3.116 \(\int \frac{1}{x^{3/2} \sqrt{a x+b x^3+c x^5}} \, dx\)

Optimal. Leaf size=330 \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{3/4} \sqrt{a x+b x^3+c x^5}}-\frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{3/4} \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt{c} x^{3/2} \left (a+b x^2+c x^4\right )}{a \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}} \]

[Out]

(Sqrt[c]*x^(3/2)*(a + b*x^2 + c*x^4))/(a*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5]) - Sqrt[a*x + b*x^3
+ c*x^5]/(a*x^(3/2)) - (c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x
^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(a^(3/4)*Sqrt[a*x + b*x^3 + c*x^
5]) + (c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2
*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(3/4)*Sqrt[a*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.184284, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1929, 12, 1914, 1139, 1103, 1195} \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{3/4} \sqrt{a x+b x^3+c x^5}}-\frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{3/4} \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt{c} x^{3/2} \left (a+b x^2+c x^4\right )}{a \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*Sqrt[a*x + b*x^3 + c*x^5]),x]

[Out]

(Sqrt[c]*x^(3/2)*(a + b*x^2 + c*x^4))/(a*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5]) - Sqrt[a*x + b*x^3
+ c*x^5]/(a*x^(3/2)) - (c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x
^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(a^(3/4)*Sqrt[a*x + b*x^3 + c*x^
5]) + (c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2
*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*a^(3/4)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 1929

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - q + 1)
*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(m + p*q + 1)), x] - Dist[1/(a*(m + p*q + 1)), Int[x^(m + n - q)*
(b*(m + p*q + (n - q)*(p + 1) + 1) + c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n
- q))^p, x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c,
0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && LtQ[m + p*q + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1139

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

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{1}{x^{3/2} \sqrt{a x+b x^3+c x^5}} \, dx &=-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}}+\frac{\int \frac{c x^{5/2}}{\sqrt{a x+b x^3+c x^5}} \, dx}{a}\\ &=-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}}+\frac{c \int \frac{x^{5/2}}{\sqrt{a x+b x^3+c x^5}} \, dx}{a}\\ &=-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}}+\frac{\left (c \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{a \sqrt{a x+b x^3+c x^5}}\\ &=-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}}+\frac{\left (\sqrt{c} \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a} \sqrt{a x+b x^3+c x^5}}-\frac{\left (\sqrt{c} \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{a}}}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a} \sqrt{a x+b x^3+c x^5}}\\ &=\frac{\sqrt{c} x^{3/2} \left (a+b x^2+c x^4\right )}{a \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{\sqrt{a x+b x^3+c x^5}}{a x^{3/2}}-\frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{3/4} \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{3/4} \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [C]  time = 0.48716, size = 303, normalized size = 0.92 \[ \frac{-4 \left (a+b x^2+c x^4\right )+\frac{i \sqrt{2} x \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \left (E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )\right )}{\sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}}}{4 a \sqrt{x} \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*Sqrt[a*x + b*x^3 + c*x^5]),x]

[Out]

(-4*(a + b*x^2 + c*x^4) + (I*Sqrt[2]*(-b + Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sq
rt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*(EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^
2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sq
rt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/Sqrt[c/(b + Sqrt[b^2 - 4*a*c])])/(4*a
*Sqrt[x]*Sqrt[x*(a + b*x^2 + c*x^4)])

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Maple [A]  time = 0.023, size = 508, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) a} \left ( -\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}\sqrt{-4\,ac+{b}^{2}}{x}^{4}c-\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}{x}^{4}bc-c\sqrt{-2\,{\frac{{x}^{2}\sqrt{-4\,ac+{b}^{2}}-b{x}^{2}-2\,a}{a}}}\sqrt{{\frac{1}{a} \left ({x}^{2}\sqrt{-4\,ac+{b}^{2}}+b{x}^{2}+2\,a \right ) }}ax{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{\sqrt{2}}{2}\sqrt{{\frac{1}{ac} \left ( b\sqrt{-4\,ac+{b}^{2}}-2\,ac+{b}^{2} \right ) }}} \right ) +c\sqrt{-2\,{\frac{{x}^{2}\sqrt{-4\,ac+{b}^{2}}-b{x}^{2}-2\,a}{a}}}\sqrt{{\frac{1}{a} \left ({x}^{2}\sqrt{-4\,ac+{b}^{2}}+b{x}^{2}+2\,a \right ) }}ax{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}},{\frac{\sqrt{2}}{2}\sqrt{{\frac{1}{ac} \left ( b\sqrt{-4\,ac+{b}^{2}}-2\,ac+{b}^{2} \right ) }}} \right ) -\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}\sqrt{-4\,ac+{b}^{2}}{x}^{2}b-\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}{x}^{2}{b}^{2}-\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}\sqrt{-4\,ac+{b}^{2}}a-\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}ab \right ) \sqrt{x \left ( c{x}^{4}+b{x}^{2}+a \right ) }{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{\frac{1}{a} \left ( -b+\sqrt{-4\,ac+{b}^{2}} \right ) }}}} \left ( b+\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{5} + b x^{3} + a x} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{5} + b x^{3} + a x} \sqrt{x}}{c x^{7} + b x^{5} + a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x)/(c*x^7 + b*x^5 + a*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{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(1/x**(3/2)/(c*x**5+b*x**3+a*x)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{5} + b x^{3} + a x} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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