\(\int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx\) [1653]

Optimal result

Integrand size = 33, antiderivative size = 339 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=-\frac {(c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{2 (b c-a d)^2 (a+b x)^2}+\frac {2 d (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3}}{(b c-a d)^3 (a+b x)}-\frac {2 d^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{b c+a d+2 b d x}}\right )}{\sqrt {3} b^{2/3} (b c-a d)^{8/3}}+\frac {2 d^2 (c+d x)^{2/3} (b c+a d+2 b d x)^{2/3} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{(b c-a d)^4}-\frac {2 d^2 \log (a+b x)}{3 b^{2/3} (b c-a d)^{8/3}}+\frac {d^2 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{b c+a d+2 b d x}\right )}{b^{2/3} (b c-a d)^{8/3}} \] Output:

-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*arctan(1/3* 
3^(1/2)+2/3*b^(2/3)*(d*x+c)^(2/3)*3^(1/2)/(-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)+2*d^2*(d*x+c)^(2/3)*(2*b*d*x+a* 
d+b*c)^(2/3)*hypergeom([1, 4/3],[5/3],2*b*(d*x+c)/(-a*d+b*c))/(-a*d+b*c)^4 
-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)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 22.68 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\frac {(c+d x)^{2/3} \left (\frac {5 (-b c+5 a d+4 b d x) (a d+b (c+2 d x))}{(a+b x)^2}-\frac {d^2 \left (20 b (c+d x) (a d+b (c+2 d x))+15\ 2^{2/3} b (b c-a d) (c+d x) \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )-2\ 2^{2/3} (b c-a d)^2 \sqrt [3]{\frac {b c+a d+2 b d x}{b c+b d x}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},\frac {b c-a d}{2 b c+2 b d x},\frac {b c-a d}{b c+b d x}\right )\right )}{b^2 (c+d x)^2}\right )}{10 (b c-a d)^3 \sqrt [3]{a d+b (c+2 d x)}} \] Input:

Integrate[1/((a + b*x)^3*(c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3)),x]
 

Output:

((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))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.16, 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

\(\displaystyle -\frac {2 d \left (-\frac {2 d \left ((b c-a d) \left (-\frac {\sqrt {3} \arctan \left (\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{a d+b c+2 b d x}}+\frac {1}{\sqrt {3}}\right )}{2 b^{2/3} (b c-a d)^{2/3}}-\frac {\log (a+b x)}{2 b^{2/3} (b c-a d)^{2/3}}+\frac {3 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{a d+b c+2 b d x}\right )}{4 b^{2/3} (b c-a d)^{2/3}}\right )-\frac {3 (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 \sqrt [3]{a d+b c+2 b d x}}\right )}{b c-a d}-\frac {3 (c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{(a+b x) (b c-a d)}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{2/3} (a d+b c+2 b d x)^{2/3}}{2 (a+b x)^2 (b c-a d)^2}\)

Input:

Int[1/((a + b*x)^3*(c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3)),x]
 

Output:

-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)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

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]))
 

rule 80

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])
 

rule 133

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]
 

rule 134

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]
 

rule 168

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]
 

rule 175

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]
 

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{3} \left (x d +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}d x\]

Input:

int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
 

Output:

int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm=" 
fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{3} \sqrt [3]{c + d x} \sqrt [3]{a d + b c + 2 b d x}}\, dx \] Input:

integrate(1/(b*x+a)**3/(d*x+c)**(1/3)/(2*b*d*x+a*d+b*c)**(1/3),x)
 

Output:

Integral(1/((a + b*x)**3*(c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)), x 
)
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm=" 
maxima")
 

Output:

integrate(1/((2*b*d*x + b*c + a*d)^(1/3)*(b*x + a)^3*(d*x + c)^(1/3)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int { \frac {1}{{\left (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (b x + a\right )}^{3} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x, algorithm=" 
giac")
 

Output:

integrate(1/((2*b*d*x + b*c + a*d)^(1/3)*(b*x + a)^3*(d*x + c)^(1/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{1/3}} \,d x \] Input:

int(1/((a + b*x)^3*(c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(1/3)),x)
 

Output:

int(1/((a + b*x)^3*(c + d*x)^(1/3)*(a*d + b*c + 2*b*d*x)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^3 \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{3}+3 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a^{2} b x +3 \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} a \,b^{2} x^{2}+\left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}} b^{3} x^{3}}d x \] Input:

int(1/(b*x+a)^3/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(1/3),x)
 

Output:

int(1/((c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**3 + 3*(c + d*x)**( 
1/3)*(a*d + b*c + 2*b*d*x)**(1/3)*a**2*b*x + 3*(c + d*x)**(1/3)*(a*d + b*c 
 + 2*b*d*x)**(1/3)*a*b**2*x**2 + (c + d*x)**(1/3)*(a*d + b*c + 2*b*d*x)**( 
1/3)*b**3*x**3),x)