\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{19/3}} \, dx\) [130]

Optimal result

Integrand size = 21, antiderivative size = 248 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\frac {(b c-a d)^2 x}{16 a b^2 \left (a+b x^3\right )^{16/3}}+\frac {(b c-a d) (15 b c+17 a d) x}{208 a^2 b^2 \left (a+b x^3\right )^{13/3}}+\frac {\left (45 b^2 c^2+6 a b c d+a^2 d^2\right ) x}{520 a^3 b^2 \left (a+b x^3\right )^{10/3}}+\frac {9 \left (45 b^2 c^2+6 a b c d+a^2 d^2\right ) x}{3640 a^4 b^2 \left (a+b x^3\right )^{7/3}}+\frac {27 \left (45 b^2 c^2+6 a b c d+a^2 d^2\right ) x}{7280 a^5 b^2 \left (a+b x^3\right )^{4/3}}+\frac {81 \left (45 b^2 c^2+6 a b c d+a^2 d^2\right ) x}{7280 a^6 b^2 \sqrt [3]{a+b x^3}} \] Output:

1/16*(-a*d+b*c)^2*x/a/b^2/(b*x^3+a)^(16/3)+1/208*(-a*d+b*c)*(17*a*d+15*b*c 
)*x/a^2/b^2/(b*x^3+a)^(13/3)+1/520*(a^2*d^2+6*a*b*c*d+45*b^2*c^2)*x/a^3/b^ 
2/(b*x^3+a)^(10/3)+9/3640*(a^2*d^2+6*a*b*c*d+45*b^2*c^2)*x/a^4/b^2/(b*x^3+ 
a)^(7/3)+27/7280*(a^2*d^2+6*a*b*c*d+45*b^2*c^2)*x/a^5/b^2/(b*x^3+a)^(4/3)+ 
81/7280*(a^2*d^2+6*a*b*c*d+45*b^2*c^2)*x/a^6/b^2/(b*x^3+a)^(1/3)
 

Mathematica [A] (verified)

Time = 2.57 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\frac {x \left (3645 b^5 c^2 x^{15}+486 a b^4 c x^{12} \left (40 c+d x^3\right )+81 a^2 b^3 x^9 \left (520 c^2+32 c d x^3+d^2 x^6\right )+520 a^5 \left (14 c^2+7 c d x^3+2 d^2 x^6\right )+144 a^3 b^2 x^6 \left (325 c^2+39 c d x^3+3 d^2 x^6\right )+156 a^4 b x^3 \left (175 c^2+40 c d x^3+6 d^2 x^6\right )\right )}{7280 a^6 \left (a+b x^3\right )^{16/3}} \] Input:

Integrate[(c + d*x^3)^2/(a + b*x^3)^(19/3),x]
 

Output:

(x*(3645*b^5*c^2*x^15 + 486*a*b^4*c*x^12*(40*c + d*x^3) + 81*a^2*b^3*x^9*( 
520*c^2 + 32*c*d*x^3 + d^2*x^6) + 520*a^5*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6) 
 + 144*a^3*b^2*x^6*(325*c^2 + 39*c*d*x^3 + 3*d^2*x^6) + 156*a^4*b*x^3*(175 
*c^2 + 40*c*d*x^3 + 6*d^2*x^6)))/(7280*a^6*(a + b*x^3)^(16/3))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {930, 910, 749, 749, 749, 746}

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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx\)

\(\Big \downarrow \) 930

\(\displaystyle \frac {\int \frac {4 d (3 b c+a d) x^3+c (15 b c+a d)}{\left (b x^3+a\right )^{16/3}}dx}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

