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

Optimal. Leaf size=380 \[ -\frac {\sqrt [4]{a} \sqrt {x} \left (\sqrt {a} b \sqrt {c}-6 a c+2 b^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 c^{7/4} \sqrt {a x+b x^3+c x^5}}+\frac {2 \sqrt [4]{a} \sqrt {x} \left (b^2-3 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 c^{7/4} \sqrt {a x+b x^3+c x^5}}-\frac {2 x^{3/2} \left (b^2-3 a c\right ) \left (a+b x^2+c x^4\right )}{15 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c} \]

[Out]

-2/15*(-3*a*c+b^2)*x^(3/2)*(c*x^4+b*x^2+a)/c^(3/2)/(a^(1/2)+x^2*c^(1/2))/(c*x^5+b*x^3+a*x)^(1/2)+1/15*(3*c*x^2
+b)*x^(1/2)*(c*x^5+b*x^3+a*x)^(1/2)/c+2/15*a^(1/4)*(-3*a*c+b^2)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos
(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(
1/2)+x^2*c^(1/2))*x^(1/2)*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^5+b*x^3+a*x)^(1/2)-1/30
*a^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^
(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*(2*b^2-6*a*c+b*a^(1/2)*c^(1/2))*x^(1/
2)*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^5+b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1919, 1953, 1197, 1103, 1195} \[ -\frac {2 x^{3/2} \left (b^2-3 a c\right ) \left (a+b x^2+c x^4\right )}{15 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} \sqrt {x} \left (\sqrt {a} b \sqrt {c}-6 a c+2 b^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 c^{7/4} \sqrt {a x+b x^3+c x^5}}+\frac {2 \sqrt [4]{a} \sqrt {x} \left (b^2-3 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 c^{7/4} \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b^2 - 3*a*c)*x^(3/2)*(a + b*x^2 + c*x^4))/(15*c^(3/2)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5])
+ (Sqrt[x]*(b + 3*c*x^2)*Sqrt[a*x + b*x^3 + c*x^5])/(15*c) + (2*a^(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/(S
qrt[a]*Sqrt[c]))/4])/(15*c^(7/4)*Sqrt[a*x + b*x^3 + c*x^5]) - (a^(1/4)*(2*b^2 + Sqrt[a]*b*Sqrt[c] - 6*a*c)*Sqr
t[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])/(30*c^(7/4)*Sqrt[a*x + b*x^3 + c*x^5])

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]

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 1919

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - n + q
+ 1)*(b*(n - q)*p + c*(m + p*q + (n - q)*(2*p - 1) + 1)*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^p)/(c*(m +
p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1)), x] + Dist[((n - q)*p)/(c*(m + p*(2*n - q) + 1)*(m + p*q +
(n - q)*(2*p - 1) + 1)), Int[x^(m - (n - 2*q))*Simp[-(a*b*(m + p*q - n + q + 1)) + (2*a*c*(m + p*q + (n - q)*
(2*p - 1) + 1) - b^2*(m + p*q + (n - q)*(p - 1) + 1))*x^(n - q), x]*(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] && GtQ[p, 0] && RationalQ[m, q] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) + 1, 0] && NeQ[m + p*
q + (n - q)*(2*p - 1) + 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]

Rubi steps

\begin {align*} \int x^{3/2} \sqrt {a x+b x^3+c x^5} \, dx &=\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}+\frac {\int \frac {\sqrt {x} \left (-a b-2 \left (b^2-3 a c\right ) x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx}{15 c}\\ &=\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}+\frac {\left (\sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {-a b-2 \left (b^2-3 a c\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{15 c \sqrt {a x+b x^3+c x^5}}\\ &=\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}+\frac {\left (2 \sqrt {a} \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}{15 c^{3/2} \sqrt {a x+b x^3+c x^5}}+\frac {\left (\sqrt {a} \left (-\sqrt {a} b \sqrt {c}-2 \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}{15 c^{3/2} \sqrt {a x+b x^3+c x^5}}\\ &=-\frac {2 \left (b^2-3 a c\right ) x^{3/2} \left (a+b x^2+c x^4\right )}{15 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt {x} \left (b+3 c x^2\right ) \sqrt {a x+b x^3+c x^5}}{15 c}+\frac {2 \sqrt [4]{a} \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 )}{15 c^{7/4} \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} \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 )}{30 c^{7/4} \sqrt {a x+b x^3+c x^5}}\\ \end {align*}

