Integrand size = 22, antiderivative size = 151 \[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x^4 \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4}{n},\frac {3}{2},\frac {3}{2},\frac {4+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{4 a \sqrt {a+b x^n+c x^{2 n}}} \] Output:
1/4*x^4*(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)*AppellF1(4/n,3/2,3/2,(4+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)))/a/(a+b*x^n+c*x^(2*n))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(398\) vs. \(2(151)=302\).
Time = 1.04 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.64 \[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {x^4 \left (-8 (4+n) \left (b^2-2 a c+b c x^n\right )-\left (b^2 (-8+n)-4 a c (-4+n)\right ) (4+n) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4}{n},\frac {1}{2},\frac {1}{2},\frac {4+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+32 b c x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4+n}{n},\frac {1}{2},\frac {1}{2},2+\frac {4}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )}{4 a \left (-b^2+4 a c\right ) n (4+n) \sqrt {a+x^n \left (b+c x^n\right )}} \] Input:
Integrate[x^3/(a + b*x^n + c*x^(2*n))^(3/2),x]
Output:
(x^4*(-8*(4 + n)*(b^2 - 2*a*c + b*c*x^n) - (b^2*(-8 + n) - 4*a*c*(-4 + n)) *(4 + n)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*S qrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[4/ n, 1/2, 1/2, (4 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + 32*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[(4 + n)/n, 1/2, 1/2, 2 + 4/n, (-2*c*x^n)/(b + Sqrt [b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]))/(4*a*(-b^2 + 4*a*c)* n*(4 + n)*Sqrt[a + x^n*(b + c*x^n)])
Time = 0.34 (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 = {1721, 1012}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1721 |
\(\displaystyle \frac {\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \int \frac {x^3}{\left (\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}}dx}{a \sqrt {a+b x^n+c x^{2 n}}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^4 \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {4}{n},\frac {3}{2},\frac {3}{2},\frac {n+4}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{4 a \sqrt {a+b x^n+c x^{2 n}}}\) |
Input:
Int[x^3/(a + b*x^n + c*x^(2*n))^(3/2),x]
Output:
(x^4*Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^n)/(b + S qrt[b^2 - 4*a*c])]*AppellF1[4/n, 3/2, 3/2, (4 + 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*a*Sqrt[a + b*x^n + c*x^(2*n)])
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] /; FreeQ[{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])
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x _Symbol] :> Simp[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]
\[\int \frac {x^{3}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}}d x\]
Input:
int(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x)
Output:
int(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x)
Exception generated. \[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**3/(a+b*x**n+c*x**(2*n))**(3/2),x)
Output:
Integral(x**3/(a + b*x**n + c*x**(2*n))**(3/2), x)
\[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="maxima")
Output:
integrate(x^3/(c*x^(2*n) + b*x^n + a)^(3/2), x)
\[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x, algorithm="giac")
Output:
integrate(x^3/(c*x^(2*n) + b*x^n + a)^(3/2), x)
Timed out. \[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {x^3}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \] Input:
int(x^3/(a + b*x^n + c*x^(2*n))^(3/2),x)
Output:
int(x^3/(a + b*x^n + c*x^(2*n))^(3/2), x)
\[ \int \frac {x^3}{\left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {\sqrt {x^{2 n} c +x^{n} b +a}\, x^{3}}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \] Input:
int(x^3/(a+b*x^n+c*x^(2*n))^(3/2),x)
Output:
int((sqrt(x**(2*n)*c + x**n*b + a)*x**3)/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)