\(\int \frac {1}{(a+b x^n+c x^{2 n})^{3/2}} \, dx\) [592]

Optimal result

Integrand size = 18, antiderivative size = 142 \[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {3}{2},\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

x*AppellF1(1/n,3/2,3/2,1+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/(a+b*x^n+c*x^(2*n))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1362, 440} \[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{n},\frac {3}{2},\frac {3}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^n+c x^{2 n}}} \]

[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 440

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 1362

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])))^Fr
acPart[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*} \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 {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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(384\) vs. \(2(142)=284\).

Time = 0.70 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.70 \[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x \left (2 b c x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )-(1+n) \left (2 \left (b^2-2 a c+b c x^n\right )+\left (b^2 (-2+n)-4 a c (-1+n)\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{a \left (-b^2+4 a c\right ) n (1+n) \sqrt {a+x^n \left (b+c x^n\right )}} \]

[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)])

Maple [F]

\[\int \frac {1}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}}d x\]

[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)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \]

[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)