Integrand size = 26, antiderivative size = 325 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (d e-c f) (e+f x)^2}-\frac {(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{6 (b e-a f) (d e-c f) (e+f x)}+\frac {(b c-a d)^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{4/3}}-\frac {(b c-a d)^2 \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{4/3}}+\frac {(b c-a d)^2 \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{4/3}} \]
1/2*(b*x+a)^(1/3)*(d*x+c)^(5/3)/(-c*f+d*e)/(f*x+e)^2-1/6*(-a*d+b*c)*(b*x+a )^(1/3)*(d*x+c)^(2/3)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)-1/18*(-a*d+b*c)^2*ln(f *x+e)/(-a*f+b*e)^(5/3)/(-c*f+d*e)^(4/3)+1/6*(-a*d+b*c)^2*ln(-(b*x+a)^(1/3) +(-a*f+b*e)^(1/3)*(d*x+c)^(1/3)/(-c*f+d*e)^(1/3))/(-a*f+b*e)^(5/3)/(-c*f+d *e)^(4/3)+1/9*(-a*d+b*c)^2*arctan(1/3*3^(1/2)+2/3*(-a*f+b*e)^(1/3)*(d*x+c) ^(1/3)/(-c*f+d*e)^(1/3)/(b*x+a)^(1/3)*3^(1/2))/(-a*f+b*e)^(5/3)/(-c*f+d*e) ^(4/3)*3^(1/2)
Time = 3.17 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\frac {1}{18} (b c-a d)^2 \left (\frac {3 \sqrt [3]{a+b x} (c+d x)^{2/3} (b (-2 c e-3 d e x+c f x)+a (-d e+3 c f+2 d f x))}{(b c-a d)^2 (b e-a f) (-d e+c f) (e+f x)^2}-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}+\frac {2 \log \left (\sqrt [3]{b e-a f}+\frac {\sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}-\frac {\log \left ((b e-a f)^{2/3}+\frac {(-d e+c f)^{2/3} (a+b x)^{2/3}}{(c+d x)^{2/3}}-\frac {\sqrt [3]{b e-a f} \sqrt [3]{-d e+c f} \sqrt [3]{a+b x}}{\sqrt [3]{c+d x}}\right )}{(b e-a f)^{5/3} (-d e+c f)^{4/3}}\right ) \]
((b*c - a*d)^2*((3*(a + b*x)^(1/3)*(c + d*x)^(2/3)*(b*(-2*c*e - 3*d*e*x + c*f*x) + a*(-(d*e) + 3*c*f + 2*d*f*x)))/((b*c - a*d)^2*(b*e - a*f)*(-(d*e) + c*f)*(e + f*x)^2) - (2*Sqrt[3]*ArcTan[(1 - (2*(-(d*e) + c*f)^(1/3)*(a + b*x)^(1/3))/((b*e - a*f)^(1/3)*(c + d*x)^(1/3)))/Sqrt[3]])/((b*e - a*f)^( 5/3)*(-(d*e) + c*f)^(4/3)) + (2*Log[(b*e - a*f)^(1/3) + ((-(d*e) + c*f)^(1 /3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)])/((b*e - a*f)^(5/3)*(-(d*e) + c*f)^( 4/3)) - Log[(b*e - a*f)^(2/3) + ((-(d*e) + c*f)^(2/3)*(a + b*x)^(2/3))/(c + d*x)^(2/3) - ((b*e - a*f)^(1/3)*(-(d*e) + c*f)^(1/3)*(a + b*x)^(1/3))/(c + d*x)^(1/3)]/((b*e - a*f)^(5/3)*(-(d*e) + c*f)^(4/3))))/18
Time = 0.32 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {105, 105, 102}
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.
