3.113 \(\int \frac{x^{3/2}}{\sqrt{a x+b x^3+c x^5}} \, dx\)

Optimal. Leaf size=82 \[ \frac{\sqrt{x} \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{c} \sqrt{a x+b x^3+c x^5}} \]

[Out]

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

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Rubi [A]  time = 0.0618114, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1914, 1107, 621, 206} \[ \frac{\sqrt{x} \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{c} \sqrt{a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

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

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.0182489, size = 82, normalized size = 1. \[ \frac{\sqrt{x} \sqrt{a+b x^2+c x^4} \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2 \sqrt{c} \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

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

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Maple [A]  time = 0.012, size = 72, normalized size = 0.9 \begin{align*}{\frac{1}{2}\sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) }\ln \left ({\frac{1}{2} \left ( 2\,c{x}^{2}+2\,\sqrt{c{x}^{4}+b{x}^{2}+a}\sqrt{c}+b \right ){\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x)

[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)*ln(1/2*(2*c*x^2+2*(c*x^4+b*x^2+a)^(1/2)*c^(1/2)+b)
/c^(1/2))/c^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\sqrt{c x^{5} + b x^{3} + a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(3/2)/sqrt(c*x^5 + b*x^3 + a*x), x)

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Fricas [A]  time = 1.46885, size = 325, normalized size = 3.96 \begin{align*} \left [\frac{\log \left (-\frac{8 \, c^{2} x^{5} + 8 \, b c x^{3} + 4 \, \sqrt{c x^{5} + b x^{3} + a x}{\left (2 \, c x^{2} + b\right )} \sqrt{c} \sqrt{x} +{\left (b^{2} + 4 \, a c\right )} x}{x}\right )}{4 \, \sqrt{c}}, -\frac{\sqrt{-c} \arctan \left (\frac{\sqrt{c x^{5} + b x^{3} + a x}{\left (2 \, c x^{2} + b\right )} \sqrt{-c} \sqrt{x}}{2 \,{\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right )}{2 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\sqrt{x \left (a + b x^{2} + c x^{4}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(c*x**5+b*x**3+a*x)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.31361, size = 54, normalized size = 0.66 \begin{align*} -\frac{\log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{1024 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

-1/1024*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(3/2)