\(\int \frac {\sqrt {x}}{\sqrt {x^3 (a+b x^2+c x^4)}} \, dx\) [136]

Optimal result

Integrand size = 26, antiderivative size = 53 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x^{3/2} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x^3+b x^5+c x^7}}\right )}{2 \sqrt {a}} \]

[Out]

-1/2*arctanh(1/2*x^(3/2)*(b*x^2+2*a)/a^(1/2)/(c*x^7+b*x^5+a*x^3)^(1/2))/a^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2022, 1927, 212} \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x^{3/2} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x^3+b x^5+c x^7}}\right )}{2 \sqrt {a}} \]

[In]

Int[Sqrt[x]/Sqrt[x^3*(a + b*x^2 + c*x^4)],x]

[Out]

-1/2*ArcTanh[(x^(3/2)*(2*a + b*x^2))/(2*Sqrt[a]*Sqrt[a*x^3 + b*x^5 + c*x^7])]/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^5+c x^7}} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x^{3/2} \left (2 a+b x^2\right )}{\sqrt {a x^3+b x^5+c x^7}}\right ) \\ & = -\frac {\tanh ^{-1}\left (\frac {x^{3/2} \left (2 a+b x^2\right )}{2 \sqrt {a} \sqrt {a x^3+b x^5+c x^7}}\right )}{2 \sqrt {a}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.55 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\frac {x^{3/2} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^3 \left (a+b x^2+c x^4\right )}} \]

[In]

Integrate[Sqrt[x]/Sqrt[x^3*(a + b*x^2 + c*x^4)],x]

[Out]

(x^(3/2)*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^3*(
a + b*x^2 + c*x^4)])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40

method	result	size
default	\(-\frac {x^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}\, \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 \sqrt {x^{3} \left (c \,x^{4}+b \,x^{2}+a \right )}\, \sqrt {a}}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.74 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\left [\frac {\log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{4} + 8 \, a^{2} x^{2} - 4 \, \sqrt {c x^{7} + b x^{5} + a x^{3}} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} \sqrt {x}}{x^{6}}\right )}{4 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {c x^{7} + b x^{5} + a x^{3}} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a} \sqrt {x}}{2 \, {\left (a c x^{6} + a b x^{4} + a^{2} x^{2}\right )}}\right )}{2 \, a}\right ] \]

[In]

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

[Out]

[1/4*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*x^4 + 8*a^2*x^2 - 4*sqrt(c*x^7 + b*x^5 + a*x^3)*(b*x^2 + 2*a)*sqrt(a)*sqr
t(x))/x^6)/sqrt(a), 1/2*sqrt(-a)*arctan(1/2*sqrt(c*x^7 + b*x^5 + a*x^3)*(b*x^2 + 2*a)*sqrt(-a)*sqrt(x)/(a*c*x^
6 + a*b*x^4 + a^2*x^2))/a]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {{\left (c x^{4} + b x^{2} + a\right )} x^{3}}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(x)/sqrt((c*x^4 + b*x^2 + a)*x^3), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\frac {\frac {\arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right )}{\sqrt {-a}}}{\mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^(1/2)/(x^3*(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))/sgn(x
)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {x^3 \left (a+b x^2+c x^4\right )}} \, dx=\int \frac {\sqrt {x}}{\sqrt {x^3\,\left (c\,x^4+b\,x^2+a\right )}} \,d x \]

[In]

int(x^(1/2)/(x^3*(a + b*x^2 + c*x^4))^(1/2),x)

[Out]

int(x^(1/2)/(x^3*(a + b*x^2 + c*x^4))^(1/2), x)