\(\int \frac {x^{14}}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\) [762]

Optimal result

Integrand size = 24, antiderivative size = 275 \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {a^4}{b^4 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{5/3}}{5 b^4 d^2}+\frac {\left (a+b x^3\right )^{8/3}}{8 b^4 d}+\frac {c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{11/3} (b c-a d)^{4/3}}-\frac {c^4 \log \left (c+d x^3\right )}{6 d^{11/3} (b c-a d)^{4/3}}+\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} (b c-a d)^{4/3}} \] Output:

-a^4/b^4/(-a*d+b*c)/(b*x^3+a)^(1/3)+1/2*(3*a^2*d^2+2*a*b*c*d+b^2*c^2)*(b*x 
^3+a)^(2/3)/b^4/d^3-1/5*(3*a*d+b*c)*(b*x^3+a)^(5/3)/b^4/d^2+1/8*(b*x^3+a)^ 
(8/3)/b^4/d+1/3*c^4*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/ 
3))*3^(1/2))*3^(1/2)/d^(11/3)/(-a*d+b*c)^(4/3)-1/6*c^4*ln(d*x^3+c)/d^(11/3 
)/(-a*d+b*c)^(4/3)+1/2*c^4*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^ 
(11/3)/(-a*d+b*c)^(4/3)
 

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.27 \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {\frac {3 d^{2/3} \left (-81 a^4 d^3+9 a^3 b d^2 \left (c-3 d x^3\right )+3 a^2 b^2 d \left (4 c^2+c d x^3+3 d^2 x^6\right )+b^4 c x^3 \left (20 c^2-8 c d x^3+5 d^2 x^6\right )+a b^3 \left (20 c^3+4 c^2 d x^3-c d^2 x^6-5 d^3 x^9\right )\right )}{b^4 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {40 \sqrt {3} c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{(b c-a d)^{4/3}}+\frac {40 c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{4/3}}-\frac {20 c^4 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{4/3}}}{120 d^{11/3}} \] Input:

Integrate[x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((3*d^(2/3)*(-81*a^4*d^3 + 9*a^3*b*d^2*(c - 3*d*x^3) + 3*a^2*b^2*d*(4*c^2 
+ c*d*x^3 + 3*d^2*x^6) + b^4*c*x^3*(20*c^2 - 8*c*d*x^3 + 5*d^2*x^6) + a*b^ 
3*(20*c^3 + 4*c^2*d*x^3 - c*d^2*x^6 - 5*d^3*x^9)))/(b^4*(b*c - a*d)*(a + b 
*x^3)^(1/3)) + (40*Sqrt[3]*c^4*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/( 
b*c - a*d)^(1/3))/Sqrt[3]])/(b*c - a*d)^(4/3) + (40*c^4*Log[(b*c - a*d)^(1 
/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(4/3) - (20*c^4*Log[(b*c - a 
*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x 
^3)^(2/3)])/(b*c - a*d)^(4/3))/(120*d^(11/3))
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.28, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 98, 2009}

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.

\(\displaystyle \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^{12}}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx^3\)

\(\Big \downarrow \) 98

\(\displaystyle \frac {1}{3} \int \left (\frac {x^6}{b d \sqrt [3]{b x^3+a}}-\frac {(b c+a d) x^3}{b^2 d^2 \sqrt [3]{b x^3+a}}+\frac {b^2 c^2+a b d c+a^2 d^2}{b^3 d^3 \sqrt [3]{b x^3+a}}+\frac {c^4}{d^3 (a d-b c) \sqrt [3]{b x^3+a} \left (d x^3+c\right )}+\frac {a^4}{b^3 (b c-a d) \left (b x^3+a\right )^{4/3}}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-\frac {3 a^4}{b^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {3 a^2 \left (a+b x^3\right )^{2/3}}{2 b^4 d}+\frac {3 \left (a+b x^3\right )^{2/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{2 b^4 d^3}+\frac {\sqrt {3} c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{11/3} (b c-a d)^{4/3}}+\frac {3 a \left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^4 d^2}-\frac {3 \left (a+b x^3\right )^{5/3} (a d+b c)}{5 b^4 d^2}-\frac {6 a \left (a+b x^3\right )^{5/3}}{5 b^4 d}+\frac {3 \left (a+b x^3\right )^{8/3}}{8 b^4 d}-\frac {c^4 \log \left (c+d x^3\right )}{2 d^{11/3} (b c-a d)^{4/3}}+\frac {3 c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{11/3} (b c-a d)^{4/3}}\right )\)

