Integrand size = 24, antiderivative size = 64 \[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {x^3 \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {3}{8},2,\frac {1}{2},\frac {11}{8},-\frac {b x^8}{a},-\frac {d x^8}{c}\right )}{3 a^2 \sqrt {c+d x^8}} \] Output:
1/3*x^3*(1+d*x^8/c)^(1/2)*AppellF1(3/8,2,1/2,11/8,-b*x^8/a,-d*x^8/c)/a^2/( d*x^8+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(170\) vs. \(2(64)=128\).
Time = 10.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.66 \[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\frac {x^3 \left (33 a b \left (c+d x^8\right )+11 (5 b c-8 a d) \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {3}{8},\frac {1}{2},1,\frac {11}{8},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )+3 b d x^8 \left (a+b x^8\right ) \sqrt {1+\frac {d x^8}{c}} \operatorname {AppellF1}\left (\frac {11}{8},\frac {1}{2},1,\frac {19}{8},-\frac {d x^8}{c},-\frac {b x^8}{a}\right )\right )}{264 a^2 (b c-a d) \left (a+b x^8\right ) \sqrt {c+d x^8}} \] Input:
Integrate[x^2/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
Output:
(x^3*(33*a*b*(c + d*x^8) + 11*(5*b*c - 8*a*d)*(a + b*x^8)*Sqrt[1 + (d*x^8) /c]*AppellF1[3/8, 1/2, 1, 11/8, -((d*x^8)/c), -((b*x^8)/a)] + 3*b*d*x^8*(a + b*x^8)*Sqrt[1 + (d*x^8)/c]*AppellF1[11/8, 1/2, 1, 19/8, -((d*x^8)/c), - ((b*x^8)/a)]))/(264*a^2*(b*c - a*d)*(a + b*x^8)*Sqrt[c + d*x^8])
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1013, 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.
\(\displaystyle \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^8}{c}+1} \int \frac {x^2}{\left (b x^8+a\right )^2 \sqrt {\frac {d x^8}{c}+1}}dx}{\sqrt {c+d x^8}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^3 \sqrt {\frac {d x^8}{c}+1} \operatorname {AppellF1}\left (\frac {3}{8},2,\frac {1}{2},\frac {11}{8},-\frac {b x^8}{a},-\frac {d x^8}{c}\right )}{3 a^2 \sqrt {c+d x^8}}\) |
Input:
Int[x^2/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]
Output:
(x^3*Sqrt[1 + (d*x^8)/c]*AppellF1[3/8, 2, 1/2, 11/8, -((b*x^8)/a), -((d*x^ 8)/c)])/(3*a^2*Sqrt[c + d*x^8])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], 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])
\[\int \frac {x^{2}}{\left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}d x\]
Input:
int(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
Output:
int(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
Timed out. \[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\text {Timed out} \] Input:
integrate(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^{2}}{\left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \] Input:
integrate(x**2/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)
Output:
Integral(x**2/((a + b*x**8)**2*sqrt(c + d*x**8)), x)
\[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {x^{2}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}} \,d x } \] Input:
integrate(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")
Output:
integrate(x^2/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)
\[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int { \frac {x^{2}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}} \,d x } \] Input:
integrate(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")
Output:
integrate(x^2/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)
Timed out. \[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {x^2}{{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \] Input:
int(x^2/((a + b*x^8)^2*(c + d*x^8)^(1/2)),x)
Output:
int(x^2/((a + b*x^8)^2*(c + d*x^8)^(1/2)), x)
\[ \int \frac {x^2}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx=\int \frac {\sqrt {d \,x^{8}+c}\, x^{2}}{b^{2} d \,x^{24}+2 a b d \,x^{16}+b^{2} c \,x^{16}+a^{2} d \,x^{8}+2 a b c \,x^{8}+a^{2} c}d x \] Input:
int(x^2/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)
Output:
int((sqrt(c + d*x**8)*x**2)/(a**2*c + a**2*d*x**8 + 2*a*b*c*x**8 + 2*a*b*d *x**16 + b**2*c*x**16 + b**2*d*x**24),x)