3.2.22 \(\int \frac {\sqrt {x}}{\sqrt {x^3 (a+b x+c x^2)}} \, dx\)

Optimal. Leaf size=49 \[ -\frac {\tanh ^{-1}\left (\frac {x^{3/2} (2 a+b x)}{2 \sqrt {a} \sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}} \]

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Rubi [A]  time = 0.09, antiderivative size = 49, 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 = {1997, 1913, 206} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {x^{3/2} (2 a+b x)}{2 \sqrt {a} \sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[x^3*(a + b*x + c*x^2)],x]

[Out]

-(ArcTanh[(x^(3/2)*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^3 + b*x^4 + c*x^5])]/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 {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx &=\int \frac {\sqrt {x}}{\sqrt {a x^3+b x^4+c x^5}} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x^{3/2} (2 a+b x)}{\sqrt {a x^3+b x^4+c x^5}}\right )\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {x^{3/2} (2 a+b x)}{2 \sqrt {a} \sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 74, normalized size = 1.51 \begin {gather*} -\frac {x^{3/2} \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^3 (a+x (b+c x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/Sqrt[x^3*(a + b*x + c*x^2)],x]

[Out]

-((x^(3/2)*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^3*(a
+ x*(b + c*x))]))

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IntegrateAlgebraic [A]  time = 0.32, size = 55, normalized size = 1.12 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {c} x^{5/2}-\sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]/Sqrt[x^3*(a + b*x + c*x^2)],x]

[Out]

(2*ArcTanh[(Sqrt[a]*x^(3/2))/(Sqrt[c]*x^(5/2) - Sqrt[a*x^3 + b*x^4 + c*x^5])])/Sqrt[a]

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fricas [A]  time = 1.15, size = 139, normalized size = 2.84 \begin {gather*} \left [\frac {\log \left (\frac {8 \, a b x^{3} + {\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a^{2} x^{2} - 4 \, \sqrt {c x^{5} + b x^{4} + a x^{3}} {\left (b x + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{4}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{4} + a x^{3}} {\left (b x + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{4} + a b x^{3} + a^{2} x^{2}\right )}}\right )}{a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.49, size = 53, normalized size = 1.08 \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(x^(1/2)/(x^3*(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 = 66, normalized size = 1.35 \begin {gather*} -\frac {\sqrt {c \,x^{2}+b x +a}\, x^{\frac {3}{2}} \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^{3}}\, \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x^3*(c*x^2+b*x+a))^(1/2),x)

[Out]

-1/(x^3*(c*x^2+b*x+a))^(1/2)*x^(3/2)*(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 {\sqrt {x}}{\sqrt {{\left (c x^{2} + b x + a\right )} x^{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x)/sqrt((c*x^2 + b*x + a)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {x}}{\sqrt {x^3\,\left (c\,x^2+b\,x+a\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x^3*(a + b*x + c*x^2))^(1/2),x)

[Out]

int(x^(1/2)/(x^3*(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(x**(1/2)/(x**3*(c*x**2+b*x+a))**(1/2),x)

[Out]

Timed out

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