3.7.60 \(\int \frac {\sqrt {a+b (c x)^n}}{x} \, dx\) [660]

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 = 0.03, 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 = {374, 12, 272, 52, 65, 214} \begin {gather*} \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 {gather*}

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 52

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 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 {\sqrt {a+b (c x)^n}}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {c \sqrt {a+b x^n}}{x} \, dx,x,c x\right )}{c}\\ &=\text {Subst}\left (\int \frac {\sqrt {a+b x^n}}{x} \, dx,x,c x\right )\\ &=\frac {\text {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 \text {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) \text {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 = 0.10, size = 46, normalized size = 0.94 \begin {gather*} \frac {2 \left (\sqrt {a+b (c x)^n}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b (c x)^n}}{\sqrt {a}}\right )\right )}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*(c*x)^n]/x,x]

[Out]

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

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Maple [A]
time = 0.57, size = 40, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {2 \sqrt {a +b \left (c x \right )^{n}}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(40\)
default	\(\frac {2 \sqrt {a +b \left (c x \right )^{n}}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{n}\)	\(40\)
risch	\(\frac {2 \sqrt {a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}{n}-\frac {2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (c x \right )}}}{\sqrt {a}}\right )}{n}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((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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Fricas [A]
time = 0.36, size = 103, normalized size = 2.10 \begin {gather*} \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 ] \end {gather*}

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

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

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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