Integrand size = 19, antiderivative size = 621 \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{5 (b c-a d) (a+b x)^{5/3}}+\frac {6 d \sqrt [3]{c+d x}}{5 (b c-a d)^2 (a+b x)^{2/3}}+\frac {2 \sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} d^{5/3} ((a+b x) (c+d x))^{2/3} \sqrt {(b c+a d+2 b d x)^2} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [3]{b} (b c-a d)^2 (a+b x)^{2/3} (c+d x)^{2/3} (b c+a d+2 b d x) \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}\right )^2}} \sqrt {(a d+b (c+2 d x))^2}} \]
-3/5*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(5/3)+6/5*d*(d*x+c)^(1/3)/(-a*d+b*c) ^2/(b*x+a)^(2/3)+2/5*2^(1/3)*3^(3/4)*d^(5/3)*((b*x+a)*(d*x+c))^(2/3)*((-a* d+b*c)^(2/3)+2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3))*EllipticF((2 ^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1-3^(1/2) ))/(2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^ (1/2))),I*3^(1/2)+2*I)*((2*b*d*x+a*d+b*c)^2)^(1/2)*(1/2*6^(1/2)+1/2*2^(1/2 ))*(((-a*d+b*c)^(4/3)-2^(2/3)*b^(1/3)*d^(1/3)*(-a*d+b*c)^(2/3)*((b*x+a)*(d *x+c))^(1/3)+2*2^(1/3)*b^(2/3)*d^(2/3)*((b*x+a)*(d*x+c))^(2/3))/(2^(2/3)*b ^(1/3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1 /2)/b^(1/3)/(-a*d+b*c)^2/(b*x+a)^(2/3)/(d*x+c)^(2/3)/(2*b*d*x+a*d+b*c)/((a *d+b*(2*d*x+c))^2)^(1/2)/((-a*d+b*c)^(2/3)*((-a*d+b*c)^(2/3)+2^(2/3)*b^(1/ 3)*d^(1/3)*((b*x+a)*(d*x+c))^(1/3))/(2^(2/3)*b^(1/3)*d^(1/3)*((b*x+a)*(d*x +c))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.12 \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=-\frac {3 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {2}{3},-\frac {2}{3},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (a+b x)^{5/3} (c+d x)^{2/3}} \]
(-3*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[-5/3, 2/3, -2/3, ( d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(a + b*x)^(5/3)*(c + d*x)^(2/3))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {80, 79}
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.
\(\displaystyle \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {\left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \int \frac {1}{(a+b x)^{8/3} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2/3}}dx}{(c+d x)^{2/3}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {3 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {2}{3},-\frac {2}{3},-\frac {d (a+b x)}{b c-a d}\right )}{5 b (a+b x)^{5/3} (c+d x)^{2/3}}\) |
(-3*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[-5/3, 2/3, -2/3, - ((d*(a + b*x))/(b*c - a*d))])/(5*b*(a + b*x)^(5/3)*(c + d*x)^(2/3))
3.17.13.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
\[\int \frac {1}{\left (b x +a \right )^{\frac {8}{3}} \left (d x +c \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {8}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
integral((b*x + a)^(1/3)*(d*x + c)^(1/3)/(b^3*d*x^4 + a^3*c + (b^3*c + 3*a *b^2*d)*x^3 + 3*(a*b^2*c + a^2*b*d)*x^2 + (3*a^2*b*c + a^3*d)*x), x)
\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {8}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {8}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {8}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b x)^{8/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{8/3}\,{\left (c+d\,x\right )}^{2/3}} \,d x \]