Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {8}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c^2 \left (a+b x^3\right )^{2/3}} \]
Leaf count is larger than twice the leaf count of optimal. \(550\) vs. \(2(62)=124\).
Time = 10.98 (sec) , antiderivative size = 550, normalized size of antiderivative = 8.87 \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\frac {\frac {b d \left (-6 b^2 c^2+21 a b c d+5 a^2 d^2\right ) x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{(-b c+a d)^3}+\frac {4 c \left (-4 a c x \left (15 a^4 d^3-6 b^4 c^2 x^3 \left (2 c+d x^3\right )+5 a^3 b d^2 \left (-9 c+4 d x^3\right )+a^2 b^2 d \left (45 c^2-21 c d x^3+5 d^2 x^6\right )+3 a b^3 c \left (-5 c^2+11 c d x^3+7 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+x^4 \left (5 a^4 d^3+10 a^3 b d^3 x^3-6 b^4 c^2 x^3 \left (c+d x^3\right )+a^2 b^2 d \left (24 c^2+24 c d x^3+5 d^2 x^6\right )+3 a b^3 c \left (-3 c^2+4 c d x^3+7 d^2 x^6\right )\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{(b c-a d)^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}}{60 a^2 c^2 \left (a+b x^3\right )^{2/3}} \]
((b*d*(-6*b^2*c^2 + 21*a*b*c*d + 5*a^2*d^2)*x^4*(1 + (b*x^3)/a)^(2/3)*Appe llF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(-(b*c) + a*d)^3 + (4* c*(-4*a*c*x*(15*a^4*d^3 - 6*b^4*c^2*x^3*(2*c + d*x^3) + 5*a^3*b*d^2*(-9*c + 4*d*x^3) + a^2*b^2*d*(45*c^2 - 21*c*d*x^3 + 5*d^2*x^6) + 3*a*b^3*c*(-5*c ^2 + 11*c*d*x^3 + 7*d^2*x^6))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -(( d*x^3)/c)] + x^4*(5*a^4*d^3 + 10*a^3*b*d^3*x^3 - 6*b^4*c^2*x^3*(c + d*x^3) + a^2*b^2*d*(24*c^2 + 24*c*d*x^3 + 5*d^2*x^6) + 3*a*b^3*c*(-3*c^2 + 4*c*d *x^3 + 7*d^2*x^6))*(3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^ 3)/c)] + 2*b*c*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))/( (b*c - a*d)^3*(a + b*x^3)*(c + d*x^3)*(4*a*c*AppellF1[1/3, 2/3, 1, 4/3, -( (b*x^3)/a), -((d*x^3)/c)] - x^3*(3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3 )/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d* x^3)/c)]))))/(60*a^2*c^2*(a + b*x^3)^(2/3))
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {937, 936}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{8/3} \left (d x^3+c\right )^2}dx}{a^2 \left (a+b x^3\right )^{2/3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {8}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 c^2 \left (a+b x^3\right )^{2/3}}\) |
(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 8/3, 2, 4/3, -((b*x^3)/a), -((d*x^3 )/c)])/(a^2*c^2*(a + b*x^3)^(2/3))
3.2.8.3.1 Defintions of rubi rules used
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])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {8}{3}} \left (d \,x^{3}+c \right )^{2}}d x\]
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {8}{3}} \left (c + d x^{3}\right )^{2}}\, dx \]
\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {8}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \]
\[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {8}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{8/3}\,{\left (d\,x^3+c\right )}^2} \,d x \]