\(\Big \downarrow \) 910

\(\displaystyle \frac {\frac {4}{13} \left (\frac {45 b c^2}{a}+\frac {a d^2}{b}+6 c d\right ) \int \frac {1}{\left (b x^3+a\right )^{13/3}}dx+\frac {x \left (\frac {15 b c^2}{a}-\frac {4 a d^2}{b}-11 c d\right )}{13 \left (a+b x^3\right )^{13/3}}}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {\frac {4}{13} \left (\frac {45 b c^2}{a}+\frac {a d^2}{b}+6 c d\right ) \left (\frac {9 \int \frac {1}{\left (b x^3+a\right )^{10/3}}dx}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )+\frac {x \left (\frac {15 b c^2}{a}-\frac {4 a d^2}{b}-11 c d\right )}{13 \left (a+b x^3\right )^{13/3}}}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {\frac {4}{13} \left (\frac {45 b c^2}{a}+\frac {a d^2}{b}+6 c d\right ) \left (\frac {9 \left (\frac {6 \int \frac {1}{\left (b x^3+a\right )^{7/3}}dx}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )+\frac {x \left (\frac {15 b c^2}{a}-\frac {4 a d^2}{b}-11 c d\right )}{13 \left (a+b x^3\right )^{13/3}}}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {\frac {4}{13} \left (\frac {45 b c^2}{a}+\frac {a d^2}{b}+6 c d\right ) \left (\frac {9 \left (\frac {6 \left (\frac {3 \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right )+\frac {x \left (\frac {15 b c^2}{a}-\frac {4 a d^2}{b}-11 c d\right )}{13 \left (a+b x^3\right )^{13/3}}}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

\(\Big \downarrow \) 746

\(\displaystyle \frac {\frac {4}{13} \left (\frac {9 \left (\frac {6 \left (\frac {3 x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )}{10 a}+\frac {x}{10 a \left (a+b x^3\right )^{10/3}}\right ) \left (\frac {45 b c^2}{a}+\frac {a d^2}{b}+6 c d\right )+\frac {x \left (\frac {15 b c^2}{a}-\frac {4 a d^2}{b}-11 c d\right )}{13 \left (a+b x^3\right )^{13/3}}}{16 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{16 a b \left (a+b x^3\right )^{16/3}}\)

Input:

Int[(c + d*x^3)^2/(a + b*x^3)^(19/3),x]
 

Output:

((b*c - a*d)*x*(c + d*x^3))/(16*a*b*(a + b*x^3)^(16/3)) + ((((15*b*c^2)/a 
- 11*c*d - (4*a*d^2)/b)*x)/(13*(a + b*x^3)^(13/3)) + (4*((45*b*c^2)/a + 6* 
c*d + (a*d^2)/b)*(x/(10*a*(a + b*x^3)^(10/3)) + (9*(x/(7*a*(a + b*x^3)^(7/ 
3)) + (6*(x/(4*a*(a + b*x^3)^(4/3)) + (3*x)/(4*a^2*(a + b*x^3)^(1/3))))/(7 
*a)))/(10*a)))/13)/(16*a*b)
 

Defintions of rubi rules used

rule 746

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) 
/a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
 

rule 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

rule 910

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - 
 b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1), x], x] /; 
FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ 
n + p, 0])
 

