\(\int \frac {(a+b x^3)^{11/3}}{(c+d x^3)^3} \, dx\) [162]

Optimal result

Integrand size = 21, antiderivative size = 458 \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\frac {b \left (18 b^2 c^2-7 a b c d-5 a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{18 c^2 d^3}-\frac {(b c-a d) x \left (a+b x^3\right )^{8/3}}{6 c d \left (c+d x^3\right )^2}-\frac {(b c-a d) (9 b c+5 a d) x \left (a+b x^3\right )^{5/3}}{18 c^2 d^2 \left (c+d x^3\right )}-\frac {b^{8/3} (9 b c-11 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} d^4}+\frac {(b c-a d)^{5/3} \left (27 b^2 c^2+12 a b c d+5 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} d^4}+\frac {(b c-a d)^{5/3} \left (27 b^2 c^2+12 a b c d+5 a^2 d^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} d^4}-\frac {(b c-a d)^{5/3} \left (27 b^2 c^2+12 a b c d+5 a^2 d^2\right ) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} d^4}+\frac {b^{8/3} (9 b c-11 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 d^4} \] Output:

1/18*b*(-5*a^2*d^2-7*a*b*c*d+18*b^2*c^2)*x*(b*x^3+a)^(2/3)/c^2/d^3-1/6*(-a 
*d+b*c)*x*(b*x^3+a)^(8/3)/c/d/(d*x^3+c)^2-1/18*(-a*d+b*c)*(5*a*d+9*b*c)*x* 
(b*x^3+a)^(5/3)/c^2/d^2/(d*x^3+c)-1/9*b^(8/3)*(-11*a*d+9*b*c)*arctan(1/3*( 
1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/d^4+1/27*(-a*d+b*c)^(5/3)* 
(5*a^2*d^2+12*a*b*c*d+27*b^2*c^2)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/ 
3)/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/c^(8/3)/d^4+1/54*(-a*d+b*c)^(5/3)*(5* 
a^2*d^2+12*a*b*c*d+27*b^2*c^2)*ln(d*x^3+c)/c^(8/3)/d^4-1/18*(-a*d+b*c)^(5/ 
3)*(5*a^2*d^2+12*a*b*c*d+27*b^2*c^2)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+ 
a)^(1/3))/c^(8/3)/d^4+1/6*b^(8/3)*(-11*a*d+9*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^ 
(1/3))/d^4
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 12.18 (sec) , antiderivative size = 908, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*x^3)^(11/3)/(c + d*x^3)^3,x]
 

Output:

((6*x*(a + b*x^3)^(2/3)*(6*b^3 - (3*(b*c - a*d)^3)/(c*(c + d*x^3)^2) + (5* 
(b*c - a*d)^2*(3*b*c + a*d))/(c^2*(c + d*x^3))))/d^3 - (81*b^4*x^4*(1 + (b 
*x^3)/a)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(d^ 
3*(a + b*x^3)^(1/3)) + (99*a*b^3*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1[4/3, 1 
/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(c*d^2*(a + b*x^3)^(1/3)) + (10*a 
^4*(2*Sqrt[3]*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^3)^(1/ 
3)))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] + 
 Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - 
 a*d)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(c^(8/3)*(b*c - a*d)^(1/3)) - (18*a*b^ 
3*c^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x 
^3)^(1/3)))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^( 
1/3)] + Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3) 
*(b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(d^3*(b*c - a*d)^(1/3)) + (16*a 
^2*b^2*(2*Sqrt[3]*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^3) 
^(1/3)))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3 
)] + Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b 
*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(c^(2/3)*d^2*(b*c - a*d)^(1/3)) + 
(4*a^3*b*(2*Sqrt[3]*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(b + a*x^ 
3)^(1/3)))/Sqrt[3]] - 2*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1 
/3)] + Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/...
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 432, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {930, 1023, 25, 1025, 27, 1026, 769, 901}

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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx\)

\(\Big \downarrow \) 930

\(\displaystyle \frac {\int \frac {\left (b x^3+a\right )^{5/3} \left (3 b (3 b c-a d) x^3+a (b c+5 a d)\right )}{\left (d x^3+c\right )^2}dx}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 1023

