\(\int \frac {1}{(a+b \sqrt [3]{x}+c x^{2/3})^{7/2}} \, dx\) [62]

Optimal result

Integrand size = 20, antiderivative size = 167 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=-\frac {6 \left (a b+\left (b^2-2 a c\right ) \sqrt [3]{x}\right )}{5 c \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}+\frac {2 \left (3 b^2+4 a c\right ) \left (b+2 c \sqrt [3]{x}\right )}{5 c \left (b^2-4 a c\right )^2 \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}-\frac {16 \left (3 b^2+4 a c\right ) \left (b+2 c \sqrt [3]{x}\right )}{5 \left (b^2-4 a c\right )^3 \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}} \] Output:

1/5*(-6*a*b-6*(-2*a*c+b^2)*x^(1/3))/c/(-4*a*c+b^2)/(a+b*x^(1/3)+c*x^(2/3)) 
^(5/2)+2/5*(4*a*c+3*b^2)*(b+2*c*x^(1/3))/c/(-4*a*c+b^2)^2/(a+b*x^(1/3)+c*x 
^(2/3))^(3/2)-16/5*(4*a*c+3*b^2)*(b+2*c*x^(1/3))/(-4*a*c+b^2)^3/(a+b*x^(1/ 
3)+c*x^(2/3))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.93 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=-\frac {2 \left (96 a^3 b c+3 b^2 x^{2/3} \left (5 b^3+30 b^2 c \sqrt [3]{x}+40 b c^2 x^{2/3}+16 c^3 x\right )+8 a^2 \left (b^3+30 b^2 c \sqrt [3]{x}+30 b c^2 x^{2/3}+20 c^3 x\right )+4 a \left (5 b^4 \sqrt [3]{x}+50 b^3 c x^{2/3}+60 b^2 c^2 x+40 b c^3 x^{4/3}+16 c^4 x^{5/3}\right )\right )}{5 \left (b^2-4 a c\right )^3 \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}} \] Input:

Integrate[(a + b*x^(1/3) + c*x^(2/3))^(-7/2),x]
 

Output:

(-2*(96*a^3*b*c + 3*b^2*x^(2/3)*(5*b^3 + 30*b^2*c*x^(1/3) + 40*b*c^2*x^(2/ 
3) + 16*c^3*x) + 8*a^2*(b^3 + 30*b^2*c*x^(1/3) + 30*b*c^2*x^(2/3) + 20*c^3 
*x) + 4*a*(5*b^4*x^(1/3) + 50*b^3*c*x^(2/3) + 60*b^2*c^2*x + 40*b*c^3*x^(4 
/3) + 16*c^4*x^(5/3))))/(5*(b^2 - 4*a*c)^3*(a + b*x^(1/3) + c*x^(2/3))^(5/ 
2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1680, 1164, 1159, 1088}

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.

\(\displaystyle \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 1680

\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{7/2}}d\sqrt [3]{x}\)

\(\Big \downarrow \) 1164

\(\displaystyle 3 \left (\frac {2 \sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{5 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}-\frac {2 \int \frac {2 a-3 b \sqrt [3]{x}}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{5/2}}d\sqrt [3]{x}}{5 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1159

\(\displaystyle 3 \left (\frac {2 \sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{5 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}-\frac {2 \left (-\frac {4 \left (4 a c+3 b^2\right ) \int \frac {1}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^{3/2}}d\sqrt [3]{x}}{3 \left (b^2-4 a c\right )}-\frac {2 \left (\sqrt [3]{x} \left (4 a c+3 b^2\right )+8 a b\right )}{3 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}\right )}{5 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1088

\(\displaystyle 3 \left (\frac {2 \sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{5 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{5/2}}-\frac {2 \left (\frac {8 \left (4 a c+3 b^2\right ) \left (b+2 c \sqrt [3]{x}\right )}{3 \left (b^2-4 a c\right )^2 \sqrt {a+b \sqrt [3]{x}+c x^{2/3}}}-\frac {2 \left (\sqrt [3]{x} \left (4 a c+3 b^2\right )+8 a b\right )}{3 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{3/2}}\right )}{5 \left (b^2-4 a c\right )}\right )\)

Input:

Int[(a + b*x^(1/3) + c*x^(2/3))^(-7/2),x]
 

Output:

