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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {12, 266, 50, 63, 208} \[ \frac {2 \sqrt {a+b x^n}}{c n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{c n} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^n]/(c*x),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 0.90 \[ \frac {2 \sqrt {a+b x^n}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{c n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^n]/(c*x),x]

[Out]

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

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fricas [A]  time = 0.42, size = 97, normalized size = 1.90 \[ \left [\frac {\sqrt {a} \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, \sqrt {b x^{n} + a}}{c n}, \frac {2 \, {\left (\sqrt {-a} \arctan \left (\frac {\sqrt {b x^{n} + a} \sqrt {-a}}{a}\right ) + \sqrt {b x^{n} + a}\right )}}{c n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^n + a)/(c*x), x)

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maple [A]  time = 0.04, size = 39, normalized size = 0.76 \[ \frac {-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b \,x^{n}+a}}{\sqrt {a}}\right )+2 \sqrt {b \,x^{n}+a}}{c n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 58, normalized size = 1.14 \[ \frac {\frac {\sqrt {a} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, \sqrt {b x^{n} + a}}{n}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(a)*log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/n + 2*sqrt(b*x^n + a)/n)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^(1/2)/(c*x),x)

[Out]

int((a + b*x^n)^(1/2)/(c*x), x)

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sympy [A]  time = 1.78, size = 78, normalized size = 1.53 \[ \frac {- \frac {2 \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{- \frac {n}{2}}}{\sqrt {b}} \right )}}{n} + \frac {2 a x^{- \frac {n}{2}}}{\sqrt {b} n \sqrt {\frac {a x^{- n}}{b} + 1}} + \frac {2 \sqrt {b} x^{\frac {n}{2}}}{n \sqrt {\frac {a x^{- n}}{b} + 1}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*sqrt(a)*asinh(sqrt(a)*x**(-n/2)/sqrt(b))/n + 2*a*x**(-n/2)/(sqrt(b)*n*sqrt(a*x**(-n)/b + 1)) + 2*sqrt(b)*x
**(n/2)/(n*sqrt(a*x**(-n)/b + 1)))/c

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