3.2.35 \(\int \frac {1}{\sqrt {x} \sqrt {x (a+b x^2+c x^4)}} \, dx\) [135]

Optimal. Leaf size=51 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*arctanh(1/2*(b*x^2+2*a)*x^(1/2)/a^(1/2)/(c*x^5+b*x^3+a*x)^(1/2))/a^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2022, 1927, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*Sqrt[x*(a + b*x^2 + c*x^4)]),x]

[Out]

-1/2*ArcTanh[(Sqrt[x]*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x + b*x^3 + c*x^5])]/Sqrt[a]

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 1927

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - q), Sub
st[Int[1/(4*a - x^2), x], x, x^(m + 1)*((2*a + b*x^(n - q))/Sqrt[a*x^q + b*x^n + c*x^r])], x] /; FreeQ[{a, b,
c, m, n, q, r}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && NeQ[b^2 - 4*a*c, 0] && EqQ[m, q/2 - 1]

Rule 2022

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
GeneralizedTrinomialQ[u, x] &&  !GeneralizedTrinomialMatchQ[u, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {x} \sqrt {x \left (a+b x^2+c x^4\right )}} \, dx &=\int \frac {1}{\sqrt {x} \sqrt {a x+b x^3+c x^5}} \, dx\\ &=-\text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {\sqrt {x} \left (2 a+b x^2\right )}{\sqrt {a x+b x^3+c x^5}}\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {x} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x+b x^3+c x^5}}\right )}{2 \sqrt {a}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 80, normalized size = 1.57 \begin {gather*} \frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x \left (a+b x^2+c x^4\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*Sqrt[x*(a + b*x^2 + c*x^4)]),x]

[Out]

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

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Maple [A]
time = 0.02, size = 72, normalized size = 1.41

method	result	size
default	\(-\frac {\sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \sqrt {a}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 137, normalized size = 2.69 \begin {gather*} \left [\frac {\log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{5} + 8 \, a b x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{5}}\right )}{4 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{5} + a b x^{3} + a^{2} x\right )}}\right )}{2 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*log(-((b^2 + 4*a*c)*x^5 + 8*a*b*x^3 + 8*a^2*x - 4*sqrt(c*x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(a)*sqrt(x)
)/x^5)/sqrt(a), 1/2*sqrt(-a)*arctan(1/2*sqrt(c*x^5 + b*x^3 + a*x)*(b*x^2 + 2*a)*sqrt(-a)*sqrt(x)/(a*c*x^5 + a*
b*x^3 + a^2*x))/a]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 4.03, size = 56, normalized size = 1.10 \begin {gather*} \frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right )}{\sqrt {-a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {x}\,\sqrt {x\,\left (c\,x^4+b\,x^2+a\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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