m/(m^3 + 3*m^2 + 3*m + 1) - 2*b^2*d^m*n*x*x^m*log(c)/(m^2 + 2*m + 1) + 2*a*b*d^m*n*x*x^m*log(x)/(m^2 + 2*m + 1
) - 2*a*b*d^m*n*x*x^m/(m^2 + 2*m + 1) + (d*x)^m*b^2*x*log(c)^2/(m + 1) + 2*(d*x)^m*a*b*x*log(c)/(m + 1) + (d*x
)^m*a^2*x/(m + 1)

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\int {\left (d\,x\right )}^m\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \]

[In]

int((d*x)^m*(a + b*log(c*x^n))^2,x)

[Out]

int((d*x)^m*(a + b*log(c*x^n))^2, x)