| \(\displaystyle \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d) \int \frac {(c+d x)^{2/3}}{(a+b x)^{2/3} (e+f x)^2}dx}{6 (d e-c f)}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}dx}{3 (b e-a f)}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x) (b e-a f)}\right )}{6 (d e-c f)}\) |
| \(\Big \downarrow \) 102 |
| \(\displaystyle \frac {\sqrt [3]{a+b x} (c+d x)^{5/3}}{2 (e+f x)^2 (d e-c f)}-\frac {(b c-a d) \left (\frac {2 (b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{(b e-a f)^{2/3} \sqrt [3]{d e-c f}}+\frac {\log (e+f x)}{2 (b e-a f)^{2/3} \sqrt [3]{d e-c f}}-\frac {3 \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{2 (b e-a f)^{2/3} \sqrt [3]{d e-c f}}\right )}{3 (b e-a f)}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x) (b e-a f)}\right )}{6 (d e-c f)}\) |
((a + b*x)^(1/3)*(c + d*x)^(5/3))/(2*(d*e - c*f)*(e + f*x)^2) - ((b*c - a* d)*(((a + b*x)^(1/3)*(c + d*x)^(2/3))/((b*e - a*f)*(e + f*x)) + (2*(b*c - a*d)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(b*e - a*f)^(1/3)*(c + d*x)^(1/3))/ (Sqrt[3]*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))])/((b*e - a*f)^(2/3)*(d*e - c* f)^(1/3))) + Log[e + f*x]/(2*(b*e - a*f)^(2/3)*(d*e - c*f)^(1/3)) - (3*Log [-(a + b*x)^(1/3) + ((b*e - a*f)^(1/3)*(c + d*x)^(1/3))/(d*e - c*f)^(1/3)] )/(2*(b*e - a*f)^(2/3)*(d*e - c*f)^(1/3))))/(3*(b*e - a*f))))/(6*(d*e - c* f))
3.31.6.3.1 Defintions of rubi rules used
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.) *(x_))), x_] :> With[{q = Rt[(d*e - c*f)/(b*e - a*f), 3]}, Simp[(-Sqrt[3])* q*(ArcTan[1/Sqrt[3] + 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))]/(d*e - c*f)), x] + (Simp[q*(Log[e + f*x]/(2*(d*e - c*f))), x] - Simp[3*q*(Log[q *(a + b*x)^(1/3) - (c + d*x)^(1/3)]/(2*(d*e - c*f))), x])] /; FreeQ[{a, b, c, d, e, f}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
\[\int \frac {\left (b x +a \right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {2}{3}}}{\left (f x +e \right )^{3}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (275) = 550\).
Time = 0.40 (sec) , antiderivative size = 3992, normalized size of antiderivative = 12.28 \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\text {Too large to display} \]
[1/18*(3*sqrt(1/3)*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^4 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^3*f + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^2 + ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^2*f^2 - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f^3 + (a*b^2*c^3 - 2*a^ 2*b*c^2*d + a^3*c*d^2)*f^4)*x^2 + 2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^ 3)*e^3*f - (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e^2*f^2 + (a*b^ 2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f^3)*x)*sqrt((-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)/(d*e - c*f))*lo g(-(3*a^2*c*f^2 + (b^2*c + 2*a*b*d)*e^2 - 2*(2*a*b*c + a^2*d)*e*f + 3*(-b^ 2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^( 1/3)*(b*e - a*f)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (3*b^2*d*e^2 - 2*(b^2*c + 2*a*b*d)*e*f + (2*a*b*c + a^2*d)*f^2)*x - 3*sqrt(1/3)*(2*(b*d*e^2 + a*c *f^2 - (b*c + a*d)*e*f)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b^2*d*e^3 + a^ 2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2 *f - (2*a*b*c + a^2*d)*e*f^2)^(1/3)*(b*c*e - a*c*f + (b*d*e - a*d*f)*x))*s qrt((-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)* e*f^2)^(1/3)/(d*e - c*f)))/(f*x + e)) - (-b^2*d*e^3 + a^2*c*f^3 + (b^2*c + 2*a*b*d)*e^2*f - (2*a*b*c + a^2*d)*e*f^2)^(2/3)*((b^2*c^2 - 2*a*b*c*d + a ^2*d^2)*f^2*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f*x + (b^2*c^2 - ...
\[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\int \frac {\sqrt [3]{a + b x} \left (c + d x\right )^{\frac {2}{3}}}{\left (e + f x\right )^{3}}\, dx \]
\[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{{\left (f x + e\right )}^{3}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{{\left (f x + e\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/3}\,{\left (c+d\,x\right )}^{2/3}}{{\left (e+f\,x\right )}^3} \,d x \]