\(\int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx\) [104]

Optimal result

Integrand size = 20, antiderivative size = 142 \[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {2 x^2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {a x+b x^3+c x^5}} \]

[Out]

2/3*x^2*AppellF1(3/4,1/2,1/2,7/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2/(
b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(c*x^5+b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1944, 1155, 524} \[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {2 x^2 \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {a x+b x^3+c x^5}} \]

[In]

Int[x/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

(2*x^2*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[3/4, 1
/2, 1/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*Sqrt[a*x + b*x^3 + c
*x^5])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1155

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 +
c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2
])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^2/(b - Sqrt[b^2 - 4*a*c]
)))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rule 1944

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {a+b x^2+c x^4}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2+c x^4}} \, dx}{\sqrt {a x+b x^3+c x^5}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {\sqrt {x}}{\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \, dx}{\sqrt {a x+b x^3+c x^5}} \\ & = \frac {2 x^2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {3}{4};\frac {1}{2},\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {a x+b x^3+c x^5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.07 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\frac {2 x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},\frac {1}{2},\frac {7}{4},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )}{3 \sqrt {x \left (a+b x^2+c x^4\right )}} \]

[In]

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

[Out]

(2*x^2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(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])]*AppellF1[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt
[b^2 - 4*a*c])])/(3*Sqrt[x*(a + b*x^2 + c*x^4)])

Maple [F]

\[\int \frac {x}{\sqrt {c \,x^{5}+b \,x^{3}+a x}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

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

[Out]

integral(sqrt(c*x^5 + b*x^3 + a*x)/(c*x^4 + b*x^2 + a), x)

Sympy [F]

\[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {x}{\sqrt {x \left (a + b x^{2} + c x^{4}\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {x}{\sqrt {c x^{5} + b x^{3} + a x}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {x}{\sqrt {c\,x^5+b\,x^3+a\,x}} \,d x \]

[In]

int(x/(a*x + b*x^3 + c*x^5)^(1/2),x)

[Out]

int(x/(a*x + b*x^3 + c*x^5)^(1/2), x)