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

Optimal. Leaf size=20 \[ -\frac {1}{a x^m+b \log ^q\left (c x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2544} \[ -\frac {1}{a x^m+b \log ^q\left (c x^n\right )} \]

Antiderivative was successfully verified.

[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)^2),x]

[Out]

-(a*x^m + b*Log[c*x^n]^q)^(-1)

Rule 2544

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

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Mathematica [A]  time = 0.04, size = 20, normalized size = 1.00 \[ -\frac {1}{a x^m+b \log ^q\left (c x^n\right )} \]

Antiderivative was successfully verified.

[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)^2),x]

[Out]

-(a*x^m + b*Log[c*x^n]^q)^(-1)

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fricas [A]  time = 0.44, size = 21, normalized size = 1.05 \[ -\frac {1}{{\left (n \log \relax (x) + \log \relax (c)\right )}^{q} b + a x^{m}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,x, algorithm="fricas")

[Out]

-1/((n*log(x) + log(c))^q*b + a*x^m)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b n q \log \left (c x^{n}\right )^{q - 1} + a m x^{m}}{{\left (a x^{m} + b \log \left (c x^{n}\right )^{q}\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,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)^2*x), x)

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maple [C]  time = 0.54, size = 68, normalized size = 3.40 \[ -\frac {1}{a \,x^{m}+b \left (-\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \left (\mathrm {csgn}\left (i x^{n}\right )-\mathrm {csgn}\left (i c \,x^{n}\right )\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2}+\ln \relax (c )+\ln \left (x^{n}\right )\right )^{q}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.22, size = 21, normalized size = 1.05 \[ -\frac {1}{a x^{m} + b {\left (\log \relax (c) + \log \left (x^{n}\right )\right )}^{q}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^2,x, algorithm="maxima")

[Out]

-1/(a*x^m + b*(log(c) + log(x^n))^q)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {a\,m\,x^m+b\,n\,q\,{\ln \left (c\,x^n\right )}^{q-1}}{x\,{\left (a\,x^m+b\,{\ln \left (c\,x^n\right )}^q\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)^2),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)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)**2,x)

[Out]

Timed out

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