3.585 \(\int \frac {1}{\sqrt {a+b x^n+c x^{2 n}}} \, dx\)

Optimal. Leaf size=139 \[ \frac {x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{n};\frac {1}{2},\frac {1}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

x*AppellF1(1/n,1/2,1/2,1+1/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)

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Rubi [A]  time = 0.08, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1348, 429} \[ \frac {x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{n};\frac {1}{2},\frac {1}{2};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^n+c x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*x^n + c*x^(2*n)],x]

[Out]

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

Rule 429

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

Rule 1348

Int[((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]))^F
racPart[p]), Int[(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, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b x^n+c x^{2 n}}} \, dx &=\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 {1}{\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 \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 {1}{n};\frac {1}{2},\frac {1}{2};1+\frac {1}{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 )}{\sqrt {a+b x^n+c x^{2 n}}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 166, normalized size = 1.19 \[ \frac {x \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^n}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{n};\frac {1}{2},\frac {1}{2};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )}{\sqrt {a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2 \, n} + b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \,x^{n}+c \,x^{2 n}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2 \, n} + b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a+b\,x^n+c\,x^{2\,n}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b x^{n} + c x^{2 n}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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