\(\displaystyle \frac {\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}-\frac {\int -\frac {\left (b x^3+a\right )^{2/3} \left (3 b \left (18 b^2 c^2-7 a b d c-5 a^2 d^2\right ) x^3+a \left (9 b^2 c^2-a b d c+10 a^2 d^2\right )\right )}{d x^3+c}dx}{3 c d}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (b x^3+a\right )^{2/3} \left (3 b \left (18 b^2 c^2-7 a b d c-5 a^2 d^2\right ) x^3+a \left (9 b^2 c^2-a b d c+10 a^2 d^2\right )\right )}{d x^3+c}dx}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {\frac {\frac {\int -\frac {6 \left (3 b^3 c^2 (9 b c-11 a d) x^3+a \left (9 b^3 c^3-8 a b^2 d c^2-2 a^2 b d^2 c-5 a^3 d^3\right )\right )}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 d}+\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{d}}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{d}-\frac {2 \int \frac {3 b^3 c^2 (9 b c-11 a d) x^3+a \left (9 b^3 c^3-8 a b^2 d c^2-2 a^2 b d^2 c-5 a^3 d^3\right )}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{d}}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 1026

\(\displaystyle \frac {\frac {\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{d}-\frac {2 \left (\frac {3 b^3 c^2 (9 b c-11 a d) \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{d}-\frac {(b c-a d)^2 \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{d}\right )}{d}}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {\frac {\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{d}-\frac {2 \left (\frac {3 b^3 c^2 (9 b c-11 a d) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{d}-\frac {(b c-a d)^2 \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{d}\right )}{d}}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {\frac {\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{d}-\frac {2 \left (\frac {3 b^3 c^2 (9 b c-11 a d) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{d}-\frac {(b c-a d)^2 \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{d}\right )}{d}}{3 c d}+\frac {x \left (a+b x^3\right )^{5/3} \left (\frac {5 a^2 d}{c}+4 a b-\frac {9 b^2 c}{d}\right )}{3 \left (c+d x^3\right )}}{6 c d}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2}\)

Input:

Int[(a + b*x^3)^(11/3)/(c + d*x^3)^3,x]
 

Output:

-1/6*((b*c - a*d)*x*(a + b*x^3)^(8/3))/(c*d*(c + d*x^3)^2) + (((4*a*b - (9 
*b^2*c)/d + (5*a^2*d)/c)*x*(a + b*x^3)^(5/3))/(3*(c + d*x^3)) + ((b*(18*b^ 
2*c^2 - 7*a*b*c*d - 5*a^2*d^2)*x*(a + b*x^3)^(2/3))/d - (2*(-(((b*c - a*d) 
^2*(27*b^2*c^2 + 12*a*b*c*d + 5*a^2*d^2)*(ArcTan[(1 + (2*(b*c - a*d)^(1/3) 
*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1/ 
3)) + Log[c + d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/3 
)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/d) + (3* 
b^3*c^2*(9*b*c - 11*a*d)*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqr 
t[3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3) 
)))/d))/d)/(3*c*d))/(6*c*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 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]
 

rule 1023

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

rule 1025

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

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.27

method	result	size
pseudoelliptic	\(\frac {-\frac {5 \left (a d -b c \right )^{2} \left (a^{2} d^{2}+\frac {12}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{54}+\frac {11 \left (a \,b^{\frac {8}{3}} d -\frac {9 b^{\frac {11}{3}} c}{11}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c^{3} \left (d \,x^{3}+c \right )^{2} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{18}+\frac {5 \left (a d -b c \right )^{2} \left (a^{2} d^{2}+\frac {12}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \left (d \,x^{3}+c \right )^{2} \sqrt {3}}{27}+\frac {5 \left (a d -b c \right )^{2} \left (a^{2} d^{2}+\frac {12}{5} a b c d +\frac {27}{5} b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{27}+\frac {4 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \left (-\frac {11 \left (a \,b^{\frac {8}{3}} d -\frac {9 b^{\frac {11}{3}} c}{11}\right ) c^{2} \left (d \,x^{3}+c \right )^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right )}{4}-\frac {11 \left (a \,b^{\frac {8}{3}} d -\frac {9 b^{\frac {11}{3}} c}{11}\right ) c^{2} \left (d \,x^{3}+c \right )^{2} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{4}+d \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {9 c^{4} b^{3}}{4}-2 d \left (-\frac {27 b \,x^{3}}{16}+a \right ) b^{2} c^{3}-\frac {\left (-\frac {3}{2} b^{2} x^{6}+\frac {25}{4} a b \,x^{3}+a^{2}\right ) d^{2} b \,c^{2}}{2}+a^{2} d^{3} \left (\frac {5 b \,x^{3}}{8}+a \right ) c +\frac {5 a^{3} d^{4} x^{3}}{8}\right ) x \right )}{9}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} d^{4} \left (d \,x^{3}+c \right )^{2} c^{3}}\)	\(580\)

