\(\int \frac {a m x^m+b n q \log ^{-1+q}(c x^n)}{x (a x^m+b \log ^q(c x^n))^3} \, dx\) [22]

Optimal result

Integrand size = 43, antiderivative size = 22 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \left (a x^m+b \log ^q\left (c x^n\right )\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2624} \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \left (a x^m+b \log ^q\left (c x^n\right )\right )^2} \]

[In]

Int[(a*m*x^m + b*n*q*Log[c*x^n]^(-1 + q))/(x*(a*x^m + b*Log[c*x^n]^q)^3),x]

[Out]

-1/2*1/(a*x^m + b*Log[c*x^n]^q)^2

Rule 2624

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \left (a x^m+b \log ^q\left (c x^n\right )\right )^2} \]

[In]

Integrate[(a*m*x^m + b*n*q*Log[c*x^n]^(-1 + q))/(x*(a*x^m + b*Log[c*x^n]^q)^3),x]

[Out]

-1/2*1/(a*x^m + b*Log[c*x^n]^q)^2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 116.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 3.09

method	result	size
risch	\(-\frac {1}{2 \left (a \,x^{m}+b {\left (\ln \left (c \right )+\ln \left (x^{n}\right )-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i x^{n}\right )\right )}{2}\right )}^{q}\right )^{2}}\)	\(68\)

[In]

int((a*m*x^m+b*n*q*ln(c*x^n)^(-1+q))/x/(a*x^m+b*ln(c*x^n)^q)^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/(a*x^m+b*(ln(c)+ln(x^n)-1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*c*x^n)+csgn(I*c))*(-csgn(I*c*x^n)+csgn(I*x^n)))^q
)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (20) = 40\).

Time = 0.34 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \, {\left (2 \, {\left (n \log \left (x\right ) + \log \left (c\right )\right )}^{q} a b x^{m} + {\left (n \log \left (x\right ) + \log \left (c\right )\right )}^{2 \, q} b^{2} + a^{2} x^{2 \, m}\right )}} \]

[In]

integrate((a*m*x^m+b*n*q*log(c*x^n)^(-1+q))/x/(a*x^m+b*log(c*x^n)^q)^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((a*m*x**m+b*n*q*ln(c*x**n)**(-1+q))/x/(a*x**m+b*ln(c*x**n)**q)**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).

Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.23 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \, {\left (a^{2} x^{2 \, m} + b^{2} {\left (\log \left (c\right ) + \log \left (x^{n}\right )\right )}^{2 \, q} + 2 \, a b e^{\left (m \log \left (x\right ) + q \log \left (\log \left (c\right ) + \log \left (x^{n}\right )\right )\right )}\right )}} \]

[In]

integrate((a*m*x^m+b*n*q*log(c*x^n)^(-1+q))/x/(a*x^m+b*log(c*x^n)^q)^3,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=\int { \frac {b n q \log \left (c x^{n}\right )^{q - 1} + a m x^{m}}{{\left (a x^{m} + b \log \left (c x^{n}\right )^{q}\right )}^{3} x} \,d x } \]

[In]

integrate((a*m*x^m+b*n*q*log(c*x^n)^(-1+q))/x/(a*x^m+b*log(c*x^n)^q)^3,x, algorithm="giac")

[Out]

integrate((b*n*q*log(c*x^n)^(q - 1) + a*m*x^m)/((a*x^m + b*log(c*x^n)^q)^3*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x \left (a x^m+b \log ^q\left (c x^n\right )\right )^3} \, dx=\int \frac {a\,m\,x^m+b\,n\,q\,{\ln \left (c\,x^n\right )}^{q-1}}{x\,{\left (a\,x^m+b\,{\ln \left (c\,x^n\right )}^q\right )}^3} \,d x \]

[In]

int((a*m*x^m + b*n*q*log(c*x^n)^(q - 1))/(x*(a*x^m + b*log(c*x^n)^q)^3),x)

[Out]

int((a*m*x^m + b*n*q*log(c*x^n)^(q - 1))/(x*(a*x^m + b*log(c*x^n)^q)^3), x)