\(\int \frac {(a+b x^3)^{11/3}}{(c+d x^3)^3} \, dx\) [110]

Optimal result

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

[Out]

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)+1/54*(-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-
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))/d^4*3^(1/2)+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))/c^(
8/3)/d^4*3^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {424, 540, 542, 544, 245, 384} \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} d^4}+\frac {b x \left (a+b x^3\right )^{2/3} \left (-5 a^2 d^2-7 a b c d+18 b^2 c^2\right )}{18 c^2 d^3}+\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \log \left (c+d x^3\right )}{54 c^{8/3} d^4}-\frac {(b c-a d)^{5/3} \left (5 a^2 d^2+12 a b c d+27 b^2 c^2\right ) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} d^4}-\frac {b^{8/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right ) (9 b c-11 a d)}{3 \sqrt {3} d^4}+\frac {b^{8/3} (9 b c-11 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 d^4}-\frac {x \left (a+b x^3\right )^{5/3} (b c-a d) (5 a d+9 b c)}{18 c^2 d^2 \left (c+d x^3\right )}-\frac {x \left (a+b x^3\right )^{8/3} (b c-a d)}{6 c d \left (c+d x^3\right )^2} \]

[In]

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

[Out]

(b*(18*b^2*c^2 - 7*a*b*c*d - 5*a^2*d^2)*x*(a + b*x^3)^(2/3))/(18*c^2*d^3) - ((b*c - a*d)*x*(a + b*x^3)^(8/3))/
(6*c*d*(c + d*x^3)^2) - ((b*c - a*d)*(9*b*c + 5*a*d)*x*(a + b*x^3)^(5/3))/(18*c^2*d^2*(c + d*x^3)) - (b^(8/3)*
(9*b*c - 11*a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*d^4) + ((b*c - a*d)^(5/3)*(
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]]
)/(9*Sqrt[3]*c^(8/3)*d^4) + ((b*c - a*d)^(5/3)*(27*b^2*c^2 + 12*a*b*c*d + 5*a^2*d^2)*Log[c + d*x^3])/(54*c^(8/
3)*d^4) - ((b*c - a*d)^(5/3)*(27*b^2*c^2 + 12*a*b*c*d + 5*a^2*d^2)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*
x^3)^(1/3)])/(18*c^(8/3)*d^4) + (b^(8/3)*(9*b*c - 11*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(6*d^4)

Rule 245

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]]/(Sq
rt[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 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[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 424

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] - Dist[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 540

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] + Dist[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 542

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] + Dist[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 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\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 {\int \frac {\left (a+b x^3\right )^{5/3} \left (a (b c+5 a d)+3 b (3 b c-a d) x^3\right )}{\left (c+d x^3\right )^2} \, dx}{6 c d} \\ & = -\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 {\int \frac {\left (a+b x^3\right )^{2/3} \left (-a \left (9 b^2 c^2-a b c d+10 a^2 d^2\right )-3 b \left (18 b^2 c^2-7 a b c d-5 a^2 d^2\right ) x^3\right )}{c+d x^3} \, dx}{18 c^2 d^2} \\ & = \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 {\int \frac {6 a \left (9 b^3 c^3-8 a b^2 c^2 d-2 a^2 b c d^2-5 a^3 d^3\right )+18 b^3 c^2 (9 b c-11 a d) x^3}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{54 c^2 d^3} \\ & = \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 {\left (b^3 (9 b c-11 a d)\right ) \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx}{3 d^4}+\frac {\left ((b c-a d)^2 \left (27 b^2 c^2+12 a b c d+5 a^2 d^2\right )\right ) \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{9 c^2 d^4} \\ & = \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) \tan ^{-1}\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 ) \tan ^{-1}\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} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 11.74 (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=\frac {1}{108} \left (\frac {6 x \left (a+b x^3\right )^{2/3} \left (6 b^3-\frac {3 (b c-a d)^3}{c \left (c+d x^3\right )^2}+\frac {5 (b c-a d)^2 (3 b c+a d)}{c^2 \left (c+d x^3\right )}\right )}{d^3}-\frac {81 b^4 x^4 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{d^3 \sqrt [3]{a+b x^3}}+\frac {99 a b^3 x^4 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c d^2 \sqrt [3]{a+b x^3}}+\frac {10 a^4 \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{b+a x^3}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )+\log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )\right )}{c^{8/3} \sqrt [3]{b c-a d}}-\frac {18 a b^3 \sqrt [3]{c} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{b+a x^3}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )+\log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )\right )}{d^3 \sqrt [3]{b c-a d}}+\frac {16 a^2 b^2 \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{b+a x^3}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )+\log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )\right )}{c^{2/3} d^2 \sqrt [3]{b c-a d}}+\frac {4 a^3 b \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{b+a x^3}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )+\log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )\right )}{c^{5/3} d \sqrt [3]{b c-a d}}\right ) \]

