Integrand size = 33, antiderivative size = 1558 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=-\frac {(c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{2 (b c-a d)^2 (a+b x)^2}+\frac {2 d (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{(b c-a d)^3 (a+b x)}-\frac {2 \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2}}{b^{2/3} (b c-a d)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}-\frac {2 d^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{b c+a d+2 b d x}}\right )}{\sqrt {3} b^{2/3} (b c-a d)^{8/3}}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} d^2 \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2 \sqrt [3]{b} (b c-a d)^{2/3} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}+4 b^{2/3} ((c+d x) (a d+b (c+2 d x)))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}\right )|-7-4 \sqrt {3}\right )}{b^{2/3} (b c-a d)^{7/3} \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}}}-\frac {2 \sqrt {2} d^2 \sqrt [3]{(c+d x) (b c+a d+2 b d x)} \sqrt {\left (d (3 b c+a d)+4 b d^2 x\right )^2} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right ) \sqrt {\frac {(b c-a d)^{4/3}-2 \sqrt [3]{b} (b c-a d)^{2/3} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}+4 b^{2/3} ((c+d x) (a d+b (c+2 d x)))^{2/3}}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}{\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} (b c-a d)^{7/3} \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x} (3 b c+a d+4 b d x) \sqrt {d^2 (3 b c+a d+4 b d x)^2} \sqrt {\frac {(b c-a d)^{2/3} \left ((b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )}{\left (\left (1+\sqrt {3}\right ) (b c-a d)^{2/3}+2 \sqrt [3]{b} \sqrt [3]{(c+d x) (a d+b (c+2 d x))}\right )^2}}}-\frac {2 d^2 \log (a+b x)}{3 b^{2/3} (b c-a d)^{8/3}}+\frac {d^2 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{b c+a d+2 b d x}\right )}{b^{2/3} (b c-a d)^{8/3}} \]
-1/2*(d*x+c)^(2/3)*(2*b*d*x+a*d+b*c)^(2/3)/(-a*d+b*c)^2/(b*x+a)^2+2*d*(d*x +c)^(2/3)*(2*b*d*x+a*d+b*c)^(2/3)/(-a*d+b*c)^3/(b*x+a)-2/3*d^2*ln(b*x+a)/b ^(2/3)/(-a*d+b*c)^(8/3)+d^2*ln(b^(2/3)*(d*x+c)^(2/3)/(-a*d+b*c)^(1/3)-(2*b *d*x+a*d+b*c)^(1/3))/b^(2/3)/(-a*d+b*c)^(8/3)-2/3*d^2*arctan(1/3*3^(1/2)+2 /3*b^(2/3)*(d*x+c)^(2/3)/(-a*d+b*c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3)*3^(1/2)) /b^(2/3)/(-a*d+b*c)^(8/3)*3^(1/2)-2*((d*x+c)*(2*b*d*x+a*d+b*c))^(1/3)*(d^2 *(4*b*d*x+a*d+3*b*c)^2)^(1/2)*((d*(a*d+3*b*c)+4*b*d^2*x)^2)^(1/2)/b^(2/3)/ (-a*d+b*c)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3)/(4*b*d*x+a*d+3*b*c)/(2* b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))-2/ 3*d^2*((d*x+c)*(2*b*d*x+a*d+b*c))^(1/3)*((-a*d+b*c)^(2/3)+2*b^(1/3)*((d*x+ c)*(a*d+b*(2*d*x+c)))^(1/3))*EllipticF((2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c )))^(1/3)+(-a*d+b*c)^(2/3)*(1-3^(1/2)))/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+ c)))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*2^(1/2)*((d*(a*d+3 *b*c)+4*b*d^2*x)^2)^(1/2)*(((-a*d+b*c)^(4/3)-2*b^(1/3)*(-a*d+b*c)^(2/3)*(( d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+4*b^(2/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(2/3 ))/(2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2 )))^2)^(1/2)*3^(3/4)/b^(2/3)/(-a*d+b*c)^(7/3)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b *c)^(1/3)/(4*b*d*x+a*d+3*b*c)/(d^2*(4*b*d*x+a*d+3*b*c)^2)^(1/2)/((-a*d+b*c )^(2/3)*((-a*d+b*c)^(2/3)+2*b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3))/(2* b^(1/3)*((d*x+c)*(a*d+b*(2*d*x+c)))^(1/3)+(-a*d+b*c)^(2/3)*(1+3^(1/2)))...
