3.660 \(\int \frac {\sqrt {a+b (c x)^n}}{x} \, dx\)

Optimal. Leaf size=49 \[ \frac {2 \sqrt {a+b (c x)^n}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {367, 12, 266, 50, 63, 208} \[ \frac {2 \sqrt {a+b (c x)^n}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x)^n]/x,x]

[Out]

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

Rule 12

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

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

Rule 266

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 367

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 {\sqrt {a+b (c x)^n}}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {c \sqrt {a+b x^n}}{x} \, dx,x,c x\right )}{c}\\ &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^n}}{x} \, dx,x,c x\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,(c x)^n\right )}{n}\\ &=\frac {2 \sqrt {a+b (c x)^n}}{n}+\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,(c x)^n\right )}{n}\\ &=\frac {2 \sqrt {a+b (c x)^n}}{n}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b (c x)^n}\right )}{b n}\\ &=\frac {2 \sqrt {a+b (c x)^n}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 0.96 \[ \frac {2 \sqrt {a+b (c x)^n}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*(c*x)^n]/x,x]

[Out]

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

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fricas [A]  time = 0.50, size = 103, normalized size = 2.10 \[ \left [\frac {\sqrt {a} \log \left (\frac {\left (c x\right )^{n} b - 2 \, \sqrt {\left (c x\right )^{n} b + a} \sqrt {a} + 2 \, a}{\left (c x\right )^{n}}\right ) + 2 \, \sqrt {\left (c x\right )^{n} b + a}}{n}, \frac {2 \, {\left (\sqrt {-a} \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b + a} \sqrt {-a}}{a}\right ) + \sqrt {\left (c x\right )^{n} b + a}\right )}}{n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((c*x)^n*b + a)/x, x)

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maple [A]  time = 0.00, size = 40, normalized size = 0.82 \[ \frac {-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \left (c x \right )^{n}+a}}{\sqrt {a}}\right )+2 \sqrt {b \left (c x \right )^{n}+a}}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*(c*x)^n+a)^(1/2)/x,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\left (c x\right )^{n} b + a}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((c*x)^n*b + a)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {a+b\,{\left (c\,x\right )}^n}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c*x)^n)^(1/2)/x,x)

[Out]

int((a + b*(c*x)^n)^(1/2)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \left (c x\right )^{n}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*(c*x)**n)/x, x)

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