\(\int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx\) [584]

Optimal result

Integrand size = 20, antiderivative size = 148 \[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\frac {x^2 \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1399, 524} \[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\frac {x^2 \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {a+b x^n+c x^{2 n}}} \]

[In]

Int[x/Sqrt[a + b*x^n + c*x^(2*n)],x]

[Out]

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

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 1399

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a +
b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^
2 - 4*a*c, 2])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[
b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.18 \[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\frac {x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {a+x^n \left (b+c x^n\right )}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {x}{\sqrt {a +b \,x^{n}+c \,x^{2 n}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\int \frac {x}{\sqrt {a + b x^{n} + c x^{2 n}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\int { \frac {x}{\sqrt {c x^{2 \, n} + b x^{n} + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\int { \frac {x}{\sqrt {c x^{2 \, n} + b x^{n} + a}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {a+b x^n+c x^{2 n}}} \, dx=\int \frac {x}{\sqrt {a+b\,x^n+c\,x^{2\,n}}} \,d x \]

[In]

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

[Out]

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