3.1.98 \(\int \frac {(a+b x^3)^{8/3}}{(c+d x^3)^2} \, dx\) [98]

Optimal. Leaf size=351 \[ \frac {b (2 b c-a d) x \left (a+b x^3\right )^{2/3}}{3 c d^2}-\frac {(b c-a d) x \left (a+b x^3\right )^{5/3}}{3 c d \left (c+d x^3\right )}-\frac {2 b^{5/3} (3 b c-4 a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} d^3}+\frac {2 (b c-a d)^{5/3} (3 b c+a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} d^3}+\frac {(b c-a d)^{5/3} (3 b c+a d) \log \left (c+d x^3\right )}{9 c^{5/3} d^3}-\frac {(b c-a d)^{5/3} (3 b c+a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} d^3}+\frac {b^{5/3} (3 b c-4 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 d^3} \]

[Out]

1/3*b*(-a*d+2*b*c)*x*(b*x^3+a)^(2/3)/c/d^2-1/3*(-a*d+b*c)*x*(b*x^3+a)^(5/3)/c/d/(d*x^3+c)+1/9*(-a*d+b*c)^(5/3)
*(a*d+3*b*c)*ln(d*x^3+c)/c^(5/3)/d^3-1/3*(-a*d+b*c)^(5/3)*(a*d+3*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^
(1/3))/c^(5/3)/d^3+1/3*b^(5/3)*(-4*a*d+3*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/d^3-2/9*b^(5/3)*(-4*a*d+3*b*c)*ar
ctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/d^3*3^(1/2)+2/9*(-a*d+b*c)^(5/3)*(a*d+3*b*c)*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^(5/3)/d^3*3^(1/2)

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Rubi [A]
time = 0.46, antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {424, 542, 544, 245, 384} \begin {gather*} -\frac {2 b^{5/3} \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right ) (3 b c-4 a d)}{3 \sqrt {3} d^3}+\frac {2 (b c-a d)^{5/3} (a d+3 b c) \text {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 )}{3 \sqrt {3} c^{5/3} d^3}+\frac {b^{5/3} (3 b c-4 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 d^3}+\frac {(b c-a d)^{5/3} (a d+3 b c) \log \left (c+d x^3\right )}{9 c^{5/3} d^3}-\frac {(b c-a d)^{5/3} (a d+3 b c) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} d^3}+\frac {b x \left (a+b x^3\right )^{2/3} (2 b c-a d)}{3 c d^2}-\frac {x \left (a+b x^3\right )^{5/3} (b c-a d)}{3 c d \left (c+d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(8/3)/(c + d*x^3)^2,x]

[Out]

(b*(2*b*c - a*d)*x*(a + b*x^3)^(2/3))/(3*c*d^2) - ((b*c - a*d)*x*(a + b*x^3)^(5/3))/(3*c*d*(c + d*x^3)) - (2*b
^(5/3)*(3*b*c - 4*a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*d^3) + (2*(b*c - a*d)
^(5/3)*(3*b*c + a*d)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c^(
5/3)*d^3) + ((b*c - a*d)^(5/3)*(3*b*c + a*d)*Log[c + d*x^3])/(9*c^(5/3)*d^3) - ((b*c - a*d)^(5/3)*(3*b*c + a*d
)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(3*c^(5/3)*d^3) + (b^(5/3)*(3*b*c - 4*a*d)*Log[-(b^(
1/3)*x) + (a + b*x^3)^(1/3)])/(3*d^3)

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 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*} \int \frac {\left (a+b x^3\right )^{8/3}}{\left (c+d x^3\right )^2} \, dx &=\frac {\left (a^2 \left (a+b x^3\right )^{2/3}\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{8/3}}{\left (c+d x^3\right )^2} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=\frac {a^2 x \left (a+b x^3\right )^{2/3} F_1\left (\frac {1}{3};-\frac {8}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^2 \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 10.70, size = 698, normalized size = 1.99 \begin {gather*} \frac {1}{18} \left (\frac {6 x \left (a+b x^3\right )^{2/3} \left (b^2+\frac {(b c-a d)^2}{c \left (c+d x^3\right )}\right )}{d^2}-\frac {9 b^3 x^4 \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{d^2 \sqrt [3]{a+b x^3}}+\frac {12 a b^2 x^4 \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c d \sqrt [3]{a+b x^3}}+\frac {2 a^3 \left (2 \sqrt {3} \tan ^{-1}\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} \sqrt [3]{b c-a d}}-\frac {2 a b^2 \sqrt [3]{c} \left (2 \sqrt {3} \tan ^{-1}\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^2 \sqrt [3]{b c-a d}}+\frac {2 a^2 b \left (2 \sqrt {3} \tan ^{-1}\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 \sqrt [3]{b c-a d}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(8/3)/(c + d*x^3)^2,x]

[Out]

((6*x*(a + b*x^3)^(2/3)*(b^2 + (b*c - a*d)^2/(c*(c + d*x^3))))/d^2 - (9*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)])/(d^2*(a + b*x^3)^(1/3)) + (12*a*b^2*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*(a + b*x^3)^(1/3)) + (2*a^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)]))/(c^(5/3)*(b*c - a*d)^(1/3)) - (2*a*b^2*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^2*(b*c - a*
d)^(1/3)) + (2*a^2*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^(2/3)*d*(b*c - a*d)^(1/3)))/18

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {8}{3}}}{\left (d \,x^{3}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(8/3)/(d*x^3+c)^2,x)

[Out]

int((b*x^3+a)^(8/3)/(d*x^3+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(8/3)/(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(8/3)/(d*x^3 + c)^2, x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 819 vs. \(2 (291) = 582\).
time = 11.01, size = 819, normalized size = 2.33 \begin {gather*} \frac {2 \, \sqrt {3} {\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2} + {\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} x^{3}\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 2 \, \sqrt {3} {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{3}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}}}{3 \, b x}\right ) - 2 \, {\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2} + {\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} x^{3}\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}{x}\right ) - 2 \, {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{3}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{3}\right )} \left (-b^{2}\right )^{\frac {1}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + {\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2} + {\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} x^{3}\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {{\left (b c - a d\right )} x^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )}}{x^{2}}\right ) + 3 \, {\left (b^{2} c d^{2} x^{4} + {\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{9 \, {\left (c d^{4} x^{3} + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(8/3)/(d*x^3+c)^2,x, algorithm="fricas")

[Out]

1/9*(2*sqrt(3)*(3*b^2*c^3 - 2*a*b*c^2*d - a^2*c*d^2 + (3*b^2*c^2*d - 2*a*b*c*d^2 - a^2*d^3)*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)) + 2*sqrt(3)*(3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*
c*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*(3*b^
2*c^3 - 2*a*b*c^2*d - a^2*c*d^2 + (3*b^2*c^2*d - 2*a*b*c*d^2 - a^2*d^3)*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) - 2*(3*b^2
*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*d^2)*x^3)*(-b^2)^(1/3)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)
/x) + (3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*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) + (3*b^2*c^3 - 2*a*b*c^2*d - a^2*c*d^2 + (3*b^2*c^2*d
- 2*a*b*c*d^2 - a^2*d^3)*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*(b^2*c*d^2*x^4 + (2*b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*(b*x^3 + a)^(2/3))/(c
*d^4*x^3 + c^2*d^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(8/3)/(d*x**3+c)**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(8/3)/(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(8/3)/(d*x^3 + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^3+a\right )}^{8/3}}{{\left (d\,x^3+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(8/3)/(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^(8/3)/(c + d*x^3)^2, x)

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