3.2.21 \(\int \frac {1}{\sqrt {x} \sqrt {x (a+b x+c x^2)}} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {x} (2 a+b x)}{2 \sqrt {a} \sqrt {a x+b x^2+c x^3}}\right )}{\sqrt {a}} \]

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Rubi [A]  time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1997, 1913, 206} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {x} (2 a+b x)}{2 \sqrt {a} \sqrt {a x+b x^2+c x^3}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 1913

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 1997

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

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Mathematica [A]  time = 0.03, size = 72, normalized size = 1.53 \begin {gather*} -\frac {\sqrt {x} \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} \sqrt {x (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 53, normalized size = 1.13 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {c} x^{3/2}-\sqrt {a x+b x^2+c x^3}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.34, size = 131, normalized size = 2.79 \begin {gather*} \left [\frac {\log \left (\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{3} + b x^{2} + a x} {\left (b x + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{3}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{3} + b x^{2} + a x} {\left (b x + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right )}{a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.49, size = 53, normalized size = 1.13 \begin {gather*} \frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, \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^2+b*x+a))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 64, normalized size = 1.36 \begin {gather*} -\frac {\sqrt {c \,x^{2}+b x +a}\, \sqrt {x}\, \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{\sqrt {\left (c \,x^{2}+b x +a \right ) x}\, \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {{\left (c x^{2} + b x + a\right )} x} \sqrt {x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2+b\,x+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 + c*x^2))^(1/2)),x)

[Out]

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

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sympy [F(-1)]  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**2+b*x+a))**(1/2),x)

[Out]

Timed out

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