3.8.72 \(\int \frac {\sqrt {a+c x^4}}{x^3} \, dx\) [772]

Optimal. Leaf size=49 \[ -\frac {\sqrt {a+c x^4}}{2 x^2}+\frac {1}{2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, 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 = {281, 283, 223, 212} \begin {gather*} \frac {1}{2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )-\frac {\sqrt {a+c x^4}}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4]/x^3,x]

[Out]

-1/2*Sqrt[a + c*x^4]/x^2 + (Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[a + c*x^4]])/2

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+c x^4}}{x^3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a+c x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+c x^4}}{2 x^2}+\frac {1}{2} c \text {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+c x^4}}{2 x^2}+\frac {1}{2} c \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {a+c x^4}}\right )\\ &=-\frac {\sqrt {a+c x^4}}{2 x^2}+\frac {1}{2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 49, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+c x^4}}{2 x^2}+\frac {1}{2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {c} x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4]/x^3,x]

[Out]

-1/2*Sqrt[a + c*x^4]/x^2 + (Sqrt[c]*ArcTanh[Sqrt[a + c*x^4]/(Sqrt[c]*x^2)])/2

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Maple [A]
time = 0.14, size = 60, normalized size = 1.22

method	result	size
risch	\(-\frac {\sqrt {x^{4} c +a}}{2 x^{2}}+\frac {\sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{2}\)	\(39\)
default	\(-\frac {\left (x^{4} c +a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {c \,x^{2} \sqrt {x^{4} c +a}}{2 a}+\frac {\sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{2}\)	\(60\)
elliptic	\(-\frac {\left (x^{4} c +a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {c \,x^{2} \sqrt {x^{4} c +a}}{2 a}+\frac {\sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{2}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/a/x^2*(c*x^4+a)^(3/2)+1/2/a*c*x^2*(c*x^4+a)^(1/2)+1/2*c^(1/2)*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))

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Maxima [A]
time = 0.50, size = 60, normalized size = 1.22 \begin {gather*} -\frac {1}{4} \, \sqrt {c} \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right ) - \frac {\sqrt {c x^{4} + a}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 96, normalized size = 1.96 \begin {gather*} \left [\frac {\sqrt {c} x^{2} \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) - 2 \, \sqrt {c x^{4} + a}}{4 \, x^{2}}, -\frac {\sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c} x^{2}}{\sqrt {c x^{4} + a}}\right ) + \sqrt {c x^{4} + a}}{2 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.69, size = 66, normalized size = 1.35 \begin {gather*} - \frac {\sqrt {a}}{2 x^{2} \sqrt {1 + \frac {c x^{4}}{a}}} + \frac {\sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{2} - \frac {c x^{2}}{2 \sqrt {a} \sqrt {1 + \frac {c x^{4}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(1/2)/x**3,x)

[Out]

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

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Giac [A]
time = 0.49, size = 60, normalized size = 1.22 \begin {gather*} -\frac {1}{4} \, \sqrt {c} \log \left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2}\right ) + \frac {a \sqrt {c}}{{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2} - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {c\,x^4+a}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2)/x^3,x)

[Out]

int((a + c*x^4)^(1/2)/x^3, x)

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