Input:

Int[x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x]
 

Output:

((-3*a^4)/(b^4*(b*c - a*d)*(a + b*x^3)^(1/3)) + (3*a^2*(a + b*x^3)^(2/3))/ 
(2*b^4*d) + (3*a*(b*c + a*d)*(a + b*x^3)^(2/3))/(2*b^4*d^2) + (3*(b^2*c^2 
+ a*b*c*d + a^2*d^2)*(a + b*x^3)^(2/3))/(2*b^4*d^3) - (6*a*(a + b*x^3)^(5/ 
3))/(5*b^4*d) - (3*(b*c + a*d)*(a + b*x^3)^(5/3))/(5*b^4*d^2) + (3*(a + b* 
x^3)^(8/3))/(8*b^4*d) + (Sqrt[3]*c^4*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1 
/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(d^(11/3)*(b*c - a*d)^(4/3)) - (c^4*Log[ 
c + d*x^3])/(2*d^(11/3)*(b*c - a*d)^(4/3)) + (3*c^4*Log[(b*c - a*d)^(1/3) 
+ d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(11/3)*(b*c - a*d)^(4/3)))/3
 

Defintions of rubi rules used

rule 98

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x 
_)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( 
e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] 
&& IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(-\frac {-\frac {243 d \left (a \left (\frac {5}{81} b^{3} x^{9}-\frac {1}{9} a \,b^{2} x^{6}+\frac {1}{3} a^{2} b \,x^{3}+a^{3}\right ) d^{3}-\frac {\left (\frac {5}{9} b^{2} x^{6}-\frac {2}{3} a b \,x^{3}+a^{2}\right ) c \left (b \,x^{3}+a \right ) b \,d^{2}}{9}-\frac {4 c^{2} \left (b \,x^{3}+a \right ) \left (-\frac {2 b \,x^{3}}{3}+a \right ) b^{2} d}{27}-\frac {20 b^{3} c^{3} \left (b \,x^{3}+a \right )}{81}\right ) \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}{20}+b^{4} c^{4} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} d^{4} \left (a d -b c \right ) b^{4}}\)	\(314\)

Input:

int(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-1/6/((a*d-b*c)/d)^(1/3)/(b*x^3+a)^(1/3)*(-243/20*d*(a*(5/81*b^3*x^9-1/9*a 
*b^2*x^6+1/3*a^2*b*x^3+a^3)*d^3-1/9*(5/9*b^2*x^6-2/3*a*b*x^3+a^2)*c*(b*x^3 
+a)*b*d^2-4/27*c^2*(b*x^3+a)*(-2/3*b*x^3+a)*b^2*d-20/81*b^3*c^3*(b*x^3+a)) 
*((a*d-b*c)/d)^(1/3)+b^4*c^4*(b*x^3+a)^(1/3)*(-2*arctan(1/3*3^(1/2)*(2*(b* 
x^3+a)^(1/3)+((a*d-b*c)/d)^(1/3))/((a*d-b*c)/d)^(1/3))*3^(1/2)+ln((b*x^3+a 
)^(2/3)+((a*d-b*c)/d)^(1/3)*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(2/3))-2*ln((b*x 
^3+a)^(1/3)-((a*d-b*c)/d)^(1/3))))/d^4/(a*d-b*c)/b^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (231) = 462\).

