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

Optimal. Leaf size=347 \[ \frac {\sqrt [4]{a} \sqrt {x} \left (2 \sqrt {a} \sqrt {c}+b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 c^{3/4} \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} b \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 c^{3/4} \sqrt {a x+b x^3+c x^5}}+\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5}+\frac {b x^{3/2} \left (a+b x^2+c x^4\right )}{3 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 347, 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 = {1921, 1953, 1197, 1103, 1195} \[ \frac {\sqrt [4]{a} \sqrt {x} \left (2 \sqrt {a} \sqrt {c}+b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 c^{3/4} \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt [4]{a} b \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 c^{3/4} \sqrt {a x+b x^3+c x^5}}+\frac {b x^{3/2} \left (a+b x^2+c x^4\right )}{3 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^(3/2)*(a + b*x^2 + c*x^4))/(3*Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[a*x + b*x^3 + c*x^5]) + (Sqrt[x]*Sqrt[
a*x + b*x^3 + c*x^5])/3 - (a^(1/4)*b*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])/(3*c^(3/4)*Sqrt[a*x + b*x^3
+ c*x^5]) + (a^(1/4)*(b + 2*Sqrt[a]*Sqrt[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])/(6*c^(3/4)*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 1921

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a x+b x^3+c x^5}}{\sqrt {x}} \, dx &=\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5}+\frac {1}{3} \int \frac {\sqrt {x} \left (2 a+b x^2\right )}{\sqrt {a x+b x^3+c x^5}} \, dx\\ &=\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5}+\frac {\left (\sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{3 \sqrt {a x+b x^3+c x^5}}\\ &=\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5}+\frac {\left (\sqrt {a} \left (2 \sqrt {a}+\frac {b}{\sqrt {c}}\right ) \sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{3 \sqrt {a x+b x^3+c x^5}}-\frac {\left (\sqrt {a} b \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}{3 \sqrt {c} \sqrt {a x+b x^3+c x^5}}\\ &=\frac {b x^{3/2} \left (a+b x^2+c x^4\right )}{3 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}+\frac {1}{3} \sqrt {x} \sqrt {a x+b x^3+c x^5}-\frac {\sqrt [4]{a} b \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 )}{3 c^{3/4} \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt [4]{a} \left (b+2 \sqrt {a} \sqrt {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 )}{6 c^{3/4} \sqrt {a x+b x^3+c x^5}}\\ \end {align*}

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Mathematica [C]  time = 0.99, size = 452, normalized size = 1.30 \[ \frac {\sqrt {x} \left (-i \left (b \sqrt {b^2-4 a c}+4 a c-b^2\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+i b \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+4 c x \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \left (a+b x^2+c x^4\right )\right )}{12 c \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {x \left (a+b x^2+c x^4\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(a + b*x^2 + c*x^4) + I*b*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[(b + S
qrt[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^2 + 4*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])]*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])]))/(12*c*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 = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{5} + b x^{3} + a x}}{\sqrt {x}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 508, normalized size = 1.46 \[ \frac {\sqrt {\left (c \,x^{4}+b \,x^{2}+a \right ) x}\, \left (\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b c \,x^{5}+\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, c \,x^{5}+\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b^{2} x^{3}+\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, b \,x^{3}+\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, a b x +\sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, a b \EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )+\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {-4 a c +b^{2}}\, a x +\sqrt {-\frac {2 \left (-b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}-2 a \right )}{a}}\, \sqrt {\frac {b \,x^{2}+\sqrt {-4 a c +b^{2}}\, x^{2}+2 a}{a}}\, \sqrt {-4 a c +b^{2}}\, a \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {2}\, \sqrt {\frac {-2 a c +b^{2}+\sqrt {-4 a c +b^{2}}\, b}{a c}}}{2}\right )\right )}{3 \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((c*x^4+b*x^2+a)*x)^(1/2)/x^(1/2)*(((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-4*a*c+b^2)^(1/2)*x^5*c+((-b+(-4*a*c
+b^2)^(1/2))/a)^(1/2)*x^5*b*c+((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-4*a*c+b^2)^(1/2)*x^3*b+((-b+(-4*a*c+b^2)^(1/
2))/a)^(1/2)*x^3*b^2+a*(-2*(-b*x^2+(-4*a*c+b^2)^(1/2)*x^2-2*a)/a)^(1/2)*((b*x^2+(-4*a*c+b^2)^(1/2)*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)*((-2*a*c+b^2+(-4*a*c+b^2)^(1/2)*b
)/a/c)^(1/2))*(-4*a*c+b^2)^(1/2)+b*a*(-2*(-b*x^2+(-4*a*c+b^2)^(1/2)*x^2-2*a)/a)^(1/2)*((b*x^2+(-4*a*c+b^2)^(1/
2)*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)*((-2*a*c+b^2+(-4*a*
c+b^2)^(1/2)*b)/a/c)^(1/2))+((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(-4*a*c+b^2)^(1/2)*x*a+((-b+(-4*a*c+b^2)^(1/2))/
a)^(1/2)*x*a*b)/(c*x^4+b*x^2+a)/((-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 \[ \int \frac {\sqrt {c x^{5} + b x^{3} + a x}}{\sqrt {x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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