\(\int \frac {a m x^m+b n q \log ^{-1+q}(c x^n)}{x} \, dx\) [19]

Optimal result

Integrand size = 25, antiderivative size = 16 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=a x^m+b \log ^q\left (c x^n\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {14, 2339, 30} \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=a x^m+b \log ^q\left (c x^n\right ) \]

[In]

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

[Out]

a*x^m + b*Log[c*x^n]^q

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=a x^m+b \log ^q\left (c x^n\right ) \]

[In]

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

[Out]

a*x^m + b*Log[c*x^n]^q

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06

method	result	size
default	\(a \,x^{m}+b \ln \left (c \,x^{n}\right )^{q}\)	\(17\)
parallelrisch	\(\ln \left (c \,x^{n}\right ) \ln \left (c \,x^{n}\right )^{-1+q} b +a \,x^{m}\)	\(25\)
risch	\(a \,x^{m}+b {\left (\ln \left (c \right )+\ln \left (x^{n}\right )-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i x^{n}\right )\right )}{2}\right )}^{-1+q} \left (\ln \left (c \right )+\ln \left (x^{n}\right )-\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \,x^{n}\right )+\operatorname {csgn}\left (i x^{n}\right )\right )}{2}\right )\)	\(119\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.75 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx={\left (b n \log \left (x\right ) + b \log \left (c\right )\right )} {\left (n \log \left (x\right ) + \log \left (c\right )\right )}^{q - 1} + a x^{m} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 49, normalized size of antiderivative = 3.06 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=- a m \left (\begin {cases} - \log {\left (x \right )} & \text {for}\: m = 0 \\- \frac {x^{m}}{m} & \text {otherwise} \end {cases}\right ) + b n q \left (\begin {cases} \log {\left (c \right )}^{q - 1} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {\begin {cases} \frac {\log {\left (c x^{n} \right )}^{q}}{q} & \text {for}\: q \neq 0 \\\log {\left (\log {\left (c x^{n} \right )} \right )} & \text {otherwise} \end {cases}}{n} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

-a*m*Piecewise((-log(x), Eq(m, 0)), (-x**m/m, True)) + b*n*q*Piecewise((log(c)**(q - 1)*log(x), Eq(n, 0)), (Pi
ecewise((log(c*x**n)**q/q, Ne(q, 0)), (log(log(c*x**n)), True))/n, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=a x^{m} + b \log \left (c x^{n}\right )^{q} \]

[In]

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

[Out]

a*x^m + b*log(c*x^n)^q

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx={\left (n \log \left (x\right ) + \log \left (c\right )\right )}^{q} b + a x^{m} \]

[In]

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

[Out]

(n*log(x) + log(c))^q*b + a*x^m

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {a m x^m+b n q \log ^{-1+q}\left (c x^n\right )}{x} \, dx=a\,x^m+b\,{\ln \left (c\,x^n\right )}^q \]

[In]

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

[Out]

a*x^m + b*log(c*x^n)^q