Time = 0.15 (sec) , antiderivative size = 1300, normalized size of antiderivative = 4.73 \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:

integrate(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

[-1/120*(60*sqrt(1/3)*(a*b^5*c^5*d - a^2*b^4*c^4*d^2 + (b^6*c^5*d - a*b^5* 
c^4*d^2)*x^3)*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 
- b*c*d + 3*a*d^2 + 3*sqrt(1/3)*(2*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(2 
/3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) + (-b*c*d^2 + a*d^3)^(1/3)*(b*c - 
a*d))*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d)) - 3*(-b*c*d^2 + a*d^3)^(2 
/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c)) + 20*(b^5*c^4*x^3 + a*b^4*c^4)*(-b*c*d 
^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b* 
x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 40*(b^5*c^4*x^3 + a*b^4*c^4 
)*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1 
/3)) - 3*(20*a*b^4*c^4*d^2 - 8*a^2*b^3*c^3*d^3 - 3*a^3*b^2*c^2*d^4 - 90*a^ 
4*b*c*d^5 + 81*a^5*d^6 + 5*(b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x^9 
 - (8*b^5*c^3*d^3 - 7*a*b^4*c^2*d^4 - 10*a^2*b^3*c*d^5 + 9*a^3*b^2*d^6)*x^ 
6 + (20*b^5*c^4*d^2 - 16*a*b^4*c^3*d^3 - a^2*b^3*c^2*d^4 - 30*a^3*b^2*c*d^ 
5 + 27*a^4*b*d^6)*x^3)*(b*x^3 + a)^(2/3))/(a*b^6*c^2*d^5 - 2*a^2*b^5*c*d^6 
 + a^3*b^4*d^7 + (b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^3), -1/120* 
(120*sqrt(1/3)*(a*b^5*c^5*d - a^2*b^4*c^4*d^2 + (b^6*c^5*d - a*b^5*c^4*d^2 
)*x^3)*sqrt(-(-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*arctan(sqrt(1/3)*(2*(b* 
x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(1/3))*sqrt(-(-b*c*d^2 + a*d^3)^(1/3 
)/(b*c - a*d))/d) + 20*(b^5*c^4*x^3 + a*b^4*c^4)*(-b*c*d^2 + a*d^3)^(2/3)* 
log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*...
 

Sympy [F]

\[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{14}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**14/(b*x**3+a)**(4/3)/(d*x**3+c),x)
 

Output:

Integral(x**14/((a + b*x**3)**(4/3)*(c + d*x**3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.55 \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {\frac {120 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{4} c^{4} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d^{5} - 2 \, \sqrt {3} a b c d^{6} + \sqrt {3} a^{2} d^{7}} - \frac {20 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{4} c^{4} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{b^{2} c^{2} d^{5} - 2 \, a b c d^{6} + a^{2} d^{7}} + \frac {40 \, b^{4} c^{4} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}} - \frac {120 \, a^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}} + \frac {3 \, {\left (20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{2} c^{2} d^{5} - 8 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b c d^{6} + 40 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b c d^{6} + 5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} d^{7} - 24 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a d^{7} + 60 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} d^{7}\right )}}{d^{8}}}{120 \, b^{4}} \] Input:

integrate(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

1/120*(120*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*arctan(1/3*sqrt(3)*(2*(b*x^3 + 
 a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/(sqrt(3)*b^2*c 
^2*d^5 - 2*sqrt(3)*a*b*c*d^6 + sqrt(3)*a^2*d^7) - 20*(-b*c*d^2 + a*d^3)^(2 
/3)*b^4*c^4*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/ 
3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d^5 - 2*a*b*c*d^6 + a^2*d^7) + 40*b^ 
4*c^4*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^ 
(1/3)))/(b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5) - 120*a^4/((b*x^3 + a)^(1/3) 
*(b*c - a*d)) + 3*(20*(b*x^3 + a)^(2/3)*b^2*c^2*d^5 - 8*(b*x^3 + a)^(5/3)* 
b*c*d^6 + 40*(b*x^3 + a)^(2/3)*a*b*c*d^6 + 5*(b*x^3 + a)^(8/3)*d^7 - 24*(b 
*x^3 + a)^(5/3)*a*d^7 + 60*(b*x^3 + a)^(2/3)*a^2*d^7)/d^8)/b^4
 