3*((-2*((-2*(8*a*b + (3*b^2 + 4*a*c)*x^(1/3)))/(3*(b^2 - 4*a*c)*(a + b*x^( 
1/3) + c*x^(2/3))^(3/2)) + (8*(3*b^2 + 4*a*c)*(b + 2*c*x^(1/3)))/(3*(b^2 - 
 4*a*c)^2*Sqrt[a + b*x^(1/3) + c*x^(2/3)])))/(5*(b^2 - 4*a*c)) + (2*(2*a + 
 b*x^(1/3))*x^(1/3))/(5*(b^2 - 4*a*c)*(a + b*x^(1/3) + c*x^(2/3))^(5/2)))
 

Defintions of rubi rules used

rule 1088

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 
2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0]
 

rule 1159

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* 
x + c*x^2)^(p + 1), x] - Simp[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 - 4*a* 
c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] & 
& LtQ[p, -1] && NeQ[p, -3/2]
 

rule 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1680

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k 
 = Denominator[n]}, Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* 
n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr 
actionQ[n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(313\) vs. \(2(139)=278\).

Time = 0.02 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.88

method	result	size
derivativedivides	\(-\frac {3 x^{\frac {1}{3}}}{4 c \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}-\frac {9 b \left (-\frac {1}{5 c \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}-\frac {b \left (\frac {\frac {2 b}{5}+\frac {4 c \,x^{\frac {1}{3}}}{5}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {2 b}{3}+\frac {4 c \,x^{\frac {1}{3}}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \,x^{\frac {1}{3}}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{2 c}\right )}{8 c}+\frac {3 a \left (\frac {\frac {2 b}{5}+\frac {4 c \,x^{\frac {1}{3}}}{5}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {2 b}{3}+\frac {4 c \,x^{\frac {1}{3}}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \,x^{\frac {1}{3}}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{4 c}\)	\(314\)
default	\(-\frac {3 x^{\frac {1}{3}}}{4 c \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}-\frac {9 b \left (-\frac {1}{5 c \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}-\frac {b \left (\frac {\frac {2 b}{5}+\frac {4 c \,x^{\frac {1}{3}}}{5}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {2 b}{3}+\frac {4 c \,x^{\frac {1}{3}}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \,x^{\frac {1}{3}}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{2 c}\right )}{8 c}+\frac {3 a \left (\frac {\frac {2 b}{5}+\frac {4 c \,x^{\frac {1}{3}}}{5}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {5}{2}}}+\frac {16 c \left (\frac {\frac {2 b}{3}+\frac {4 c \,x^{\frac {1}{3}}}{3}}{\left (4 a c -b^{2}\right ) \left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{\frac {3}{2}}}+\frac {16 c \left (b +2 c \,x^{\frac {1}{3}}\right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}}}\right )}{5 \left (4 a c -b^{2}\right )}\right )}{4 c}\)	\(314\)

Input:

int(1/(a+b*x^(1/3)+c*x^(2/3))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-3/4*x^(1/3)/c/(a+b*x^(1/3)+c*x^(2/3))^(5/2)-9/8*b/c*(-1/5/c/(a+b*x^(1/3)+ 
c*x^(2/3))^(5/2)-1/2*b/c*(2/5*(b+2*c*x^(1/3))/(4*a*c-b^2)/(a+b*x^(1/3)+c*x 
^(2/3))^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*(b+2*c*x^(1/3))/(4*a*c-b^2)/(a+b*x^( 
1/3)+c*x^(2/3))^(3/2)+16/3*c/(4*a*c-b^2)^2*(b+2*c*x^(1/3))/(a+b*x^(1/3)+c* 
x^(2/3))^(1/2))))+3/4*a/c*(2/5*(b+2*c*x^(1/3))/(4*a*c-b^2)/(a+b*x^(1/3)+c* 
x^(2/3))^(5/2)+16/5*c/(4*a*c-b^2)*(2/3*(b+2*c*x^(1/3))/(4*a*c-b^2)/(a+b*x^ 
(1/3)+c*x^(2/3))^(3/2)+16/3*c/(4*a*c-b^2)^2*(b+2*c*x^(1/3))/(a+b*x^(1/3)+c 
*x^(2/3))^(1/2)))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^(7/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \sqrt [3]{x} + c x^{\frac {2}{3}}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(a+b*x**(1/3)+c*x**(2/3))**(7/2),x)
 

Output:

Integral((a + b*x**(1/3) + c*x**(2/3))**(-7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^(7/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (137) = 274\).

