3.5.92 \(\int \frac {1}{x^9 (a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=351 \[ \frac {\left (a+b x^3\right )^{2/3} (9 b c-4 a d) (a d+3 b c)}{20 a^3 c^2 x^5 (b c-a d)}-\frac {\left (a+b x^3\right )^{2/3} (9 b c-a d)}{8 a^2 c x^8 (b c-a d)}-\frac {\left (a+b x^3\right )^{2/3} \left (-20 a^3 d^3-12 a^2 b c d^2-9 a b^2 c^2 d+81 b^3 c^3\right )}{40 a^4 c^3 x^2 (b c-a d)}+\frac {d^4 \log \left (c+d x^3\right )}{6 c^{11/3} (b c-a d)^{4/3}}-\frac {d^4 \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{11/3} (b c-a d)^{4/3}}+\frac {d^4 \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{11/3} (b c-a d)^{4/3}}+\frac {b}{a x^8 \sqrt [3]{a+b x^3} (b c-a d)} \]

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Rubi [C]  time = 8.30, antiderivative size = 1486, normalized size of antiderivative = 4.23, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {511, 510}

result too large to display

Warning: Unable to verify antiderivative.

[In]

Int[1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(280*c^6*(a + b*x^3)^2 - 672*c^5*d*x^3*(a + b*x^3)^2 + 3024*c^4*d^2*x^6*(a + b*x^3)^2 + 18144*c^3*d^3*x^9*(a +
b*x^3)^2 + 13608*c^2*d^4*x^12*(a + b*x^3)^2 - 280*c^6*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*
d)*x^3)/(c*(a + b*x^3))] + 672*c^5*d*x^3*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a
+ b*x^3))] - 3024*c^4*d^2*x^6*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))]
- 18144*c^3*d^3*x^9*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 13608*c^
2*d^4*x^12*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 66*c^4*(b*c - a*d
)^2*x^6*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 312*c^3*d*(b*c - a*d)^2*x^9*Hyper
geometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 2268*c^2*d^2*(b*c - a*d)^2*x^12*Hypergeometric
2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 6696*c*d^3*(b*c - a*d)^2*x^15*Hypergeometric2F1[2, 7/3,
10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 4050*d^4*(b*c - a*d)^2*x^18*Hypergeometric2F1[2, 7/3, 10/3, ((b*c
- a*d)*x^3)/(c*(a + b*x^3))] + 189*c^4*(b*c - a*d)^2*x^6*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d
)*x^3)/(c*(a + b*x^3))] - 108*c^3*d*(b*c - a*d)^2*x^9*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x
^3)/(c*(a + b*x^3))] - 3618*c^2*d^2*(b*c - a*d)^2*x^12*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*
x^3)/(c*(a + b*x^3))] - 6156*c*d^3*(b*c - a*d)^2*x^15*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x
^3)/(c*(a + b*x^3))] - 2835*d^4*(b*c - a*d)^2*x^18*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)
/(c*(a + b*x^3))] + 54*c^4*(b*c - a*d)^2*x^6*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 10/3}, ((b*c - a*d)*x^3)
/(c*(a + b*x^3))] - 648*c^3*d*(b*c - a*d)^2*x^9*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 10/3}, ((b*c - a*d)*x
^3)/(c*(a + b*x^3))] - 2268*c^2*d^2*(b*c - a*d)^2*x^12*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 10/3}, ((b*c -
a*d)*x^3)/(c*(a + b*x^3))] - 2376*c*d^3*(b*c - a*d)^2*x^15*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 10/3}, ((
b*c - a*d)*x^3)/(c*(a + b*x^3))] - 810*d^4*(b*c - a*d)^2*x^18*HypergeometricPFQ[{2, 2, 2, 7/3}, {1, 1, 10/3},
((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 81*c^4*(b*c - a*d)^2*x^6*HypergeometricPFQ[{2, 2, 2, 2, 7/3}, {1, 1, 1, 1
0/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 324*c^3*d*(b*c - a*d)^2*x^9*HypergeometricPFQ[{2, 2, 2, 2, 7/3}, {1
, 1, 1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 486*c^2*d^2*(b*c - a*d)^2*x^12*HypergeometricPFQ[{2, 2, 2,
2, 7/3}, {1, 1, 1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 324*c*d^3*(b*c - a*d)^2*x^15*HypergeometricPFQ
[{2, 2, 2, 2, 7/3}, {1, 1, 1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 81*d^4*(b*c - a*d)^2*x^18*Hypergeome
tricPFQ[{2, 2, 2, 2, 7/3}, {1, 1, 1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(560*c^6*(b*c - a*d)*x^11*(a +
b*x^3)^(7/3))

