\(\int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx\) [671]

Optimal result

Integrand size = 17, antiderivative size = 30 \[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b (c x)^m}}{\sqrt {a}}\right )}{\sqrt {a} m} \]

[Out]

-2*arctanh((a+b*(c*x)^m)^(1/2)/a^(1/2))/m/a^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {374, 12, 272, 65, 214} \[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b (c x)^m}}{\sqrt {a}}\right )}{\sqrt {a} m} \]

[In]

Int[1/(x*Sqrt[a + b*(c*x)^m]),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*(c*x)^m]/Sqrt[a]])/(Sqrt[a]*m)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 374

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {c}{x \sqrt {a+b x^m}} \, dx,x,c x\right )}{c} \\ & = \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x^m}} \, dx,x,c x\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,(c x)^m\right )}{m} \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b (c x)^m}\right )}{b m} \\ & = -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^m}}{\sqrt {a}}\right )}{\sqrt {a} m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b (c x)^m}}{\sqrt {a}}\right )}{\sqrt {a} m} \]

[In]

Integrate[1/(x*Sqrt[a + b*(c*x)^m]),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*(c*x)^m]/Sqrt[a]])/(Sqrt[a]*m)

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \left (c x \right )^{m}}}{\sqrt {a}}\right )}{m \sqrt {a}}\)	\(25\)
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \left (c x \right )^{m}}}{\sqrt {a}}\right )}{m \sqrt {a}}\)	\(25\)

[In]

int(1/x/(a+b*(c*x)^m)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*arctanh((a+b*(c*x)^m)^(1/2)/a^(1/2))/m/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.60 \[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=\left [\frac {\log \left (\frac {\left (c x\right )^{m} b - 2 \, \sqrt {\left (c x\right )^{m} b + a} \sqrt {a} + 2 \, a}{\left (c x\right )^{m}}\right )}{\sqrt {a} m}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {\left (c x\right )^{m} b + a} \sqrt {-a}}{a}\right )}{a m}\right ] \]

[In]

integrate(1/x/(a+b*(c*x)^m)^(1/2),x, algorithm="fricas")

[Out]

[log(((c*x)^m*b - 2*sqrt((c*x)^m*b + a)*sqrt(a) + 2*a)/(c*x)^m)/(sqrt(a)*m), 2*sqrt(-a)*arctan(sqrt((c*x)^m*b
+ a)*sqrt(-a)/a)/(a*m)]

Sympy [F]

\[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=\int \frac {1}{x \sqrt {a + b \left (c x\right )^{m}}}\, dx \]

[In]

integrate(1/x/(a+b*(c*x)**m)**(1/2),x)

[Out]

Integral(1/(x*sqrt(a + b*(c*x)**m)), x)

Maxima [F]

\[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=\int { \frac {1}{\sqrt {\left (c x\right )^{m} b + a} x} \,d x } \]

[In]

integrate(1/x/(a+b*(c*x)^m)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt((c*x)^m*b + a)*x), x)

Giac [F]

\[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=\int { \frac {1}{\sqrt {\left (c x\right )^{m} b + a} x} \,d x } \]

[In]

integrate(1/x/(a+b*(c*x)^m)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt((c*x)^m*b + a)*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {a+b (c x)^m}} \, dx=\int \frac {1}{x\,\sqrt {a+b\,{\left (c\,x\right )}^m}} \,d x \]

[In]

int(1/(x*(a + b*(c*x)^m)^(1/2)),x)

[Out]

int(1/(x*(a + b*(c*x)^m)^(1/2)), x)