\(\int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx\) [588]

Optimal result

Integrand size = 22, antiderivative size = 151 \[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=-\frac {\sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {1}{2},\frac {1}{2},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {a+b x^n+c x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 151, 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} \[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=-\frac {\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {1}{2},\frac {1}{2},-\frac {2-n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {a+b x^n+c x^{2 n}}} \]

[In]

Int[1/(x^3*Sqrt[a + b*x^n + c*x^(2*n)]),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, 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])])/(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 \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 {\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 x^2 \sqrt {a+b x^n+c x^{2 n}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=-\frac {\sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {-2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {a+x^n \left (b+c x^n\right )}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{x^{3} \sqrt {a +b \,x^{n}+c \,x^{2 n}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

integrate(1/x^3/(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]

\[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=\int \frac {1}{x^{3} \sqrt {a + b x^{n} + c x^{2 n}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=\int { \frac {1}{\sqrt {c x^{2 \, n} + b x^{n} + a} x^{3}} \,d x } \]

[In]

integrate(1/x^3/(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^3), x)

Giac [F]

\[ \int \frac {1}{x^3 \sqrt {a+b x^n+c x^{2 n}}} \, dx=\int { \frac {1}{\sqrt {c x^{2 \, n} + b x^{n} + a} x^{3}} \,d x } \]

[In]

integrate(1/x^3/(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^3), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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