3.7.74 \(\int \frac {1}{x \sqrt {a+b (c (d (e (f x)^m)^n)^p)^q}} \, dx\) [674]

Optimal. Leaf size=51 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}}{\sqrt {a}}\right )}{\sqrt {a} m n p q} \]

[Out]

-2*arctanh((a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2)/a^(1/2))/m/n/p/q/a^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.46, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {374, 12, 272, 65, 214} \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}}{\sqrt {a}}\right )}{\sqrt {a} m n p q} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a + b*(c*(d*(e*(f*x)^m)^n)^p)^q]),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*(c*(d*(e*(f*x)^m)^n)^p)^q]/Sqrt[a]])/(Sqrt[a]*m*n*p*q)

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*} \int \frac {1}{x \sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b \left (c \left (d (e x)^n\right )^p\right )^q}} \, dx,x,(f x)^m\right )}{m}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b \left (c (d x)^p\right )^q}} \, dx,x,\left (e (f x)^m\right )^n\right )}{m n}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b (c x)^q}} \, dx,x,\left (d \left (e (f x)^m\right )^n\right )^p\right )}{m n p}\\ &=\frac {\text {Subst}\left (\int \frac {c}{x \sqrt {a+b x^q}} \, dx,x,c \left (d \left (e (f x)^m\right )^n\right )^p\right )}{c m n p}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x^q}} \, dx,x,c \left (d \left (e (f x)^m\right )^n\right )^p\right )}{m n p}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q\right )}{m n p q}\\ &=\frac {2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}\right )}{b m n p q}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}}{\sqrt {a}}\right )}{\sqrt {a} m n p q}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.23, size = 51, normalized size = 1.00 \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c \left (d \left (e (f x)^m\right )^n\right )^p\right )^q}}{\sqrt {a}}\right )}{\sqrt {a} m n p q} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[a + b*(c*(d*(e*(f*x)^m)^n)^p)^q]),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*(c*(d*(e*(f*x)^m)^n)^p)^q]/Sqrt[a]])/(Sqrt[a]*m*n*p*q)

________________________________________________________________________________________

Maple [A]
time = 0.44, size = 46, normalized size = 0.90

method	result	size
derivativedivides	\(-\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (c \left (d \left (e \left (f x \right )^{m}\right )^{n}\right )^{p}\right )^{q}}}{\sqrt {a}}\right )}{m n p q \sqrt {a}}\)	\(46\)
default	\(-\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (c \left (d \left (e \left (f x \right )^{m}\right )^{n}\right )^{p}\right )^{q}}}{\sqrt {a}}\right )}{m n p q \sqrt {a}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*arctanh((a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2)/a^(1/2))/m/n/p/q/a^(1/2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(((((f*x)^m*e)^n*d)^p*c)^q*b + a)*x), x)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 174, normalized size = 3.41 \begin {gather*} \left [\frac {\log \left ({\left (b e^{\left (m n p q \log \left (f x\right ) + n p q + p q \log \left (d\right ) + q \log \left (c\right )\right )} - 2 \, \sqrt {b e^{\left (m n p q \log \left (f x\right ) + n p q + p q \log \left (d\right ) + q \log \left (c\right )\right )} + a} \sqrt {a} + 2 \, a\right )} e^{\left (-m n p q \log \left (f x\right ) - n p q - p q \log \left (d\right ) - q \log \left (c\right )\right )}\right )}{\sqrt {a} m n p q}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b e^{\left (m n p q \log \left (f x\right ) + n p q + p q \log \left (d\right ) + q \log \left (c\right )\right )} + a} \sqrt {-a}}{a}\right )}{a m n p q}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2),x, algorithm="fricas")

[Out]

[log((b*e^(m*n*p*q*log(f*x) + n*p*q + p*q*log(d) + q*log(c)) - 2*sqrt(b*e^(m*n*p*q*log(f*x) + n*p*q + p*q*log(
d) + q*log(c)) + a)*sqrt(a) + 2*a)*e^(-m*n*p*q*log(f*x) - n*p*q - p*q*log(d) - q*log(c)))/(sqrt(a)*m*n*p*q), 2
*sqrt(-a)*arctan(sqrt(b*e^(m*n*p*q*log(f*x) + n*p*q + p*q*log(d) + q*log(c)) + a)*sqrt(-a)/a)/(a*m*n*p*q)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {a + b \left (c \left (d \left (e \left (f x\right )^{m}\right )^{n}\right )^{p}\right )^{q}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*(d*(e*(f*x)**m)**n)**p)**q)**(1/2),x)

[Out]

Integral(1/(x*sqrt(a + b*(c*(d*(e*(f*x)**m)**n)**p)**q)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(a+b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(((((f*x)^m*e)^n*d)^p*c)^q*b + a)*x), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x\,\sqrt {a+b\,{\left (c\,{\left (d\,{\left (e\,{\left (f\,x\right )}^m\right )}^n\right )}^p\right )}^q}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2)),x)

[Out]

int(1/(x*(a + b*(c*(d*(e*(f*x)^m)^n)^p)^q)^(1/2)), x)

________________________________________________________________________________________