\(\int \frac {\sqrt {x}}{\sqrt {x^3 (a+b x+c x^2)}} \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 49 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x^{3/2} (2 a+b x)}{2 \sqrt {a} \sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}} \]

[Out]

-arctanh(1/2*x^(3/2)*(b*x+2*a)/a^(1/2)/(c*x^5+b*x^4+a*x^3)^(1/2))/a^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2022, 1927, 212} \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x^{3/2} (2 a+b x)}{2 \sqrt {a} \sqrt {a x^3+b x^4+c x^5}}\right )}{\sqrt {a}} \]

[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 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*} \text {integral}& = \int \frac {\sqrt {x}}{\sqrt {a x^3+b x^4+c x^5}} \, dx \\ & = -\left (2 \text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\frac {2 x^{3/2} \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^3 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {x^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {x^{3} \left (c \,x^{2}+b x +a \right )}\, \sqrt {a}}\)	\(66\)

[In]

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

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.84 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\left [\frac {\log \left (\frac {8 \, a b x^{3} + {\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a^{2} x^{2} - 4 \, \sqrt {c x^{5} + b x^{4} + a x^{3}} {\left (b x + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{4}}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{5} + b x^{4} + a x^{3}} {\left (b x + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{4} + a b x^{3} + a^{2} x^{2}\right )}}\right )}{a}\right ] \]

[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]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {{\left (c x^{2} + b x + a\right )} x^{3}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\frac {2 \, {\left (\frac {\arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right )}{\sqrt {-a}}\right )}}{\mathrm {sgn}\left (x\right )} \]

[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) - arctan(sqrt(a)/sqrt(-a))/sqrt(-a))/sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x+c x^2\right )}} \, dx=\int \frac {\sqrt {x}}{\sqrt {x^3\,\left (c\,x^2+b\,x+a\right )}} \,d x \]

[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)