Mupad [B] (verification not implemented)

Time = 4.39 (sec) , antiderivative size = 564, normalized size of antiderivative = 2.05 \[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\left (\frac {3\,a^2}{b^4\,d}+\frac {\left (\frac {4\,a}{b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{b^8\,d^2}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{2\,b^4\,d}\right )\,{\left (b\,x^3+a\right )}^{2/3}-\left (\frac {4\,a}{5\,b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{5\,b^8\,d^2}\right )\,{\left (b\,x^3+a\right )}^{5/3}+\frac {{\left (b\,x^3+a\right )}^{8/3}}{8\,b^4\,d}+\frac {a^4}{b^4\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}+\frac {c^4\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {c^8\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{9\,d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {{\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{36\,d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}{6\,d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {c^4\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^8\,d^5-b\,c^9\,d^4\right )-\frac {c^8\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\,\left (9\,a^4\,d^{15}-36\,a^3\,b\,c\,d^{14}+54\,a^2\,b^2\,c^2\,d^{13}-36\,a\,b^3\,c^3\,d^{12}+9\,b^4\,c^4\,d^{11}\right )}{d^{22/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{11/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \] Input:

int(x^14/((a + b*x^3)^(4/3)*(c + d*x^3)),x)
 

Output:

((3*a^2)/(b^4*d) + (((4*a)/(b^4*d) + (b^5*c - a*b^4*d)/(b^8*d^2))*(b^5*c - 
 a*b^4*d))/(2*b^4*d))*(a + b*x^3)^(2/3) - ((4*a)/(5*b^4*d) + (b^5*c - a*b^ 
4*d)/(5*b^8*d^2))*(a + b*x^3)^(5/3) + (a + b*x^3)^(8/3)/(8*b^4*d) + a^4/(b 
^4*(a + b*x^3)^(1/3)*(a*d - b*c)) + (c^4*log((a + b*x^3)^(1/3)*(a*c^8*d^5 
- b*c^9*d^4) - (c^8*(9*a^4*d^15 + 9*b^4*c^4*d^11 - 36*a*b^3*c^3*d^12 + 54* 
a^2*b^2*c^2*d^13 - 36*a^3*b*c*d^14))/(9*d^(22/3)*(a*d - b*c)^(8/3))))/(3*d 
^(11/3)*(a*d - b*c)^(4/3)) - (log((a + b*x^3)^(1/3)*(a*c^8*d^5 - b*c^9*d^4 
) - ((3^(1/2)*c^4*1i + c^4)^2*(9*a^4*d^15 + 9*b^4*c^4*d^11 - 36*a*b^3*c^3* 
d^12 + 54*a^2*b^2*c^2*d^13 - 36*a^3*b*c*d^14))/(36*d^(22/3)*(a*d - b*c)^(8 
/3)))*(3^(1/2)*c^4*1i + c^4))/(6*d^(11/3)*(a*d - b*c)^(4/3)) + (c^4*log((a 
 + b*x^3)^(1/3)*(a*c^8*d^5 - b*c^9*d^4) - (c^8*((3^(1/2)*1i)/6 - 1/6)^2*(9 
*a^4*d^15 + 9*b^4*c^4*d^11 - 36*a*b^3*c^3*d^12 + 54*a^2*b^2*c^2*d^13 - 36* 
a^3*b*c*d^14))/(d^(22/3)*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(1 
1/3)*(a*d - b*c)^(4/3))
 

Reduce [F]

\[ \int \frac {x^{14}}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{14}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b c \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b d \,x^{6}}d x \] Input:

int(x^14/(b*x^3+a)^(4/3)/(d*x^3+c),x)
 

Output:

int(x**14/((a + b*x**3)**(1/3)*a*c + (a + b*x**3)**(1/3)*a*d*x**3 + (a + b 
*x**3)**(1/3)*b*c*x**3 + (a + b*x**3)**(1/3)*b*d*x**6),x)