\(\int \frac {a+b \log (c x^n)}{x^2} \, dx\) [48]

Optimal result

Integrand size = 14, antiderivative size = 23 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {b n}{x}-\frac {a+b \log \left (c x^n\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2341} \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x} \]

[In]

Int[(a + b*Log[c*x^n])/x^2,x]

[Out]

-((b*n)/x) - (a + b*Log[c*x^n])/x

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
]

Rubi steps \begin{align*} \text {integral}& = -\frac {b n}{x}-\frac {a+b \log \left (c x^n\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {a}{x}-\frac {b n}{x}-\frac {b \log \left (c x^n\right )}{x} \]

[In]

Integrate[(a + b*Log[c*x^n])/x^2,x]

[Out]

-(a/x) - (b*n)/x - (b*Log[c*x^n])/x

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(-\frac {b \ln \left (c \,x^{n}\right )+b n +a}{x}\)	\(19\)
risch	\(-\frac {b \ln \left (x^{n}\right )}{x}-\frac {-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 b n +2 a}{2 x}\)	\(112\)

[In]

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

[Out]

-1/x*(b*ln(c*x^n)+b*n+a)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {b n \log \left (x\right ) + b n + b \log \left (c\right ) + a}{x} \]

[In]

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

[Out]

-(b*n*log(x) + b*n + b*log(c) + a)/x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=- \frac {a}{x} - \frac {b n}{x} - \frac {b \log {\left (c x^{n} \right )}}{x} \]

[In]

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

[Out]

-a/x - b*n/x - b*log(c*x**n)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {b n}{x} - \frac {b \log \left (c x^{n}\right )}{x} - \frac {a}{x} \]

[In]

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

[Out]

-b*n/x - b*log(c*x^n)/x - a/x

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {b n \log \left (x\right )}{x} - \frac {b n + b \log \left (c\right ) + a}{x} \]

[In]

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

[Out]

-b*n*log(x)/x - (b*n + b*log(c) + a)/x

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2} \, dx=-\frac {a+b\,n}{x}-\frac {b\,\ln \left (c\,x^n\right )}{x} \]

[In]

int((a + b*log(c*x^n))/x^2,x)

[Out]

- (a + b*n)/x - (b*log(c*x^n))/x