\(\int \frac {1}{x^3 (a+b x^n+c x^{2 n})^{3/2}} \, dx\) [595]

Optimal result

Integrand size = 22, antiderivative size = 154 \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {3}{2},\frac {3}{2},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a x^2 \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1399, 524} \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=-\frac {\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {3}{2},\frac {3}{2},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a x^2 \sqrt {a+b x^n+c x^{2 n}}} \]

[In]

Int[1/(x^3*(a + b*x^n + c*x^(2*n))^(3/2)),x]

[Out]

-1/2*(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, 3
/2, 3/2, -((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])])/(a*x^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 {1}{x^3 \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 {\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 {3}{2},\frac {3}{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 a x^2 \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. \(399\) vs. \(2(154)=308\).

Time = 0.61 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.59 \[ \int \frac {1}{x^3 \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {(-2+n) \left (-4 a c (2+n)+b^2 (4+n)\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {-2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )-4 \left ((-2+n) \left (b^2-2 a c+b c x^n\right )+2 b c x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {-2+n}{n},\frac {1}{2},\frac {1}{2},2-\frac {2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \left (-b^2+4 a c\right ) (-2+n) n x^2 \sqrt {a+x^n \left (b+c x^n\right )}} \]

[In]

Integrate[1/(x^3*(a + b*x^n + c*x^(2*n))^(3/2)),x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

integrate(1/x^3/(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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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