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

Optimal. Leaf size=38 \[ \frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}-\frac {b n (f x)^m}{f m^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2304} \[ \frac {(f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}-\frac {b n (f x)^m}{f m^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 0.76 \[ \frac {(f x)^m \left (a m+b m \log \left (c x^n\right )-b n\right )}{f m^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 42, normalized size = 1.11 \[ \frac {{\left (b m n x \log \relax (x) + b m x \log \relax (c) + {\left (a m - b n\right )} x\right )} e^{\left ({\left (m - 1\right )} \log \relax (f) + {\left (m - 1\right )} \log \relax (x)\right )}}{m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.32, size = 80, normalized size = 2.11 \[ \frac {b \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m} \log \relax (c)}{f m} + \frac {b f^{m} n x^{m} \log \relax (x)}{f m} + \frac {a \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m}}{f m} - \frac {b f^{m} n x^{m}}{f m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 281, normalized size = 7.39 \[ \frac {b x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )+i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i f x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )^{2}-i \pi \mathrm {csgn}\left (i f x \right )^{3}+2 \ln \relax (f )+2 \ln \relax (x )\right )}{2}} \ln \left (x^{n}\right )}{m}+\frac {\left (-i \pi b m \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b m \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b m \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b m \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 b m \ln \relax (c )+2 a m -2 b n \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )+i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i f x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )^{2}-i \pi \mathrm {csgn}\left (i f x \right )^{3}+2 \ln \relax (f )+2 \ln \relax (x )\right )}{2}}}{2 m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/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)+1/2*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*m-I*Pi*b*csgn(I*x^
n)*csgn(I*c*x^n)*csgn(I*c)*m-I*Pi*b*csgn(I*c*x^n)^3*m+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)*m+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)*csg
n(I*f*x)^2-I*Pi*csgn(I*f*x)^3+2*ln(f)+2*ln(x)))

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maxima [A]  time = 0.72, size = 48, normalized size = 1.26 \[ -\frac {b f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a}{f m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (f\,x\right )}^{m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 22.25, size = 148, normalized size = 3.89 \[ \begin {cases} \tilde {\infty } \left (a x + b n x \log {\relax (x )} - b n x + b x \log {\relax (c )}\right ) & \text {for}\: f = 0 \wedge m = 0 \\\frac {\begin {cases} a \log {\relax (x )} & \text {for}\: b = 0 \\- \left (- a - b \log {\relax (c )}\right ) \log {\relax (x )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}}{f} & \text {for}\: m = 0 \\0^{m - 1} \left (a x + b n x \log {\relax (x )} - b n x + b x \log {\relax (c )}\right ) & \text {for}\: f = 0 \\\frac {a f^{m} x^{m}}{f m} + \frac {b f^{m} n x^{m} \log {\relax (x )}}{f m} + \frac {b f^{m} x^{m} \log {\relax (c )}}{f m} - \frac {b f^{m} n x^{m}}{f m^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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