3.767 \(\int \frac {\sqrt {a+c x^4}}{x} \, dx\)

Optimal. Leaf size=43 \[ \frac {1}{2} \sqrt {a+c x^4}-\frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \]

[Out]

-1/2*arctanh((c*x^4+a)^(1/2)/a^(1/2))*a^(1/2)+1/2*(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ \frac {1}{2} \sqrt {a+c x^4}-\frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/x,x]

[Out]

Sqrt[a + c*x^4]/2 - (Sqrt[a]*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/2

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+c x^4}}{x} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x} \, dx,x,x^4\right )\\ &=\frac {1}{2} \sqrt {a+c x^4}+\frac {1}{4} a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^4\right )\\ &=\frac {1}{2} \sqrt {a+c x^4}+\frac {a \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^4}\right )}{2 c}\\ &=\frac {1}{2} \sqrt {a+c x^4}-\frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.00 \[ \frac {1}{2} \sqrt {a+c x^4}-\frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4]/x,x]

[Out]

Sqrt[a + c*x^4]/2 - (Sqrt[a]*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/2

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fricas [A]  time = 0.77, size = 84, normalized size = 1.95 \[ \left [\frac {1}{4} \, \sqrt {a} \log \left (\frac {c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + a}, \frac {1}{2} \, \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + a} \sqrt {-a}}{a}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/4*sqrt(a)*log((c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) + 1/2*sqrt(c*x^4 + a), 1/2*sqrt(-a)*arctan(sqr
t(c*x^4 + a)*sqrt(-a)/a) + 1/2*sqrt(c*x^4 + a)]

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giac [A]  time = 0.15, size = 36, normalized size = 0.84 \[ \frac {a \arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} + \frac {1}{2} \, \sqrt {c x^{4} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x,x, algorithm="giac")

[Out]

1/2*a*arctan(sqrt(c*x^4 + a)/sqrt(-a))/sqrt(-a) + 1/2*sqrt(c*x^4 + a)

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maple [A]  time = 0.01, size = 41, normalized size = 0.95 \[ -\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {c \,x^{4}+a}\, \sqrt {a}}{x^{2}}\right )}{2}+\frac {\sqrt {c \,x^{4}+a}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(1/2)/x,x)

[Out]

1/2*(c*x^4+a)^(1/2)-1/2*a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^4+a)^(1/2))/x^2)

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maxima [A]  time = 3.04, size = 49, normalized size = 1.14 \[ \frac {1}{4} \, \sqrt {a} \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x,x, algorithm="maxima")

[Out]

1/4*sqrt(a)*log((sqrt(c*x^4 + a) - sqrt(a))/(sqrt(c*x^4 + a) + sqrt(a))) + 1/2*sqrt(c*x^4 + a)

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mupad [B]  time = 1.20, size = 31, normalized size = 0.72 \[ \frac {\sqrt {c\,x^4+a}}{2}-\frac {\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)/x,x)

[Out]

(a + c*x^4)^(1/2)/2 - (a^(1/2)*atanh((a + c*x^4)^(1/2)/a^(1/2)))/2

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sympy [A]  time = 2.36, size = 66, normalized size = 1.53 \[ - \frac {\sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x^{2}} \right )}}{2} + \frac {a}{2 \sqrt {c} x^{2} \sqrt {\frac {a}{c x^{4}} + 1}} + \frac {\sqrt {c} x^{2}}{2 \sqrt {\frac {a}{c x^{4}} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(1/2)/x,x)

[Out]

-sqrt(a)*asinh(sqrt(a)/(sqrt(c)*x**2))/2 + a/(2*sqrt(c)*x**2*sqrt(a/(c*x**4) + 1)) + sqrt(c)*x**2/(2*sqrt(a/(c
*x**4) + 1))

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