Time = 0.17 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.11 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (2 \, {\left (4 \, x^{\frac {1}{3}} {\left (\frac {2 \, {\left (3 \, b^{2} c^{3} + 4 \, a c^{4}\right )} x^{\frac {1}{3}}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}} + \frac {5 \, {\left (3 \, b^{3} c^{2} + 4 \, a b c^{3}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} + \frac {5 \, {\left (9 \, b^{4} c + 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x^{\frac {1}{3}} + \frac {5 \, {\left (3 \, b^{5} + 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x^{\frac {1}{3}} + \frac {20 \, {\left (a b^{4} + 12 \, a^{2} b^{2} c\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )} x^{\frac {1}{3}} + \frac {8 \, {\left (a^{2} b^{3} + 12 \, a^{3} b c\right )}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )}}{5 \, {\left (c x^{\frac {2}{3}} + b x^{\frac {1}{3}} + a\right )}^{\frac {5}{2}}} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^(7/2),x, algorithm="giac")
 

Output:

-2/5*(((2*(4*x^(1/3)*(2*(3*b^2*c^3 + 4*a*c^4)*x^(1/3)/(b^6 - 12*a*b^4*c + 
48*a^2*b^2*c^2 - 64*a^3*c^3) + 5*(3*b^3*c^2 + 4*a*b*c^3)/(b^6 - 12*a*b^4*c 
 + 48*a^2*b^2*c^2 - 64*a^3*c^3)) + 5*(9*b^4*c + 24*a*b^2*c^2 + 16*a^2*c^3) 
/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))*x^(1/3) + 5*(3*b^5 + 40 
*a*b^3*c + 48*a^2*b*c^2)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)) 
*x^(1/3) + 20*(a*b^4 + 12*a^2*b^2*c)/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 
64*a^3*c^3))*x^(1/3) + 8*(a^2*b^3 + 12*a^3*b*c)/(b^6 - 12*a*b^4*c + 48*a^2 
*b^2*c^2 - 64*a^3*c^3))/(c*x^(2/3) + b*x^(1/3) + a)^(5/2)
 

Mupad [B] (verification not implemented)

Time = 22.25 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.08 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx=\frac {\frac {2\,c^2\,x^{1/3}\,\left (56\,b^2+32\,a\,c\right )}{5\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}+\frac {b\,c\,\left (56\,b^2+32\,a\,c\right )}{5\,\left (4\,a\,c^2-b^2\,c\right )\,{\left (4\,a\,c-b^2\right )}^2}}{\sqrt {a+b\,x^{1/3}+c\,x^{2/3}}}+\frac {x^{1/3}\,\left (\frac {6\,b^2}{5\,\left (4\,a\,c^2-b^2\,c\right )}-\frac {12\,a\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )}\right )+\frac {6\,a\,b}{5\,\left (4\,a\,c^2-b^2\,c\right )}}{{\left (a+b\,x^{1/3}+c\,x^{2/3}\right )}^{5/2}}+\frac {\frac {16\,c^2\,x^{1/3}}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {8\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}}{\sqrt {a+b\,x^{1/3}+c\,x^{2/3}}}-\frac {\frac {4\,x^{1/3}}{5\,\left (4\,a\,c-b^2\right )}-\frac {2\,b}{5\,c\,\left (4\,a\,c-b^2\right )}}{{\left (a+b\,x^{1/3}+c\,x^{2/3}\right )}^{3/2}}+\frac {x^{1/3}\,\left (\frac {2\,c\,\left (8\,b^2+8\,a\,c\right )}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}+\frac {16\,a\,c^2}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {8\,b^2\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}\right )+\frac {b\,\left (8\,b^2+8\,a\,c\right )}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}-\frac {8\,a\,b\,c}{5\,\left (4\,a\,c^2-b^2\,c\right )\,\left (4\,a\,c-b^2\right )}}{{\left (a+b\,x^{1/3}+c\,x^{2/3}\right )}^{3/2}} \] Input:

int(1/(a + b*x^(1/3) + c*x^(2/3))^(7/2),x)
 

Output:

