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

Optimal. Leaf size=468 \[ \frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a} b \sqrt{c}-6 a c+2 b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{2 \sqrt{c} x^{3/2} \left (b^2-3 a c\right ) \left (a+b x^2+c x^4\right )}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 x^{3/2} \left (b^2-4 a c\right )}-\frac{2 \sqrt [4]{c} \sqrt{x} \left (b^2-3 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{-2 a c+b^2+b c x^2}{a \sqrt{x} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*Sqrt[x]*Sqrt[a*x + b*x^3 + c*x^5]) + (2*Sqrt[c]*(b^2 - 3*a*c)*x^(3/2)
*(a + b*x^2 + c*x^4))/(a^2*(b^2 - 4*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5]) - (2*(b^2 - 3*a*c)
*Sqrt[a*x + b*x^3 + c*x^5])/(a^2*(b^2 - 4*a*c)*x^(3/2)) - (2*c^(1/4)*(b^2 - 3*a*c)*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^(7/4)*(b^2 - 4*a*c)*Sqrt[a*x + b*x^3 + c*x^5]) + (c^(1/4)*(2*b^2 + Sqrt[a]*b*Sqrt[c] - 6*
a*c)*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^(7/4)*(b^2 - 4*a*c)*Sqrt[a*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.408659, antiderivative size = 468, 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 = {1924, 1951, 1953, 1197, 1103, 1195} \[ \frac{2 \sqrt{c} x^{3/2} \left (b^2-3 a c\right ) \left (a+b x^2+c x^4\right )}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 x^{3/2} \left (b^2-4 a c\right )}+\frac{\sqrt [4]{c} \sqrt{x} \left (\sqrt{a} b \sqrt{c}-6 a c+2 b^2\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{2 a^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}-\frac{2 \sqrt [4]{c} \sqrt{x} \left (b^2-3 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{\frac{a+b x^2+c x^4}{\left (\sqrt{a}+\sqrt{c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac{1}{4} \left (2-\frac{b}{\sqrt{a} \sqrt{c}}\right )\right )}{a^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{-2 a c+b^2+b c x^2}{a \sqrt{x} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*Sqrt[x]*Sqrt[a*x + b*x^3 + c*x^5]) + (2*Sqrt[c]*(b^2 - 3*a*c)*x^(3/2)
*(a + b*x^2 + c*x^4))/(a^2*(b^2 - 4*a*c)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5]) - (2*(b^2 - 3*a*c)
*Sqrt[a*x + b*x^3 + c*x^5])/(a^2*(b^2 - 4*a*c)*x^(3/2)) - (2*c^(1/4)*(b^2 - 3*a*c)*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^(7/4)*(b^2 - 4*a*c)*Sqrt[a*x + b*x^3 + c*x^5]) + (c^(1/4)*(2*b^2 + Sqrt[a]*b*Sqrt[c] - 6*
a*c)*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^(7/4)*(b^2 - 4*a*c)*Sqrt[a*x + b*x^3 + c*x^5])

Rule 1924

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(m - q + 1
)*(b^2 - 2*a*c + b*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), x]
+ Dist[1/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - q)*(b^2*(m + p*q + (n - q)*(p + 1) + 1) - 2*a*c*(m + p
*q + 2*(n - q)*(p + 1) + 1) + b*c*(m + p*q + (n - q)*(2*p + 3) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))
^(p + 1), 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] && LtQ[p, -1] && RationalQ[m, q] && LtQ[m + p*q + 1, n - q]

Rule 1951

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(A*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)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p + 1) + 1) - A*c*(m + p*q + 2*(
n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && Eq
Q[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] &&
((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*
q + 1, 0]

Rule 1953

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(j_.)))/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Sym
bol] :> 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)*(A + B*x^(n - q)))/Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, A, B, m, n, q}, x
] && EqQ[j, n - q] && EqQ[r, 2*n - q] && PosQ[n - q] && (EqQ[m, 1/2] || EqQ[m, -2^(-1)]) && EqQ[n, 3] && EqQ[q
, 1]