rule 930

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

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(\frac {\left (\left (\frac {1}{7} d^{2} x^{6}+\frac {1}{2} c d \,x^{3}+c^{2}\right ) a^{5}+\frac {15 b \left (\frac {6}{175} d^{2} x^{6}+\frac {8}{35} c d \,x^{3}+c^{2}\right ) x^{3} a^{4}}{4}+\frac {45 \left (\frac {3}{325} d^{2} x^{6}+\frac {3}{25} c d \,x^{3}+c^{2}\right ) b^{2} x^{6} a^{3}}{7}+\frac {81 \left (\frac {1}{520} d^{2} x^{6}+\frac {4}{65} c d \,x^{3}+c^{2}\right ) b^{3} x^{9} a^{2}}{14}+\frac {243 c \left (\frac {d \,x^{3}}{40}+c \right ) b^{4} x^{12} a}{91}+\frac {729 b^{5} c^{2} x^{15}}{1456}\right ) x}{\left (b \,x^{3}+a \right )^{\frac {16}{3}} a^{6}}\)	\(156\)
gosper	\(\frac {x \left (81 a^{2} b^{3} d^{2} x^{15}+486 a \,b^{4} c d \,x^{15}+3645 b^{5} c^{2} x^{15}+432 a^{3} b^{2} d^{2} x^{12}+2592 a^{2} b^{3} c d \,x^{12}+19440 a \,b^{4} c^{2} x^{12}+936 a^{4} b \,d^{2} x^{9}+5616 a^{3} b^{2} c d \,x^{9}+42120 a^{2} b^{3} c^{2} x^{9}+1040 a^{5} d^{2} x^{6}+6240 a^{4} b c d \,x^{6}+46800 a^{3} b^{2} c^{2} x^{6}+3640 a^{5} c d \,x^{3}+27300 a^{4} b \,c^{2} x^{3}+7280 c^{2} a^{5}\right )}{7280 \left (b \,x^{3}+a \right )^{\frac {16}{3}} a^{6}}\)	\(197\)
trager	\(\frac {x \left (81 a^{2} b^{3} d^{2} x^{15}+486 a \,b^{4} c d \,x^{15}+3645 b^{5} c^{2} x^{15}+432 a^{3} b^{2} d^{2} x^{12}+2592 a^{2} b^{3} c d \,x^{12}+19440 a \,b^{4} c^{2} x^{12}+936 a^{4} b \,d^{2} x^{9}+5616 a^{3} b^{2} c d \,x^{9}+42120 a^{2} b^{3} c^{2} x^{9}+1040 a^{5} d^{2} x^{6}+6240 a^{4} b c d \,x^{6}+46800 a^{3} b^{2} c^{2} x^{6}+3640 a^{5} c d \,x^{3}+27300 a^{4} b \,c^{2} x^{3}+7280 c^{2} a^{5}\right )}{7280 \left (b \,x^{3}+a \right )^{\frac {16}{3}} a^{6}}\)	\(197\)
orering	\(\frac {x \left (81 a^{2} b^{3} d^{2} x^{15}+486 a \,b^{4} c d \,x^{15}+3645 b^{5} c^{2} x^{15}+432 a^{3} b^{2} d^{2} x^{12}+2592 a^{2} b^{3} c d \,x^{12}+19440 a \,b^{4} c^{2} x^{12}+936 a^{4} b \,d^{2} x^{9}+5616 a^{3} b^{2} c d \,x^{9}+42120 a^{2} b^{3} c^{2} x^{9}+1040 a^{5} d^{2} x^{6}+6240 a^{4} b c d \,x^{6}+46800 a^{3} b^{2} c^{2} x^{6}+3640 a^{5} c d \,x^{3}+27300 a^{4} b \,c^{2} x^{3}+7280 c^{2} a^{5}\right )}{7280 \left (b \,x^{3}+a \right )^{\frac {16}{3}} a^{6}}\)	\(197\)

Input:

int((d*x^3+c)^2/(b*x^3+a)^(19/3),x,method=_RETURNVERBOSE)
 

Output:

((1/7*d^2*x^6+1/2*c*d*x^3+c^2)*a^5+15/4*b*(6/175*d^2*x^6+8/35*c*d*x^3+c^2) 
*x^3*a^4+45/7*(3/325*d^2*x^6+3/25*c*d*x^3+c^2)*b^2*x^6*a^3+81/14*(1/520*d^ 
2*x^6+4/65*c*d*x^3+c^2)*b^3*x^9*a^2+243/91*c*(1/40*d*x^3+c)*b^4*x^12*a+729 
/1456*b^5*c^2*x^15)/(b*x^3+a)^(16/3)*x/a^6
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.99 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\frac {{\left (81 \, {\left (45 \, b^{5} c^{2} + 6 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{16} + 432 \, {\left (45 \, a b^{4} c^{2} + 6 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x^{13} + 936 \, {\left (45 \, a^{2} b^{3} c^{2} + 6 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{10} + 7280 \, a^{5} c^{2} x + 1040 \, {\left (45 \, a^{3} b^{2} c^{2} + 6 \, a^{4} b c d + a^{5} d^{2}\right )} x^{7} + 1820 \, {\left (15 \, a^{4} b c^{2} + 2 \, a^{5} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{7280 \, {\left (a^{6} b^{6} x^{18} + 6 \, a^{7} b^{5} x^{15} + 15 \, a^{8} b^{4} x^{12} + 20 \, a^{9} b^{3} x^{9} + 15 \, a^{10} b^{2} x^{6} + 6 \, a^{11} b x^{3} + a^{12}\right )}} \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(19/3),x, algorithm="fricas")
 