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 23.11 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.20 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\frac {(c+d x)^{2/3} \left (\frac {5 (-b c+5 a d+4 b d x) (a d+b (c+2 d x))}{(a+b x)^2}-\frac {d^2 \left (20 b (c+d x) (a d+b (c+2 d x))+15\ 2^{2/3} b (b c-a d) (c+d x) \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )-2\ 2^{2/3} (b c-a d)^2 \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )\right )}{b^2 (c+d x)^2}\right )}{10 (b c-a d)^3 \sqrt [3]{a d+b (c+2 d x)}} \]
((c + d*x)^(2/3)*((5*(-(b*c) + 5*a*d + 4*b*d*x)*(a*d + b*(c + 2*d*x)))/(a + b*x)^2 - (d^2*(20*b*(c + d*x)*(a*d + b*(c + 2*d*x)) + 15*2^(2/3)*b*(b*c - a*d)*(c + d*x)*((b*c + a*d + 2*b*d*x)/(b*c + b*d*x))^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, (b*c - a*d)/(2*b*c + 2*b*d*x), (b*c - a*d)/(b*c + b*d*x)] - 2*2^(2/3)*(b*c - a*d)^2*((b*c + a*d + 2*b*d*x)/(b*c + b*d*x))^(1/3)*Appell F1[5/3, 1/3, 1, 8/3, (b*c - a*d)/(2*b*c + 2*b*d*x), (b*c - a*d)/(b*c + b*d *x)]))/(b^2*(c + d*x)^2)))/(10*(b*c - a*d)^3*(a*d + b*(c + 2*d*x))^(1/3))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.44 (sec) , antiderivative size = 393, normalized size of antiderivative = 0.25, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {134, 27, 168, 27, 175, 80, 79, 133}
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.
| \(\displaystyle \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{a d+b c+2 b d x}} \, dx\) |
| \(\Big \downarrow \) 134 |
| \(\displaystyle -\frac {d \int \frac {4 (3 b c-2 a d+b d x)}{(a+b x)^2 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{6 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 d \int \frac {3 b c-2 a d+b d x}{(a+b x)^2 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {2 d \left (-\frac {\int \frac {2 b d (b c-a d) (b c-2 a d-b d x)}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{b (b c-a d)^2}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 d \left (-\frac {2 d \int \frac {b c-2 a d-b d x}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle -\frac {2 d \left (-\frac {2 d \left ((b c-a d) \int \frac {1}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx-d \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx\right )}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle -\frac {2 d \left (-\frac {2 d \left ((b c-a d) \int \frac {1}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx-\frac {d \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \int \frac {1}{\sqrt [3]{c+d x} \sqrt [3]{-\frac {b c+a d}{b c-a d}-\frac {2 b d x}{b c-a d}}}dx}{\sqrt [3]{a d+b c+2 b d x}}\right )}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle -\frac {2 d \left (-\frac {2 d \left ((b c-a d) \int \frac {1}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}}dx-\frac {3 (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 \sqrt [3]{a d+b c+2 b d x}}\right )}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
| \(\Big \downarrow \) 133 |
| \(\displaystyle -\frac {2 d \left (-\frac {2 d \left ((b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{a d+b c+2 b d x}}+\frac {1}{\sqrt {3}}\right )}{2 b^{2/3} (b c-a d)^{2/3}}-\frac {\log (a+b x)}{2 b^{2/3} (b c-a d)^{2/3}}+\frac {3 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{a d+b c+2 b d x}\right )}{4 b^{2/3} (b c-a d)^{2/3}}\right )-\frac {3 (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 \sqrt [3]{a d+b c+2 b d x}}\right )}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\) |
-1/2*((c + d*x)^(2/3)*(b*c + a*d + 2*b*d*x)^(2/3))/((b*c - a*d)^2*(a + b*x )^2) - (2*d*((-3*(c + d*x)^(2/3)*(b*c + a*d + 2*b*d*x)^(2/3))/((b*c - a*d) *(a + b*x)) - (2*d*((-3*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b*d*x)/(b*c - a* d)))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])/ (2*(b*c + a*d + 2*b*d*x)^(1/3)) + (b*c - a*d)*(-1/2*(Sqrt[3]*ArcTan[1/Sqrt [3] + (2*b^(2/3)*(c + d*x)^(2/3))/(Sqrt[3]*(b*c - a*d)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3))])/(b^(2/3)*(b*c - a*d)^(2/3)) - Log[a + b*x]/(2*b^(2/3)*(b *c - a*d)^(2/3)) + (3*Log[(b^(2/3)*(c + d*x)^(2/3))/(b*c - a*d)^(1/3) - (b *c + a*d + 2*b*d*x)^(1/3)])/(4*b^(2/3)*(b*c - a*d)^(2/3)))))/(b*c - a*d))) /(3*(b*c - a*d)^2)
3.31.28.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x_)) ^(1/3)), x_] :> With[{q = Rt[b*((b*e - a*f)/(b*c - a*d)^2), 3]}, Simp[-Log[ a + b*x]/(2*q*(b*c - a*d)), x] + (-Simp[Sqrt[3]*(ArcTan[1/Sqrt[3] + 2*q*((c + d*x)^(2/3)/(Sqrt[3]*(e + f*x)^(1/3)))]/(2*q*(b*c - a*d))), x] + Simp[3*( Log[q*(c + d*x)^(2/3) - (e + f*x)^(1/3)]/(4*q*(b*c - a*d))), x])] /; FreeQ[ {a, b, c, d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0]
Int[((a_.) + (b_.)*(x_))^(m_)/(((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x _))^(1/3)), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(2/3)*((e + f*x)^(2/3 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[f/(6*(m + 1)*(b*c - a*d)*(b *e - a*f)) Int[(a + b*x)^(m + 1)*((a*d*(3*m + 1) - 3*b*c*(3*m + 5) - 2*b* d*(3*m + 7)*x)/((c + d*x)^(1/3)*(e + f*x)^(1/3))), x], x] /; FreeQ[{a, b, c , d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0] && ILtQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
\[\int \frac {1}{\left (b x +a \right )^{3} \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{1/3}} \,d x \]