Rule 1197

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]
/; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, 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}{\sqrt{x} \left (a x+b x^3+c x^5\right )^{3/2}} \, dx &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{x} \sqrt{a x+b x^3+c x^5}}-\frac{\int \frac{-2 b^2+6 a c-b c x^2}{x^{3/2} \sqrt{a x+b x^3+c x^5}} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{x} \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 \left (b^2-4 a c\right ) x^{3/2}}+\frac{\int \frac{\sqrt{x} \left (a b c+2 c \left (b^2-3 a c\right ) x^2\right )}{\sqrt{a x+b x^3+c x^5}} \, dx}{a^2 \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{x} \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 \left (b^2-4 a c\right ) x^{3/2}}+\frac{\left (\sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{a b c+2 c \left (b^2-3 a c\right ) x^2}{\sqrt{a+b x^2+c x^4}} \, dx}{a^2 \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{x} \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 \left (b^2-4 a c\right ) x^{3/2}}-\frac{\left (2 \sqrt{c} \left (b^2-3 a c\right ) \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}{a^{3/2} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{\left (\left (\sqrt{a} b c^{3/2}+2 c \left (b^2-3 a c\right )\right ) \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{1}{\sqrt{a+b x^2+c x^4}} \, dx}{a^{3/2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}\\ &=\frac{b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{x} \sqrt{a x+b x^3+c x^5}}+\frac{2 \sqrt{c} \left (b^2-3 a c\right ) x^{3/2} \left (a+b x^2+c x^4\right )}{a^2 \left (b^2-4 a c\right ) \left (\sqrt{a}+\sqrt{c} x^2\right ) \sqrt{a x+b x^3+c x^5}}-\frac{2 \left (b^2-3 a c\right ) \sqrt{a x+b x^3+c x^5}}{a^2 \left (b^2-4 a c\right ) x^{3/2}}-\frac{2 \sqrt [4]{c} \left (b^2-3 a c\right ) \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^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}+\frac{\sqrt [4]{c} \left (2 b^2+\sqrt{a} b \sqrt{c}-6 a c\right ) \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^{7/4} \left (b^2-4 a c\right ) \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [C]  time = 1.35104, size = 519, normalized size = 1.11 \[ -\frac{i x \left (b^2 \sqrt{b^2-4 a c}-3 a c \sqrt{b^2-4 a c}+4 a b c-b^3\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} x \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}}\right ),\frac{\sqrt{b^2-4 a c}+b}{b-\sqrt{b^2-4 a c}}\right )+2 \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \left (-4 a^2 c+a \left (b^2-7 b c x^2-6 c^2 x^4\right )+2 b^2 x^2 \left (b+c x^2\right )\right )-i x \left (b^2-3 a c\right ) \left (\sqrt{b^2-4 a c}-b\right ) \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \sqrt{\frac{-2 \sqrt{b^2-4 a c}+2 b+4 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{c}{b+\sqrt{b^2-4 a c}}} x\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 \sqrt{x} \left (b^2-4 a c\right ) \sqrt{\frac{c}{\sqrt{b^2-4 a c}+b}} \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(-4*a^2*c + 2*b^2*x^2*(b + c*x^2) + a*(b^2 - 7*b*c*x^2 - 6*c^2*x^4)) - I*(
b^2 - 3*a*c)*(-b + Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(
2*b - 2*Sqrt[b^2 - 4*a*c] + 4*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])] + I*(-b^3 + 4*a*b*c + b^2*Sqrt[b^2 - 4*a*c] -
3*a*c*Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt
[b^2 - 4*a*c] + 4*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[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])])/(2*a^2*(b^2 - 4*a*c)*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqr
t[x]*Sqrt[x*(a + b*x^2 + c*x^4)])

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Maple [B]  time = 0.029, size = 1136, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x*(c*x^4+b*x^2+a))^(1/2)/x^(3/2)*(12*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*x^4*a*c^2-4*
(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*x^4*b^2*c+12*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*x^4*a*
b*c^2-4*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*x^4*b^3*c+a*b*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)*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))*x*(-4*a*c+b^2)^(1/2)+12*(-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)*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))*x*a^2*c^2-3*a*b^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)*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))*x-12*(-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)*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))*x*a^2*c^2+4*(-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)*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))*x*a*b^2*c+14*(1/a*(-b+(-4*a*c+b^2)
^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*x^2*a*b*c-4*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*x^2*b^3+1
4*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*x^2*a*b^2*c-4*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*x^2*b^4+8*(1/a*(-b+(-4
*a*c+b^2)^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*a^2*c-2*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*(-4*a*c+b^2)^(1/2)*a*b^
2+8*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*a^2*b*c-2*(1/a*(-b+(-4*a*c+b^2)^(1/2)))^(1/2)*a*b^3)/(c*x^4+b*x^2+a)/a
^2/(4*a*c-b^2)/(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}{{\left (c x^{5} + b x^{3} + a x\right )}^{\frac{3}{2}} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^5 + b*x^3 + a*x)^(3/2)*sqrt(x)), 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^{2} x^{11} + 2 \, b c x^{9} +{\left (b^{2} + 2 \, a c\right )} x^{7} + 2 \, a b x^{5} + a^{2} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x)/(c^2*x^11 + 2*b*c*x^9 + (b^2 + 2*a*c)*x^7 + 2*a*b*x^5 + a^2*x^3), x
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x} \left (x \left (a + b x^{2} + c x^{4}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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