((2*c^2*x^(1/3)*(32*a*c + 56*b^2))/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2) + 
 (b*c*(32*a*c + 56*b^2))/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)^2))/(a + b*x^( 
1/3) + c*x^(2/3))^(1/2) + (x^(1/3)*((6*b^2)/(5*(4*a*c^2 - b^2*c)) - (12*a* 
c)/(5*(4*a*c^2 - b^2*c))) + (6*a*b)/(5*(4*a*c^2 - b^2*c)))/(a + b*x^(1/3) 
+ c*x^(2/3))^(5/2) + ((16*c^2*x^(1/3))/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) 
 + (8*b*c)/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)))/(a + b*x^(1/3) + c*x^(2/3) 
)^(1/2) - ((4*x^(1/3))/(5*(4*a*c - b^2)) - (2*b)/(5*c*(4*a*c - b^2)))/(a + 
 b*x^(1/3) + c*x^(2/3))^(3/2) + (x^(1/3)*((2*c*(8*a*c + 8*b^2))/(5*(4*a*c^ 
2 - b^2*c)*(4*a*c - b^2)) + (16*a*c^2)/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) 
 - (8*b^2*c)/(5*(4*a*c^2 - b^2*c)*(4*a*c - b^2))) + (b*(8*a*c + 8*b^2))/(5 
*(4*a*c^2 - b^2*c)*(4*a*c - b^2)) - (8*a*b*c)/(5*(4*a*c^2 - b^2*c)*(4*a*c 
- b^2)))/(a + b*x^(1/3) + c*x^(2/3))^(3/2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 869, normalized size of antiderivative = 5.20 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^{7/2}} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*x^(1/3)+c*x^(2/3))^(7/2),x)
 

Output:

(2*(240*x**(2/3)*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a**2*b*c**2 + 200*x**(2 
/3)*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a*b**3*c + 64*x**(2/3)*sqrt(x**(2/3) 
*c + x**(1/3)*b + a)*a*c**4*x + 15*x**(2/3)*sqrt(x**(2/3)*c + x**(1/3)*b + 
 a)*b**5 + 48*x**(2/3)*sqrt(x**(2/3)*c + x**(1/3)*b + a)*b**2*c**3*x + 240 
*x**(1/3)*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a**2*b**2*c + 20*x**(1/3)*sqrt 
(x**(2/3)*c + x**(1/3)*b + a)*a*b**4 + 160*x**(1/3)*sqrt(x**(2/3)*c + x**( 
1/3)*b + a)*a*b*c**3*x + 120*x**(1/3)*sqrt(x**(2/3)*c + x**(1/3)*b + a)*b* 
*3*c**2*x + 96*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a**3*b*c + 8*sqrt(x**(2/3 
)*c + x**(1/3)*b + a)*a**2*b**3 + 160*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a* 
*2*c**3*x + 240*sqrt(x**(2/3)*c + x**(1/3)*b + a)*a*b**2*c**2*x + 90*sqrt( 
x**(2/3)*c + x**(1/3)*b + a)*b**4*c*x - 192*x**(2/3)*sqrt(c)*a**3*c**2 - 3 
36*x**(2/3)*sqrt(c)*a**2*b**2*c - 144*x**(2/3)*sqrt(c)*a*b**4 - 192*x**(2/ 
3)*sqrt(c)*a*b*c**3*x - 144*x**(2/3)*sqrt(c)*b**3*c**2*x - 192*x**(1/3)*sq 
rt(c)*a**3*b*c - 144*x**(1/3)*sqrt(c)*a**2*b**3 - 192*x**(1/3)*sqrt(c)*a** 
2*c**3*x - 336*x**(1/3)*sqrt(c)*a*b**2*c**2*x - 144*x**(1/3)*sqrt(c)*b**4* 
c*x - 64*sqrt(c)*a**4*c - 48*sqrt(c)*a**3*b**2 - 384*sqrt(c)*a**2*b*c**2*x 
 - 352*sqrt(c)*a*b**3*c*x - 64*sqrt(c)*a*c**4*x**2 - 48*sqrt(c)*b**5*x - 4 
8*sqrt(c)*b**2*c**3*x**2))/(5*(192*x**(2/3)*a**5*c**4 + 48*x**(2/3)*a**4*b 
**2*c**3 - 108*x**(2/3)*a**3*b**4*c**2 + 192*x**(2/3)*a**3*b*c**5*x + 33*x 
**(2/3)*a**2*b**6*c - 144*x**(2/3)*a**2*b**3*c**4*x - 3*x**(2/3)*a*b**8...