Integrand size = 19, antiderivative size = 216 \[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=-\frac {5 (b c-a d) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 d^2}+\frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 d}-\frac {5 (b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt {3} \sqrt [3]{b} d^{8/3}}-\frac {5 (b c-a d)^2 \log (c+d x)}{18 \sqrt [3]{b} d^{8/3}}-\frac {5 (b c-a d)^2 \log \left (-1+\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 \sqrt [3]{b} d^{8/3}} \]
-5/6*(-a*d+b*c)*(b*x+a)^(2/3)*(d*x+c)^(1/3)/d^2+1/2*(b*x+a)^(5/3)*(d*x+c)^ (1/3)/d-5/18*(-a*d+b*c)^2*ln(d*x+c)/b^(1/3)/d^(8/3)-5/6*(-a*d+b*c)^2*ln(-1 +d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3))/b^(1/3)/d^(8/3)-5/9*(-a*d+b* c)^2*arctan(1/3*3^(1/2)+2/3*d^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(d*x+c)^(1/3)*3^ (1/2))/b^(1/3)/d^(8/3)*3^(1/2)
Time = 0.50 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\frac {3 \sqrt [3]{b} d^{2/3} (a+b x)^{2/3} \sqrt [3]{c+d x} (-5 b c+8 a d+3 b d x)+10 \sqrt {3} (b c-a d)^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{d} \sqrt [3]{a+b x}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )-10 (b c-a d)^2 \log \left (\sqrt [3]{d} \sqrt [3]{a+b x}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+5 (b c-a d)^2 \log \left (d^{2/3} (a+b x)^{2/3}+\sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{a+b x} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}\right )}{18 \sqrt [3]{b} d^{8/3}} \]
(3*b^(1/3)*d^(2/3)*(a + b*x)^(2/3)*(c + d*x)^(1/3)*(-5*b*c + 8*a*d + 3*b*d *x) + 10*Sqrt[3]*(b*c - a*d)^2*ArcTan[(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))/(2 *d^(1/3)*(a + b*x)^(1/3) + b^(1/3)*(c + d*x)^(1/3))] - 10*(b*c - a*d)^2*Lo g[d^(1/3)*(a + b*x)^(1/3) - b^(1/3)*(c + d*x)^(1/3)] + 5*(b*c - a*d)^2*Log [d^(2/3)*(a + b*x)^(2/3) + b^(1/3)*d^(1/3)*(a + b*x)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3)])/(18*b^(1/3)*d^(8/3))
Time = 0.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {60, 60, 71}
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.
\(\displaystyle \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{2/3}}{(c+d x)^{2/3}}dx}{6 d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{2/3} \sqrt [3]{c+d x}}{d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt [3]{a+b x} (c+d x)^{2/3}}dx}{3 d}\right )}{6 d}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{2/3} \sqrt [3]{c+d x}}{d}-\frac {2 (b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{b} d^{2/3}}-\frac {3 \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{2 \sqrt [3]{b} d^{2/3}}-\frac {\log (c+d x)}{2 \sqrt [3]{b} d^{2/3}}\right )}{3 d}\right )}{6 d}\) |
((a + b*x)^(5/3)*(c + d*x)^(1/3))/(2*d) - (5*(b*c - a*d)*(((a + b*x)^(2/3) *(c + d*x)^(1/3))/d - (2*(b*c - a*d)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*d^( 1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(b^(1/3)*d^(2/3) )) - Log[c + d*x]/(2*b^(1/3)*d^(2/3)) - (3*Log[-1 + (d^(1/3)*(a + b*x)^(1/ 3))/(b^(1/3)*(c + d*x)^(1/3))])/(2*b^(1/3)*d^(2/3))))/(3*d)))/(6*d)
3.17.1.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/( Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[d/b]
\[\int \frac {\left (b x +a \right )^{\frac {5}{3}}}{\left (d x +c \right )^{\frac {2}{3}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (166) = 332\).
Time = 0.24 (sec) , antiderivative size = 741, normalized size of antiderivative = 3.43 \[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\left [\frac {15 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \log \left (-3 \, b d^{2} x - 2 \, b c d - a d^{2} - 3 \, \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} d - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}\right ) - 10 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + 5 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right ) + 3 \, {\left (3 \, b^{2} d^{3} x - 5 \, b^{2} c d^{2} + 8 \, a b d^{3}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{18 \, b d^{4}}, \frac {30 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}}{b d^{2} x + a d^{2}}\right ) - 10 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) + 5 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-b d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right ) + 3 \, {\left (3 \, b^{2} d^{3} x - 5 \, b^{2} c d^{2} + 8 \, a b d^{3}\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{18 \, b d^{4}}\right ] \]
[1/18*(15*sqrt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*sqrt((-b*d^2)^ (1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 - 3*(-b*d^2)^(1/3)*(b*x + a)^(2/ 3)*(d*x + c)^(1/3)*d - 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt((-b*d^2)^(1/3)/b)) - 10*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(-b*d^ 2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + 5*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(-b*d^2)^(2/3)*log(((b* x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c )^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*x - 5*b^ 2*c*d^2 + 8*a*b*d^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b*d^4), 1/18*(30*sq rt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*sqrt(-(-b*d^2)^(1/3)/b)*ar ctan(sqrt(1/3)*(2*(-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2 )^(1/3)*(b*d*x + a*d))*sqrt(-(-b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2)) - 10*(b^ 2*c^2 - 2*a*b*c*d + a^2*d^2)*(-b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c) ^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + 5*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(-b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (-b* d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d)) /(b*x + a)) + 3*(3*b^2*d^3*x - 5*b^2*c*d^2 + 8*a*b*d^3)*(b*x + a)^(2/3)*(d *x + c)^(1/3))/(b*d^4)]
\[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{3}}}{\left (c + d x\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{3}}}{{\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {5}{3}}}{{\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {(a+b x)^{5/3}}{(c+d x)^{2/3}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/3}}{{\left (c+d\,x\right )}^{2/3}} \,d x \]