Integrand size = 26, antiderivative size = 386 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=-\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 (b e-a f) (d e-c f) (e+f x)^2}+\frac {(3 b d e-b c f-2 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (b e-a f) (d e-c f)^2 (e+f x)}+\frac {(b c-a d) (3 b d e-b c f-2 a d f) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} (b e-a f)^{5/3} (d e-c f)^{7/3}}-\frac {(b c-a d) (3 b d e-b c f-2 a d f) \log (e+f x)}{18 (b e-a f)^{5/3} (d e-c f)^{7/3}}+\frac {(b c-a d) (3 b d e-b c f-2 a d f) \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{6 (b e-a f)^{5/3} (d e-c f)^{7/3}} \]
-1/2*f*(b*x+a)^(4/3)*(d*x+c)^(2/3)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)^2+1/3*(-2 *a*d*f-b*c*f+3*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)/(-a*f+b*e)/(-c*f+d*e)^2/ (f*x+e)-1/18*(-a*d+b*c)*(-2*a*d*f-b*c*f+3*b*d*e)*ln(f*x+e)/(-a*f+b*e)^(5/3 )/(-c*f+d*e)^(7/3)+1/6*(-a*d+b*c)*(-2*a*d*f-b*c*f+3*b*d*e)*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)^(7/3)+1/9*(-a*d+b*c)*(-2*a*d*f-b*c*f+3*b*d*e)*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)^(7/3)*3^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\frac {\sqrt [3]{a+b x} \left (-3 f (a+b x) (c+d x)+\frac {2 (3 b d e-b c f-2 a d f) (e+f x) \left ((b e-a f) (c+d x)-(b c-a d) (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},1,\frac {4}{3},\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(b e-a f) (d e-c f)}\right )}{6 (b e-a f) (d e-c f) \sqrt [3]{c+d x} (e+f x)^2} \]
((a + b*x)^(1/3)*(-3*f*(a + b*x)*(c + d*x) + (2*(3*b*d*e - b*c*f - 2*a*d*f )*(e + f*x)*((b*e - a*f)*(c + d*x) - (b*c - a*d)*(e + f*x)*Hypergeometric2 F1[1/3, 1, 4/3, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/((b*e - a*f)*(d*e - c*f))))/(6*(b*e - a*f)*(d*e - c*f)*(c + d*x)^(1/3)*(e + f*x)^ 2)
Time = 0.34 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {107, 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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {(-2 a d f-b c f+3 b d e) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^2}dx}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 (e+f x)^2 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {(-2 a d f-b c f+3 b d e) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x) (d e-c f)}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}dx}{3 (d e-c f)}\right )}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 (e+f x)^2 (b e-a f) (d e-c f)}\) |
\(\Big \downarrow \) 102 |
\(\displaystyle \frac {(-2 a d f-b c f+3 b d e) \left (\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{(e+f x) (d e-c f)}-\frac {(b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{(b e-a f)^{2/3} \sqrt [3]{d e-c f}}+\frac {\log (e+f x)}{2 (b e-a f)^{2/3} \sqrt [3]{d e-c f}}-\frac {3 \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{2 (b e-a f)^{2/3} \sqrt [3]{d e-c f}}\right )}{3 (d e-c f)}\right )}{3 (b e-a f) (d e-c f)}-\frac {f (a+b x)^{4/3} (c+d x)^{2/3}}{2 (e+f x)^2 (b e-a f) (d e-c f)}\) |
-1/2*(f*(a + b*x)^(4/3)*(c + d*x)^(2/3))/((b*e - a*f)*(d*e - c*f)*(e + f*x )^2) + ((3*b*d*e - b*c*f - 2*a*d*f)*(((a + b*x)^(1/3)*(c + d*x)^(2/3))/((d *e - c*f)*(e + f*x)) - ((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*(d*e - c*f))))/(3*(b*e - a*f)*(d*e - c*f))
3.31.13.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
\[\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} \left (f x +e \right )^{3}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 2307 vs. \(2 (336) = 672\).
Time = 0.73 (sec) , antiderivative size = 4776, normalized size of antiderivative = 12.37 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\text {Too large to display} \]
[1/18*(3*sqrt(1/3)*(3*(b^3*c*d^2 - a*b^2*d^3)*e^5 - (4*b^3*c^2*d + a*b^2*c *d^2 - 5*a^2*b*d^3)*e^4*f + (b^3*c^3 + 5*a*b^2*c^2*d - 4*a^2*b*c*d^2 - 2*a ^3*d^3)*e^3*f^2 - (a*b^2*c^3 + a^2*b*c^2*d - 2*a^3*c*d^2)*e^2*f^3 + (3*(b^ 3*c*d^2 - a*b^2*d^3)*e^3*f^2 - (4*b^3*c^2*d + a*b^2*c*d^2 - 5*a^2*b*d^3)*e ^2*f^3 + (b^3*c^3 + 5*a*b^2*c^2*d - 4*a^2*b*c*d^2 - 2*a^3*d^3)*e*f^4 - (a* b^2*c^3 + a^2*b*c^2*d - 2*a^3*c*d^2)*f^5)*x^2 + 2*(3*(b^3*c*d^2 - a*b^2*d^ 3)*e^4*f - (4*b^3*c^2*d + a*b^2*c*d^2 - 5*a^2*b*d^3)*e^3*f^2 + (b^3*c^3 + 5*a*b^2*c^2*d - 4*a^2*b*c*d^2 - 2*a^3*d^3)*e^2*f^3 - (a*b^2*c^3 + a^2*b*c^ 2*d - 2*a^3*c*d^2)*e*f^4)*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))*log(-(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))*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)))/(f*x + e)) - (b^2*d*e^3 - a^2*c*f^3 - (b^2*c + 2*a*b*d)*e^2*f +...
\[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\int \frac {\sqrt [3]{a + b x}}{\sqrt [3]{c + d x} \left (e + f x\right )^{3}}\, dx \]
\[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{3}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^3} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/3}}{{\left (e+f\,x\right )}^3\,{\left (c+d\,x\right )}^{1/3}} \,d x \]