3.592 \(\int \frac{1}{(a+b x^n+c x^{2 n})^{3/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac{x \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{n};\frac{3}{2},\frac{3}{2};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a \sqrt{a+b x^n+c x^{2 n}}} \]

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

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Rubi [A]  time = 0.0822772, antiderivative size = 142, normalized size of antiderivative = 1., 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} \[ \frac{x \sqrt{\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^n}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{n};\frac{3}{2},\frac{3}{2};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a \sqrt{a+b x^n+c x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

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

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]

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])

Rubi steps

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

Mathematica [B]  time = 0.973431, size = 384, normalized size = 2.7 \[ \frac{x \left (2 b c x^n \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (1+\frac{1}{n};\frac{1}{2},\frac{1}{2};2+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )-(n+1) \left (\left (b^2 (n-2)-4 a c (n-1)\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^n}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^n}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{1}{n};\frac{1}{2},\frac{1}{2};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}},\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )+2 \left (-2 a c+b^2+b c x^n\right )\right )\right )}{a n (n+1) \left (4 a c-b^2\right ) \sqrt{a+x^n \left (b+c x^n\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.012, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^(2*n) + b*x^n + a)^(-3/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**n + c*x**(2*n))**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^(2*n) + b*x^n + a)^(-3/2), x)