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {1}{x^9 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \int \frac {1}{x^9 \left (1+\frac {b x^3}{a}\right )^{4/3} \left (c+d x^3\right )} \, dx}{a \sqrt [3]{a+b x^3}}\\ &=\frac {280 c^6 \left (a+b x^3\right )^2-672 c^5 d x^3 \left (a+b x^3\right )^2+3024 c^4 d^2 x^6 \left (a+b x^3\right )^2+18144 c^3 d^3 x^9 \left (a+b x^3\right )^2+13608 c^2 d^4 x^{12} \left (a+b x^3\right )^2-280 c^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+672 c^5 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-3024 c^4 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-18144 c^3 d^3 x^9 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-13608 c^2 d^4 x^{12} \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-66 c^4 (b c-a d)^2 x^6 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+312 c^3 d (b c-a d)^2 x^9 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-2268 c^2 d^2 (b c-a d)^2 x^{12} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-6696 c d^3 (b c-a d)^2 x^{15} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-4050 d^4 (b c-a d)^2 x^{18} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+189 c^4 (b c-a d)^2 x^6 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-108 c^3 d (b c-a d)^2 x^9 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-3618 c^2 d^2 (b c-a d)^2 x^{12} \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-6156 c d^3 (b c-a d)^2 x^{15} \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-2835 d^4 (b c-a d)^2 x^{18} \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+54 c^4 (b c-a d)^2 x^6 \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-648 c^3 d (b c-a d)^2 x^9 \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-2268 c^2 d^2 (b c-a d)^2 x^{12} \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-2376 c d^3 (b c-a d)^2 x^{15} \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-810 d^4 (b c-a d)^2 x^{18} \, _4F_3\left (2,2,2,\frac {7}{3};1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-81 c^4 (b c-a d)^2 x^6 \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-324 c^3 d (b c-a d)^2 x^9 \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-486 c^2 d^2 (b c-a d)^2 x^{12} \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-324 c d^3 (b c-a d)^2 x^{15} \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-81 d^4 (b c-a d)^2 x^{18} \, _5F_4\left (2,2,2,2,\frac {7}{3};1,1,1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{560 c^6 (b c-a d) x^{11} \left (a+b x^3\right )^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 5.86, size = 278, normalized size = 0.79 \begin {gather*} \frac {\left (a+b x^3\right )^{2/3} \left (-\frac {x^6 \left (20 a^2 d^2+32 a b c d+41 b^2 c^2\right )}{c^3}-\frac {5 a^2}{c}+\frac {40 b^4 x^9}{\left (a+b x^3\right ) (a d-b c)}+\frac {2 a x^3 (4 a d+7 b c)}{c^2}\right )}{40 a^4 x^8}+\frac {d^4 \left (\log \left (\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+\frac {x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+c^{2/3}\right )-2 \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt {3}}\right )\right )}{6 c^{11/3} (b c-a d)^{4/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(2/3)*((-5*a^2)/c + (2*a*(7*b*c + 4*a*d)*x^3)/c^2 - ((41*b^2*c^2 + 32*a*b*c*d + 20*a^2*d^2)*x^6)/
c^3 + (40*b^4*x^9)/((-(b*c) + a*d)*(a + b*x^3))))/(40*a^4*x^8) + (d^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)]))/(6*c^
(11/3)*(b*c - a*d)^(4/3))

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IntegrateAlgebraic [C]  time = 11.24, size = 528, normalized size = 1.50 \begin {gather*} \frac {-5 a^4 c^2 d+8 a^4 c d^2 x^3-20 a^4 d^3 x^6+5 a^3 b c^3+a^3 b c^2 d x^3-4 a^3 b c d^2 x^6-20 a^3 b d^3 x^9-9 a^2 b^2 c^3 x^3-3 a^2 b^2 c^2 d x^6-12 a^2 b^2 c d^2 x^9+27 a b^3 c^3 x^6-9 a b^3 c^2 d x^9+81 b^4 c^3 x^9}{40 a^4 c^3 x^8 \sqrt [3]{a+b x^3} (a d-b c)}+\frac {\left (d^4+i \sqrt {3} d^4\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 c^{11/3} (b c-a d)^{4/3}}-\frac {\sqrt {\frac {1}{6} \left (-1+i \sqrt {3}\right )} d^4 \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{11/3} (b c-a d)^{4/3}}-\frac {i \left (\sqrt {3} d^4-i d^4\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 c^{11/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(5*a^3*b*c^3 - 5*a^4*c^2*d - 9*a^2*b^2*c^3*x^3 + a^3*b*c^2*d*x^3 + 8*a^4*c*d^2*x^3 + 27*a*b^3*c^3*x^6 - 3*a^2*
b^2*c^2*d*x^6 - 4*a^3*b*c*d^2*x^6 - 20*a^4*d^3*x^6 + 81*b^4*c^3*x^9 - 9*a*b^3*c^2*d*x^9 - 12*a^2*b^2*c*d^2*x^9
- 20*a^3*b*d^3*x^9)/(40*a^4*c^3*(-(b*c) + a*d)*x^8*(a + b*x^3)^(1/3)) - (Sqrt[(-1 + I*Sqrt[3])/6]*d^4*ArcTan[
(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1/3) - Sqrt[3]*c^(1/3)*(a +
b*x^3)^(1/3))])/(c^(11/3)*(b*c - a*d)^(4/3)) + ((d^4 + I*Sqrt[3]*d^4)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[
3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*c^(11/3)*(b*c - a*d)^(4/3)) - ((I/12)*((-I)*d^4 + Sqrt[3]*d^4)*Log[(-2*I)*(
b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (I + Sqrt[3])*c^(2/3)*(
a + b*x^3)^(2/3)])/(c^(11/3)*(b*c - a*d)^(4/3))

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fricas [F(-1)]  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(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^9), x)

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right ) x^{9}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^9), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^9\,{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

int(1/(x^9*(a + b*x^3)^(4/3)*(c + d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{9} \left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(1/(x**9*(a + b*x**3)**(4/3)*(c + d*x**3)), x)

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