[In]

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

[Out]

((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)) - (1
8*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^(1/3)*(b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)]))/(c^(5/3)*d*(b*c -
a*d)^(1/3)))/108

Maple [A] (verified)

Time = 4.97 (sec) , antiderivative size = 609, normalized size of antiderivative = 1.33

method	result	size
pseudoelliptic	\(-\frac {-3 b^{\frac {8}{3}} \left (11 a d -9 b c \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 )-2 \left (5 a^{2} d^{2}+12 a b c d +27 b^{2} c^{2}\right ) \left (d \,x^{3}+c \right )^{2} \left (a d -b 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 )+6 b^{\frac {8}{3}} \left (11 a d -9 b c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \sqrt {3}\, c^{3} \left (d \,x^{3}+c \right )^{2} \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+6 b^{\frac {8}{3}} \left (11 a d -9 b c \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 {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-3 x d c \left (6 b^{3} c^{2} d^{2} x^{6}+5 a^{3} d^{4} x^{3}+5 a^{2} b c \,d^{3} x^{3}-25 a \,b^{2} c^{2} d^{2} x^{3}+27 b^{3} c^{3} d \,x^{3}+8 a^{3} c \,d^{3}-4 a^{2} b \,c^{2} d^{2}-16 a \,b^{2} c^{3} d +18 c^{4} b^{3}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (5 a^{2} d^{2}+12 a b c d +27 b^{2} c^{2}\right ) \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\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 )\right ) \left (d \,x^{3}+c \right )^{2} \left (a d -b c \right )^{2}}{54 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )^{2} d^{4} c^{3}}\)	\(609\)

[In]

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

[Out]

-1/54*(-3*b^(8/3)*(11*a*d-9*b*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)-2*(5*a^2*d^2+12*a*b*c*d+27*b^2*c^2)*(d*x^3+c)^2*(a*d-b*c)^2*ln((((a*d-b*c)/c)^(1/3)*x+(
b*x^3+a)^(1/3))/x)+6*b^(8/3)*(11*a*d-9*b*c)*((a*d-b*c)/c)^(1/3)*3^(1/2)*c^3*(d*x^3+c)^2*arctan(1/3*3^(1/2)*(b^
(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+6*b^(8/3)*(11*a*d-9*b*c)*((a*d-b*c)/c)^(1/3)*c^3*(d*x^3+c)^2*ln((-b^(1/3
)*x+(b*x^3+a)^(1/3))/x)-3*x*d*c*(6*b^3*c^2*d^2*x^6+5*a^3*d^4*x^3+5*a^2*b*c*d^3*x^3-25*a*b^2*c^2*d^2*x^3+27*b^3
*c^3*d*x^3+8*a^3*c*d^3-4*a^2*b*c^2*d^2-16*a*b^2*c^3*d+18*b^3*c^4)*(b*x^3+a)^(2/3)*((a*d-b*c)/c)^(1/3)+(5*a^2*d
^2+12*a*b*c*d+27*b^2*c^2)*(-2*arctan(1/3*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3))/((a*d-b*c)/c)^(1/3)
/x)*3^(1/2)+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))*(d*x^3+c)
^2*(a*d-b*c)^2)/((a*d-b*c)/c)^(1/3)/(d*x^3+c)^2/d^4/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 = 18.18 (sec) , antiderivative size = 1246, normalized size of antiderivative = 2.72 \[ \int \frac {\left (a+b x^3\right )^{11/3}}{\left (c+d x^3\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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)*log(-((-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^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(-((b*c - a*d)*x^2*((b^2*c^2 - 2*a*b*c*d + a^2*d
^2)/c^2)^(1/3) + (b*x^3 + a)^(1/3)*c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(2/3) + (b*x^3 + a)^(2/3)*(b*c -
a*d))/x^2) + 3*(6*b^3*c^2*d^3*x^7 + (27*b^3*c^3*d^2 - 25*a*b^2*c^2*d^3 + 5*a^2*b*c*d^4 + 5*a^3*d^5)*x^4 + 2*(9
*b^3*c^4*d - 8*a*b^2*c^3*d^2 - 2*a^2*b*c^2*d^3 + 4*a^3*c*d^4)*x)*(b*x^3 + a)^(2/3))/(c^2*d^6*x^6 + 2*c^3*d^5*x
^3 + c^4*d^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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