Integrand size = 33, antiderivative size = 790 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {32 d^2 (c+d x)^{2/3}}{(b c-a d)^4 \sqrt [3]{b c+a d+2 b d x}}-\frac {(c+d x)^{2/3}}{2 (b c-a d)^2 (a+b x)^2 \sqrt [3]{b c+a d+2 b d x}}+\frac {3 d (c+d x)^{2/3}}{(b c-a d)^3 (a+b x) \sqrt [3]{b c+a d+2 b d x}}-\frac {22 d^2 \arctan \left (\frac {\sqrt [3]{b}-\frac {2 \left (\sqrt [3]{b c-a d}+\sqrt [3]{b c+a d+2 b d x}\right )}{\sqrt [3]{c+d x}}}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} b^{2/3} (b c-a d)^{11/3}}-\frac {11 d^2 \arctan \left (\frac {1+\frac {\sqrt [3]{b c-a d} \left (1+\frac {\sqrt [3]{b c+a d+2 b d x}}{\sqrt [3]{b c-a d}}\right )}{\sqrt [3]{b} \sqrt [3]{c+d x}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3} (b c-a d)^{11/3}}-\frac {8 d^2 (b c+a d+2 b d x)^{2/3} \sqrt [3]{1+\frac {b c+a d+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-\frac {b c+a d+2 b d x}{b c-a d}\right )}{b (b c-a d)^4 \sqrt [3]{c+d x}}-\frac {11 d^2 \log \left (\left (\sqrt [3]{b c-a d}-\sqrt [3]{b c+a d+2 b d x}\right )^2 \left (\sqrt [3]{b c-a d}+\sqrt [3]{b c+a d+2 b d x}\right )\right )}{6 b^{2/3} (b c-a d)^{11/3}}+\frac {11 d^2 \log \left (\sqrt [3]{b c-a d}-2 \sqrt [3]{b} \sqrt [3]{c+d x}+\sqrt [3]{b c+a d+2 b d x}\right )}{2 b^{2/3} (b c-a d)^{11/3}}+\frac {22 d^2 \log \left (\sqrt [3]{b}+\frac {\sqrt [3]{b c-a d}+\sqrt [3]{b c+a d+2 b d x}}{\sqrt [3]{c+d x}}\right )}{3 b^{2/3} (b c-a d)^{11/3}}-\frac {11 d^2 \log \left (b^{2/3}-\frac {\sqrt [3]{b} \sqrt [3]{b c-a d} \left (1+\frac {\sqrt [3]{b c+a d+2 b d x}}{\sqrt [3]{b c-a d}}\right )}{\sqrt [3]{c+d x}}+\frac {(b c-a d)^{2/3} \left (1+\frac {\sqrt [3]{b c+a d+2 b d x}}{\sqrt [3]{b c-a d}}\right )^2}{(c+d x)^{2/3}}\right )}{3 b^{2/3} (b c-a d)^{11/3}} \] Output:
32*d^2*(d*x+c)^(2/3)/(-a*d+b*c)^4/(2*b*d*x+a*d+b*c)^(1/3)-1/2*(d*x+c)^(2/3 )/(-a*d+b*c)^2/(b*x+a)^2/(2*b*d*x+a*d+b*c)^(1/3)+3*d*(d*x+c)^(2/3)/(-a*d+b *c)^3/(b*x+a)/(2*b*d*x+a*d+b*c)^(1/3)-22/3*d^2*arctan(1/3*(b^(1/3)-2*((-a* d+b*c)^(1/3)+(2*b*d*x+a*d+b*c)^(1/3))/(d*x+c)^(1/3))*3^(1/2)/b^(1/3))*3^(1 /2)/b^(2/3)/(-a*d+b*c)^(11/3)-11/3*d^2*arctan(1/3*(1+(-a*d+b*c)^(1/3)*(1+( 2*b*d*x+a*d+b*c)^(1/3)/(-a*d+b*c)^(1/3))/b^(1/3)/(d*x+c)^(1/3))*3^(1/2))*3 ^(1/2)/b^(2/3)/(-a*d+b*c)^(11/3)-8*d^2*(2*b*d*x+a*d+b*c)^(2/3)*(1+(2*b*d*x +a*d+b*c)/(-a*d+b*c))^(1/3)*hypergeom([1/3, 2/3],[5/3],-(2*b*d*x+a*d+b*c)/ (-a*d+b*c))/b/(-a*d+b*c)^4/(d*x+c)^(1/3)-11/6*d^2*ln(((-a*d+b*c)^(1/3)-(2* b*d*x+a*d+b*c)^(1/3))^2*((-a*d+b*c)^(1/3)+(2*b*d*x+a*d+b*c)^(1/3)))/b^(2/3 )/(-a*d+b*c)^(11/3)+11/2*d^2*ln((-a*d+b*c)^(1/3)-2*b^(1/3)*(d*x+c)^(1/3)+( 2*b*d*x+a*d+b*c)^(1/3))/b^(2/3)/(-a*d+b*c)^(11/3)+22/3*d^2*ln(b^(1/3)+((-a *d+b*c)^(1/3)+(2*b*d*x+a*d+b*c)^(1/3))/(d*x+c)^(1/3))/b^(2/3)/(-a*d+b*c)^( 11/3)-11/3*d^2*ln(b^(2/3)-b^(1/3)*(-a*d+b*c)^(1/3)*(1+(2*b*d*x+a*d+b*c)^(1 /3)/(-a*d+b*c)^(1/3))/(d*x+c)^(1/3)+(-a*d+b*c)^(2/3)*(1+(2*b*d*x+a*d+b*c)^ (1/3)/(-a*d+b*c)^(1/3))^2/(d*x+c)^(2/3))/b^(2/3)/(-a*d+b*c)^(11/3)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 22.83 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.41 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {(c+d x)^{2/3} \left (\frac {5 \left (57 a^2 d^2+2 a b d (4 c+61 d x)+b^2 \left (-c^2+6 c d x+64 d^2 x^2\right )\right )}{(a+b x)^2}-\frac {d^2 \left (160 b (c+d x) (a d+b (c+2 d x))+95\ 2^{2/3} b (b c-a d) (c+d x) \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )-16\ 2^{2/3} (b c-a d)^2 \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )\right )}{b^2 (c+d x)^2}\right )}{10 (b c-a d)^4 \sqrt [3]{a d+b (c+2 d x)}} \] Input:
Integrate[1/((a + b*x)^3*(c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]
Output:
((c + d*x)^(2/3)*((5*(57*a^2*d^2 + 2*a*b*d*(4*c + 61*d*x) + b^2*(-c^2 + 6* c*d*x + 64*d^2*x^2)))/(a + b*x)^2 - (d^2*(160*b*(c + d*x)*(a*d + b*(c + 2* d*x)) + 95*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)] - 16*2^(2/3)*(b*c - a*d)^2*((b*c + a*d + 2*b*d*x)/ (b*c + b*d*x))^(1/3)*AppellF1[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)^4*(a *d + b*(c + 2*d*x))^(1/3))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.22 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {154, 153}