Output:

1/7280*(81*(45*b^5*c^2 + 6*a*b^4*c*d + a^2*b^3*d^2)*x^16 + 432*(45*a*b^4*c 
^2 + 6*a^2*b^3*c*d + a^3*b^2*d^2)*x^13 + 936*(45*a^2*b^3*c^2 + 6*a^3*b^2*c 
*d + a^4*b*d^2)*x^10 + 7280*a^5*c^2*x + 1040*(45*a^3*b^2*c^2 + 6*a^4*b*c*d 
 + a^5*d^2)*x^7 + 1820*(15*a^4*b*c^2 + 2*a^5*c*d)*x^4)*(b*x^3 + a)^(2/3)/( 
a^6*b^6*x^18 + 6*a^7*b^5*x^15 + 15*a^8*b^4*x^12 + 20*a^9*b^3*x^9 + 15*a^10 
*b^2*x^6 + 6*a^11*b*x^3 + a^12)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\text {Timed out} \] Input:

integrate((d*x**3+c)**2/(b*x**3+a)**(19/3),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=-\frac {{\left (455 \, b^{3} - \frac {1680 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {2184 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {1040 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} d^{2} x^{16}}{7280 \, {\left (b x^{3} + a\right )}^{\frac {16}{3}} a^{4}} + \frac {{\left (455 \, b^{4} - \frac {2240 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {4368 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {4160 \, {\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac {1820 \, {\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} c d x^{16}}{3640 \, {\left (b x^{3} + a\right )}^{\frac {16}{3}} a^{5}} - \frac {{\left (91 \, b^{5} - \frac {560 \, {\left (b x^{3} + a\right )} b^{4}}{x^{3}} + \frac {1456 \, {\left (b x^{3} + a\right )}^{2} b^{3}}{x^{6}} - \frac {2080 \, {\left (b x^{3} + a\right )}^{3} b^{2}}{x^{9}} + \frac {1820 \, {\left (b x^{3} + a\right )}^{4} b}{x^{12}} - \frac {1456 \, {\left (b x^{3} + a\right )}^{5}}{x^{15}}\right )} c^{2} x^{16}}{1456 \, {\left (b x^{3} + a\right )}^{\frac {16}{3}} a^{6}} \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(19/3),x, algorithm="maxima")
 

Output:

-1/7280*(455*b^3 - 1680*(b*x^3 + a)*b^2/x^3 + 2184*(b*x^3 + a)^2*b/x^6 - 1 
040*(b*x^3 + a)^3/x^9)*d^2*x^16/((b*x^3 + a)^(16/3)*a^4) + 1/3640*(455*b^4 
 - 2240*(b*x^3 + a)*b^3/x^3 + 4368*(b*x^3 + a)^2*b^2/x^6 - 4160*(b*x^3 + a 
)^3*b/x^9 + 1820*(b*x^3 + a)^4/x^12)*c*d*x^16/((b*x^3 + a)^(16/3)*a^5) - 1 
/1456*(91*b^5 - 560*(b*x^3 + a)*b^4/x^3 + 1456*(b*x^3 + a)^2*b^3/x^6 - 208 
0*(b*x^3 + a)^3*b^2/x^9 + 1820*(b*x^3 + a)^4*b/x^12 - 1456*(b*x^3 + a)^5/x 
^15)*c^2*x^16/((b*x^3 + a)^(16/3)*a^6)
 

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {19}{3}}} \,d x } \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(19/3),x, algorithm="giac")
 

Output:

integrate((d*x^3 + c)^2/(b*x^3 + a)^(19/3), x)
 