Input:

int((b*x^3+a)^(11/3)/(d*x^3+c)^3,x,method=_RETURNVERBOSE)
 

Output:

5/27/((a*d-b*c)/c)^(1/3)*(-1/2*(a*d-b*c)^2*(a^2*d^2+12/5*a*b*c*d+27/5*b^2* 
c^2)*(d*x^3+c)^2*ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^3+a) 
^(1/3)*x+(b*x^3+a)^(2/3))/x^2)+33/10*(a*b^(8/3)*d-9/11*b^(11/3)*c)*((a*d-b 
*c)/c)^(1/3)*c^3*(d*x^3+c)^2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b* 
x^3+a)^(2/3))/x^2)+(a*d-b*c)^2*(a^2*d^2+12/5*a*b*c*d+27/5*b^2*c^2)*arctan( 
1/3*3^(1/2)*(-2/((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/x)*(d*x^3+c)^2*3^(1 
/2)+(a*d-b*c)^2*(a^2*d^2+12/5*a*b*c*d+27/5*b^2*c^2)*(d*x^3+c)^2*ln((((a*d- 
b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)+12/5*((a*d-b*c)/c)^(1/3)*c*(-11/4*(a*b 
^(8/3)*d-9/11*b^(11/3)*c)*c^2*(d*x^3+c)^2*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b 
*x^3+a)^(1/3)/b^(1/3)+x)/x)-11/4*(a*b^(8/3)*d-9/11*b^(11/3)*c)*c^2*(d*x^3+ 
c)^2*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)+d*(b*x^3+a)^(2/3)*(9/4*c^4*b^3-2*d 
*(-27/16*b*x^3+a)*b^2*c^3-1/2*(-3/2*b^2*x^6+25/4*a*b*x^3+a^2)*d^2*b*c^2+a^ 
2*d^3*(5/8*b*x^3+a)*c+5/8*a^3*d^4*x^3)*x))/d^4/(d*x^3+c)^2/c^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1246 vs. \(2 (394) = 788\).

Time = 19.59 (sec) , antiderivative size = 1246, normalized size of antiderivative = 2.72 \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((b*x^3+a)^(11/3)/(d*x^3+c)^3,x, algorithm="fricas")
 

Output:

1/54*(2*sqrt(3)*(27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2 - 5*a^3*c^2 
*d^3 + (27*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5)*x^6 
 + 2*(27*b^3*c^4*d - 15*a*b^2*c^3*d^2 - 7*a^2*b*c^2*d^3 - 5*a^3*c*d^4)*x^3 
)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*arctan(-1/3*(sqrt(3)*(b*c - 
a*d)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*c*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^ 
2)^(1/3))/((b*c - a*d)*x)) + 6*sqrt(3)*(9*b^3*c^5 - 11*a*b^2*c^4*d + (9*b^ 
3*c^3*d^2 - 11*a*b^2*c^2*d^3)*x^6 + 2*(9*b^3*c^4*d - 11*a*b^2*c^3*d^2)*x^3 
)*(-b^2)^(1/3)*arctan(-1/3*(sqrt(3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^ 
2)^(1/3))/(b*x)) - 2*(27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2 - 5*a^ 
3*c^2*d^3 + (27*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5 
)*x^6 + 2*(27*b^3*c^4*d - 15*a*b^2*c^3*d^2 - 7*a^2*b*c^2*d^3 - 5*a^3*c*d^4 
)*x^3)*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log((c*x*((b^2*c^2 - 2* 
a*b*c*d + a^2*d^2)/c^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d))/x) - 6*(9*b 
^3*c^5 - 11*a*b^2*c^4*d + (9*b^3*c^3*d^2 - 11*a*b^2*c^2*d^3)*x^6 + 2*(9*b^ 
3*c^4*d - 11*a*b^2*c^3*d^2)*x^3)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)*x - (b*x^ 
3 + a)^(1/3)*b)/x) + 3*(9*b^3*c^5 - 11*a*b^2*c^4*d + (9*b^3*c^3*d^2 - 11*a 
*b^2*c^2*d^3)*x^6 + 2*(9*b^3*c^4*d - 11*a*b^2*c^3*d^2)*x^3)*(-b^2)^(1/3)*l 
og(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x - (b*x^3 + a)^( 
2/3)*b)/x^2) + (27*b^3*c^5 - 15*a*b^2*c^4*d - 7*a^2*b*c^3*d^2 - 5*a^3*c^2* 
d^3 + (27*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 - 7*a^2*b*c*d^4 - 5*a^3*d^5)*x...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**3+a)**(11/3)/(d*x**3+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {11}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^3+a)^(11/3)/(d*x^3+c)^3,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(11/3)/(d*x^3 + c)^3, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {11}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^3+a)^(11/3)/(d*x^3+c)^3,x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(11/3)/(d*x^3 + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{11/3}}{{\left (d\,x^3+c\right )}^3} \,d x \] Input:

int((a + b*x^3)^(11/3)/(c + d*x^3)^3,x)
 

Output:

int((a + b*x^3)^(11/3)/(c + d*x^3)^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\text {too large to display} \] Input:

int((b*x^3+a)^(11/3)/(d*x^3+c)^3,x)
 

Output:

( - 24*(a + b*x**3)**(2/3)*a**4*b*d**2*x - 54*(a + b*x**3)**(2/3)*a**3*b** 
2*d**2*x**4 + 28*(a + b*x**3)**(2/3)*a**2*b**3*c**2*x + 89*(a + b*x**3)**( 
2/3)*a**2*b**3*c*d*x**4 + 10*(a + b*x**3)**(2/3)*a**2*b**3*d**2*x**7 - 21* 
(a + b*x**3)**(2/3)*a*b**4*c**2*x**4 - 24*(a + b*x**3)**(2/3)*a*b**4*c*d*x 
**7 + 18*(a + b*x**3)**(2/3)*b**5*c**2*x**7 + 150*int((a + b*x**3)**(2/3)/ 
(5*a**3*c**3*d**2 + 15*a**3*c**2*d**3*x**3 + 15*a**3*c*d**4*x**6 + 5*a**3* 
d**5*x**9 - 12*a**2*b*c**4*d - 31*a**2*b*c**3*d**2*x**3 - 21*a**2*b*c**2*d 
**3*x**6 + 3*a**2*b*c*d**4*x**9 + 5*a**2*b*d**5*x**12 + 9*a*b**2*c**5 + 15 
*a*b**2*c**4*d*x**3 - 9*a*b**2*c**3*d**2*x**6 - 27*a*b**2*c**2*d**3*x**9 - 
 12*a*b**2*c*d**4*x**12 + 9*b**3*c**5*x**3 + 27*b**3*c**4*d*x**6 + 27*b**3 
*c**3*d**2*x**9 + 9*b**3*c**2*d**3*x**12),x)*a**8*c**2*d**5 + 300*int((a + 
 b*x**3)**(2/3)/(5*a**3*c**3*d**2 + 15*a**3*c**2*d**3*x**3 + 15*a**3*c*d** 
4*x**6 + 5*a**3*d**5*x**9 - 12*a**2*b*c**4*d - 31*a**2*b*c**3*d**2*x**3 - 
21*a**2*b*c**2*d**3*x**6 + 3*a**2*b*c*d**4*x**9 + 5*a**2*b*d**5*x**12 + 9* 
a*b**2*c**5 + 15*a*b**2*c**4*d*x**3 - 9*a*b**2*c**3*d**2*x**6 - 27*a*b**2* 
c**2*d**3*x**9 - 12*a*b**2*c*d**4*x**12 + 9*b**3*c**5*x**3 + 27*b**3*c**4* 
d*x**6 + 27*b**3*c**3*d**2*x**9 + 9*b**3*c**2*d**3*x**12),x)*a**8*c*d**6*x 
**3 + 150*int((a + b*x**3)**(2/3)/(5*a**3*c**3*d**2 + 15*a**3*c**2*d**3*x* 
*3 + 15*a**3*c*d**4*x**6 + 5*a**3*d**5*x**9 - 12*a**2*b*c**4*d - 31*a**2*b 
*c**3*d**2*x**3 - 21*a**2*b*c**2*d**3*x**6 + 3*a**2*b*c*d**4*x**9 + 5*a...