3.362 \(\int (f x)^{-1+m} (a+b \log (c x^n))^2 \, dx\)

Optimal. Leaf size=69 \[ -\frac {2 b n (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m^2}+\frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )^2}{f m}+\frac {2 b^2 n^2 (f x)^m}{f m^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2305, 2304} \[ -\frac {2 b n (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m^2}+\frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )^2}{f m}+\frac {2 b^2 n^2 (f x)^m}{f m^3} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^(-1 + m)*(a + b*Log[c*x^n])^2,x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 67, normalized size = 0.97 \[ \frac {(f x)^m \left (a^2 m^2+2 b m (a m-b n) \log \left (c x^n\right )-2 a b m n+b^2 m^2 \log ^2\left (c x^n\right )+2 b^2 n^2\right )}{f m^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^(-1 + m)*(a + b*Log[c*x^n])^2,x]

[Out]

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

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fricas [A]  time = 0.46, size = 124, normalized size = 1.80 \[ \frac {{\left (b^{2} m^{2} n^{2} x \log \relax (x)^{2} + b^{2} m^{2} x \log \relax (c)^{2} + 2 \, {\left (a b m^{2} - b^{2} m n\right )} x \log \relax (c) + {\left (a^{2} m^{2} - 2 \, a b m n + 2 \, b^{2} n^{2}\right )} x + 2 \, {\left (b^{2} m^{2} n x \log \relax (c) + {\left (a b m^{2} n - b^{2} m n^{2}\right )} x\right )} \log \relax (x)\right )} e^{\left ({\left (m - 1\right )} \log \relax (f) + {\left (m - 1\right )} \log \relax (x)\right )}}{m^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.59, size = 222, normalized size = 3.22 \[ \frac {b^{2} f^{m} n^{2} x^{m} \log \relax (x)^{2}}{f m} + \frac {b^{2} \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m} \log \relax (c)^{2}}{f m} + \frac {2 \, b^{2} f^{m} n x^{m} \log \relax (c) \log \relax (x)}{f m} + \frac {2 \, a b \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m} \log \relax (c)}{f m} + \frac {2 \, a b f^{m} n x^{m} \log \relax (x)}{f m} - \frac {2 \, b^{2} f^{m} n^{2} x^{m} \log \relax (x)}{f m^{2}} + \frac {a^{2} \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m}}{f m} - \frac {2 \, b^{2} f^{m} n x^{m} \log \relax (c)}{f m^{2}} - \frac {2 \, a b f^{m} n x^{m}}{f m^{2}} + \frac {2 \, b^{2} f^{m} n^{2} x^{m}}{f m^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.23, size = 1008, normalized size = 14.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(m-1)*(b*ln(c*x^n)+a)^2,x)

[Out]

b^2/m*x*exp(1/2*(m-1)*(-I*Pi*csgn(I*f)*csgn(I*x)*csgn(I*f*x)+I*Pi*csgn(I*f)*csgn(I*f*x)^2+I*Pi*csgn(I*x)*csgn(
I*f*x)^2-I*Pi*csgn(I*f*x)^3+2*ln(f)+2*ln(x)))*ln(x^n)^2+b*(-I*Pi*b*m*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*Pi*
b*m*csgn(I*c)*csgn(I*c*x^n)^2+I*Pi*b*m*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*m*csgn(I*c*x^n)^3+2*b*m*ln(c)+2*a*m-
2*b*n)/m^2*x*exp(1/2*(m-1)*(-I*Pi*csgn(I*f)*csgn(I*x)*csgn(I*f*x)+I*Pi*csgn(I*f)*csgn(I*f*x)^2+I*Pi*csgn(I*x)*
csgn(I*f*x)^2-I*Pi*csgn(I*f*x)^3+2*ln(f)+2*ln(x)))*ln(x^n)+1/4*(4*b^2*m^2*ln(c)^2-Pi^2*b^2*m^2*csgn(I*c*x^n)^6
+8*b^2*n^2-8*a*b*m*n+4*a^2*m^2-Pi^2*b^2*m^2*csgn(I*c*x^n)^4*csgn(I*c)^2-4*I*Pi*ln(c)*b^2*m^2*csgn(I*x^n)*csgn(
I*c*x^n)*csgn(I*c)-4*I*Pi*a*b*m^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+4*I*Pi*b^2*m*n*csgn(I*x^n)*csgn(I*c*x^n)
*csgn(I*c)+4*I*Pi*a*b*m^2*csgn(I*c*x^n)^2*csgn(I*c)-4*I*Pi*b^2*m*n*csgn(I*x^n)*csgn(I*c*x^n)^2+8*a*b*m^2*ln(c)
-8*b^2*m*n*ln(c)-4*I*Pi*b^2*m*n*csgn(I*c*x^n)^2*csgn(I*c)-Pi^2*b^2*m^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+2*Pi^2*b^
2*m^2*csgn(I*x^n)*csgn(I*c*x^n)^5+2*Pi^2*b^2*m^2*csgn(I*c*x^n)^5*csgn(I*c)+2*Pi^2*b^2*m^2*csgn(I*x^n)^2*csgn(I
*c*x^n)^3*csgn(I*c)-Pi^2*b^2*m^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2*csgn(I*c)^2-4*Pi^2*b^2*m^2*csgn(I*x^n)*csgn(I*c
*x^n)^4*csgn(I*c)+2*Pi^2*b^2*m^2*csgn(I*x^n)*csgn(I*c*x^n)^3*csgn(I*c)^2-4*I*Pi*ln(c)*b^2*m^2*csgn(I*c*x^n)^3-
4*I*Pi*a*b*m^2*csgn(I*c*x^n)^3+4*I*Pi*b^2*m*n*csgn(I*c*x^n)^3+4*I*Pi*ln(c)*b^2*m^2*csgn(I*x^n)*csgn(I*c*x^n)^2
+4*I*Pi*ln(c)*b^2*m^2*csgn(I*c*x^n)^2*csgn(I*c)+4*I*Pi*a*b*m^2*csgn(I*x^n)*csgn(I*c*x^n)^2)/m^3*x*exp(1/2*(m-1
)*(-I*Pi*csgn(I*f)*csgn(I*x)*csgn(I*f*x)+I*Pi*csgn(I*f)*csgn(I*f*x)^2+I*Pi*csgn(I*x)*csgn(I*f*x)^2-I*Pi*csgn(I
*f*x)^3+2*ln(f)+2*ln(x)))