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\frac {x\,\left (\frac {c^2}{16\,a}+\frac {a\,\left (\frac {d^2}{16\,b}-\frac {c\,d}{8\,a}\right )}{b}\right )}{{\left (b\,x^3+a\right )}^{16/3}}-\frac {x\,\left (\frac {d^2}{13\,b^2}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+15\,b^2\,c^2}{208\,a^2\,b^2}\right )}{{\left (b\,x^3+a\right )}^{13/3}}+\frac {x\,\left (a^2\,d^2+6\,a\,b\,c\,d+45\,b^2\,c^2\right )}{520\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{10/3}}+\frac {x\,\left (9\,a^2\,d^2+54\,a\,b\,c\,d+405\,b^2\,c^2\right )}{3640\,a^4\,b^2\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (27\,a^2\,d^2+162\,a\,b\,c\,d+1215\,b^2\,c^2\right )}{7280\,a^5\,b^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (81\,a^2\,d^2+486\,a\,b\,c\,d+3645\,b^2\,c^2\right )}{7280\,a^6\,b^2\,{\left (b\,x^3+a\right )}^{1/3}} \] Input:

int((c + d*x^3)^2/(a + b*x^3)^(19/3),x)
 

Output:

(x*(c^2/(16*a) + (a*(d^2/(16*b) - (c*d)/(8*a)))/b))/(a + b*x^3)^(16/3) - ( 
x*(d^2/(13*b^2) - (15*b^2*c^2 - a^2*d^2 + 2*a*b*c*d)/(208*a^2*b^2)))/(a + 
b*x^3)^(13/3) + (x*(a^2*d^2 + 45*b^2*c^2 + 6*a*b*c*d))/(520*a^3*b^2*(a + b 
*x^3)^(10/3)) + (x*(9*a^2*d^2 + 405*b^2*c^2 + 54*a*b*c*d))/(3640*a^4*b^2*( 
a + b*x^3)^(7/3)) + (x*(27*a^2*d^2 + 1215*b^2*c^2 + 162*a*b*c*d))/(7280*a^ 
5*b^2*(a + b*x^3)^(4/3)) + (x*(81*a^2*d^2 + 3645*b^2*c^2 + 486*a*b*c*d))/( 
7280*a^6*b^2*(a + b*x^3)^(1/3))
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{19/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{6}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5} b \,x^{3}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b^{2} x^{6}+20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{3} x^{9}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{4} x^{12}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{5} x^{15}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{6} x^{18}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{6}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5} b \,x^{3}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b^{2} x^{6}+20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{3} x^{9}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{4} x^{12}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{5} x^{15}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{6} x^{18}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{6}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5} b \,x^{3}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b^{2} x^{6}+20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{3} x^{9}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{4} x^{12}+6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{5} x^{15}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{6} x^{18}}d x \right ) c^{2} \] Input:

int((d*x^3+c)^2/(b*x^3+a)^(19/3),x)
 

Output:

int(x**6/((a + b*x**3)**(1/3)*a**6 + 6*(a + b*x**3)**(1/3)*a**5*b*x**3 + 1 
5*(a + b*x**3)**(1/3)*a**4*b**2*x**6 + 20*(a + b*x**3)**(1/3)*a**3*b**3*x* 
*9 + 15*(a + b*x**3)**(1/3)*a**2*b**4*x**12 + 6*(a + b*x**3)**(1/3)*a*b**5 
*x**15 + (a + b*x**3)**(1/3)*b**6*x**18),x)*d**2 + 2*int(x**3/((a + b*x**3 
)**(1/3)*a**6 + 6*(a + b*x**3)**(1/3)*a**5*b*x**3 + 15*(a + b*x**3)**(1/3) 
*a**4*b**2*x**6 + 20*(a + b*x**3)**(1/3)*a**3*b**3*x**9 + 15*(a + b*x**3)* 
*(1/3)*a**2*b**4*x**12 + 6*(a + b*x**3)**(1/3)*a*b**5*x**15 + (a + b*x**3) 
**(1/3)*b**6*x**18),x)*c*d + int(1/((a + b*x**3)**(1/3)*a**6 + 6*(a + b*x* 
*3)**(1/3)*a**5*b*x**3 + 15*(a + b*x**3)**(1/3)*a**4*b**2*x**6 + 20*(a + b 
*x**3)**(1/3)*a**3*b**3*x**9 + 15*(a + b*x**3)**(1/3)*a**2*b**4*x**12 + 6* 
(a + b*x**3)**(1/3)*a*b**5*x**15 + (a + b*x**3)**(1/3)*b**6*x**18),x)*c**2