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.
| \(\displaystyle \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (a d+b c+2 b d x)^{4/3}} \, dx\) |
| \(\Big \downarrow \) 154 |
| \(\displaystyle -\frac {\sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \left (-\frac {b c+a d}{b c-a d}-\frac {2 b d x}{b c-a d}\right )^{4/3}}dx}{(b c-a d) \sqrt [3]{a d+b c+2 b d x}}\) |
| \(\Big \downarrow \) 153 |
| \(\displaystyle \frac {3 d^2 (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {4}{3},3,\frac {5}{3},\frac {2 b (c+d x)}{b c-a d},\frac {b (c+d x)}{b c-a d}\right )}{2 (b c-a d)^4 \sqrt [3]{a d+b c+2 b d x}}\) |
Input:
Int[1/((a + b*x)^3*(c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]
Output:
(3*d^2*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b*d*x)/(b*c - a*d)))^(1/3)*Appell F1[2/3, 4/3, 3, 5/3, (2*b*(c + d*x))/(b*c - a*d), (b*(c + d*x))/(b*c - a*d )])/(2*(b*c - a*d)^4*(b*c + a*d + 2*b*d*x)^(1/3))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( b*c - a*d)], 0] && !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[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*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && !G tQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x]
\[\int \frac {1}{\left (b x +a \right )^{3} \left (x d +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}d x\]
Input:
int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x)
Output:
int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x)
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm=" fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)**3/(d*x+c)**(1/3)/(2*b*d*x+a*d+b*c)**(4/3),x)
Output:
Timed out
\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {4}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm=" maxima")
Output:
integrate(1/((2*b*d*x + b*c + a*d)^(4/3)*(b*x + a)^3*(d*x + c)^(1/3)), x)
\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {4}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm=" giac")
Output:
integrate(1/((2*b*d*x + b*c + a*d)^(4/3)*(b*x + a)^3*(d*x + c)^(1/3)), x)
Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{4/3}} \,d x \] Input:
int(1/((a + b*x)^3*(c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(4/3)),x)
Output:
int(1/((a + b*x)^3*(c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(4/3)), x)
\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{4} d +\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{3} b c +5 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{3} b d x +3 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{2} b^{2} c x +9 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{2} b^{2} d \,x^{2}+3 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a \,b^{3} c \,x^{2}+7 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a \,b^{3} d \,x^{3}+\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} b^{4} c \,x^{3}+2 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} b^{4} d \,x^{4}}d x \] Input:
int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x)
Output:
int(1/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**4*d + (c + d*x)**( 1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**3*b*c + 5*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**3*b*d*x + 3*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)** (1/3)*a**2*b**2*c*x + 9*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**2 *b**2*d*x**2 + 3*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a*b**3*c*x* *2 + 7*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a*b**3*d*x**3 + (c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*b**4*c*x**3 + 2*(c + d*x)**(1/3)* (a*d + b*c + 2*b*d*x)**(1/3)*b**4*d*x**4),x)