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Mathematica [C]  time = 1.48, size = 486, normalized size = 1.28 \[ \frac {\sqrt {x} \left (-i \left (b^2-3 a c\right ) \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+2 c x \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )+i \left (b^2 \sqrt {b^2-4 a c}-3 a c \sqrt {b^2-4 a c}+4 a b c-b^3\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{30 c^2 \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(b + 3*c*x^2)*(a + b*x^2 + c*x^4) - I*(b^2 - 3*a*c)*(-b + Sqrt
[b^2 - 4*a*c])*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])
*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])]))/(30*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x*(a + b*x^2 + c*x^4)])

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}, x\right ) \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}\,{d x} \]

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]

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

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maple [B]  time = 0.05, size = 1042, normalized size = 2.74 \[ -\frac {\sqrt {\left (c \,x^{4}+b \,x^{2}+a \right ) x}\, \left (-6 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b \,c^{2} x^{7}-6 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, c^{2} x^{7}-8 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b^{2} c \,x^{5}-8 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, b c \,x^{5}-6 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, a b c \,x^{3}-2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b^{3} x^{3}-6 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, a c \,x^{3}-2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, b^{2} x^{3}-12 \sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, a^{2} c \EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )+12 \sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, a^{2} c \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )-2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, a \,b^{2} x +4 \sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, a \,b^{2} \EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )-3 \sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, a \,b^{2} \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )-2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, a b x +\sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, \sqrt {-4 a c +b^{2}}\, a b \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )\right )}{30 \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) c \sqrt {x}} \]

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]

-1/30*(x*(c*x^4+b*x^2+a))^(1/2)*(-6*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x^7*b*c^2-6*((-b+(-4*a*c+b^2)^(1/2))/a)^
(1/2)*(-4*a*c+b^2)^(1/2)*x^7*c^2-8*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x^5*b^2*c-8*((-b+(-4*a*c+b^2)^(1/2))/a)^(
1/2)*(-4*a*c+b^2)^(1/2)*x^5*b*c-6*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x^3*a*b*c-6*((-b+(-4*a*c+b^2)^(1/2))/a)^(1
/2)*(-4*a*c+b^2)^(1/2)*x^3*a*c-2*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x^3*b^3-2*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)
*(-4*a*c+b^2)^(1/2)*x^3*b^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*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*2^(1/2)*(((-4*a*c+b^2)^(1/2)*b-2*a
*c+b^2)/a/c)^(1/2))*a^2*c-3*b^2*a*(-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*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*2^(1/2)*(((-4*a*c+b^2)^(1/2)*b-
2*a*c+b^2)/a/c)^(1/2))+b*a*(-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*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*2^(1/2)*(((-4*a*c+b^2)^(1/2)*b-2*a*c+b
^2)/a/c)^(1/2))*(-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)*EllipticE(1/2*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*2^(1/2)*(((-4*a*c+b^2)^(1/2
)*b-2*a*c+b^2)/a/c)^(1/2))*a^2*c+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*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*2^(1/2)*(((-4*a*c+b^2)^(1/2)*b
-2*a*c+b^2)/a/c)^(1/2))*a*b^2-2*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x*a*b^2-2*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*
(-4*a*c+b^2)^(1/2)*x*a*b)/x^(1/2)/(c*x^4+b*x^2+a)/c/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)/(b+(-4*a*c+b^2)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^{3/2}\,\sqrt {c\,x^5+b\,x^3+a\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \sqrt {x \left (a + b x^{2} + c x^{4}\right )}\, dx \]

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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