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maxima [A]  time = 1.07, size = 117, normalized size = 1.70 \[ -2 \, {\left (\frac {f^{m - 1} n x^{m} \log \left (c x^{n}\right )}{m^{2}} - \frac {f^{m - 1} n^{2} x^{m}}{m^{3}}\right )} b^{2} - \frac {2 \, a b f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b^{2} \log \left (c x^{n}\right )^{2}}{f m} + \frac {2 \, \left (f x\right )^{m} a b \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a^{2}}{f m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(m - 1)*(a + b*log(c*x^n))^2,x)

[Out]

int((f*x)^(m - 1)*(a + b*log(c*x^n))^2, x)

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sympy [A]  time = 68.77, size = 490, normalized size = 7.10 \[ \begin {cases} \tilde {\infty } \left (a^{2} x + 2 a b n x \log {\relax (x )} - 2 a b n x + 2 a b x \log {\relax (c )} + b^{2} n^{2} x \log {\relax (x )}^{2} - 2 b^{2} n^{2} x \log {\relax (x )} + 2 b^{2} n^{2} x + 2 b^{2} n x \log {\relax (c )} \log {\relax (x )} - 2 b^{2} n x \log {\relax (c )} + b^{2} x \log {\relax (c )}^{2}\right ) & \text {for}\: f = 0 \wedge m = 0 \\\frac {\begin {cases} \frac {a^{2} \log {\left (c x^{n} \right )} + a b \log {\left (c x^{n} \right )}^{2} + \frac {b^{2} \log {\left (c x^{n} \right )}^{3}}{3}}{n} & \text {for}\: n \neq 0 \\\left (a^{2} + 2 a b \log {\relax (c )} + b^{2} \log {\relax (c )}^{2}\right ) \log {\relax (x )} & \text {otherwise} \end {cases}}{f} & \text {for}\: m = 0 \\0^{m - 1} \left (a^{2} x + 2 a b n x \log {\relax (x )} - 2 a b n x + 2 a b x \log {\relax (c )} + b^{2} n^{2} x \log {\relax (x )}^{2} - 2 b^{2} n^{2} x \log {\relax (x )} + 2 b^{2} n^{2} x + 2 b^{2} n x \log {\relax (c )} \log {\relax (x )} - 2 b^{2} n x \log {\relax (c )} + b^{2} x \log {\relax (c )}^{2}\right ) & \text {for}\: f = 0 \\\frac {a^{2} f^{m} x^{m}}{f m} + \frac {2 a b f^{m} n x^{m} \log {\relax (x )}}{f m} + \frac {2 a b f^{m} x^{m} \log {\relax (c )}}{f m} - \frac {2 a b f^{m} n x^{m}}{f m^{2}} + \frac {b^{2} f^{m} n^{2} x^{m} \log {\relax (x )}^{2}}{f m} + \frac {2 b^{2} f^{m} n x^{m} \log {\relax (c )} \log {\relax (x )}}{f m} + \frac {b^{2} f^{m} x^{m} \log {\relax (c )}^{2}}{f m} - \frac {2 b^{2} f^{m} n^{2} x^{m} \log {\relax (x )}}{f m^{2}} - \frac {2 b^{2} f^{m} n x^{m} \log {\relax (c )}}{f m^{2}} + \frac {2 b^{2} f^{m} n^{2} x^{m}}{f m^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**(-1+m)*(a+b*ln(c*x**n))**2,x)

[Out]

Piecewise((zoo*(a**2*x + 2*a*b*n*x*log(x) - 2*a*b*n*x + 2*a*b*x*log(c) + b**2*n**2*x*log(x)**2 - 2*b**2*n**2*x
*log(x) + 2*b**2*n**2*x + 2*b**2*n*x*log(c)*log(x) - 2*b**2*n*x*log(c) + b**2*x*log(c)**2), Eq(f, 0) & Eq(m, 0
)), (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*l
og(c) + b**2*log(c)**2)*log(x), True))/f, Eq(m, 0)), (0**(m - 1)*(a**2*x + 2*a*b*n*x*log(x) - 2*a*b*n*x + 2*a*
b*x*log(c) + b**2*n**2*x*log(x)**2 - 2*b**2*n**2*x*log(x) + 2*b**2*n**2*x + 2*b**2*n*x*log(c)*log(x) - 2*b**2*
n*x*log(c) + b**2*x*log(c)**2), Eq(f, 0)), (a**2*f**m*x**m/(f*m) + 2*a*b*f**m*n*x**m*log(x)/(f*m) + 2*a*b*f**m
*x**m*log(c)/(f*m) - 2*a*b*f**m*n*x**m/(f*m**2) + b**2*f**m*n**2*x**m*log(x)**2/(f*m) + 2*b**2*f**m*n*x**m*log
(c)*log(x)/(f*m) + b**2*f**m*x**m*log(c)**2/(f*m) - 2*b**2*f**m*n**2*x**m*log(x)/(f*m**2) - 2*b**2*f**m*n*x**m
*log(c)/(f*m**2) + 2*b**2*f**m*n**2*x**m/(f*m**3), True))

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