3.439 \(\int \frac{x^3 (c+d x+e x^2)}{(a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=542 \[ \frac{4 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \left (\sqrt [3]{b} c-2 \left (1-\sqrt{3}\right ) \sqrt [3]{a} d\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{4 \sqrt{2-\sqrt{3}} \sqrt [3]{a} d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{3^{3/4} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{8 d \sqrt{a+b x^3}}{3 b^{5/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac{4 e \sqrt{a+b x^3}}{3 b^2}-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}} \]

[Out]

(-2*x*(c + d*x + e*x^2))/(3*b*Sqrt[a + b*x^3]) + (4*e*Sqrt[a + b*x^3])/(3*b^2) + (8*d*Sqrt[a + b*x^3])/(3*b^(5
/3)*((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)) - (4*Sqrt[2 - Sqrt[3]]*a^(1/3)*d*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3)
- a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticE[ArcSin[((1 - Sqrt[3])*a^(1
/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(3^(3/4)*b^(5/3)*Sqrt[(a^(1/3)*(a^(1/3
) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3]) + (4*Sqrt[2 + Sqrt[3]]*(b^(1/3)*c - 2*
(1 - Sqrt[3])*a^(1/3)*d)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])
*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)
*x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*b^(5/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3
)*x)^2]*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.327242, antiderivative size = 542, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1828, 1886, 261, 1878, 218, 1877} \[ \frac{4 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \left (\sqrt [3]{b} c-2 \left (1-\sqrt{3}\right ) \sqrt [3]{a} d\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{4 \sqrt{2-\sqrt{3}} \sqrt [3]{a} d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{3^{3/4} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{8 d \sqrt{a+b x^3}}{3 b^{5/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac{4 e \sqrt{a+b x^3}}{3 b^2}-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(c + d*x + e*x^2))/(3*b*Sqrt[a + b*x^3]) + (4*e*Sqrt[a + b*x^3])/(3*b^2) + (8*d*Sqrt[a + b*x^3])/(3*b^(5
/3)*((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)) - (4*Sqrt[2 - Sqrt[3]]*a^(1/3)*d*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3)
- a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticE[ArcSin[((1 - Sqrt[3])*a^(1
/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(3^(3/4)*b^(5/3)*Sqrt[(a^(1/3)*(a^(1/3
) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3]) + (4*Sqrt[2 + Sqrt[3]]*(b^(1/3)*c - 2*
(1 - Sqrt[3])*a^(1/3)*d)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])
*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)
*x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*b^(5/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3
)*x)^2]*Sqrt[a + b*x^3])

Rule 1828

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = m + Expon[Pq, x]}, Module[{Q = Pol
ynomialQuotient[b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] +
1)*x^m*Pq, a + b*x^n, x]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[
a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q
- 1)/n] + 1)), x]] /; GeQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 0]

Rule 1886

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[Coeff[Pq, x, n - 1], Int[x^(n - 1)*(a + b*x^n)^p, x
], x] + Int[ExpandToSum[Pq - Coeff[Pq, x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && Pol
yQ[Pq, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1

Rule 261

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

Rule 1878

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a,
3]]}, Dist[(c*r - (1 - Sqrt[3])*d*s)/r, Int[1/Sqrt[a + b*x^3], x], x] + Dist[d/r, Int[((1 - Sqrt[3])*s + r*x)
/Sqrt[a + b*x^3], x], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && NeQ[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin{align*} \int \frac{x^3 \left (c+d x+e x^2\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}}-\frac{2 \int \frac{-a b c-2 a b d x-3 a b e x^2}{\sqrt{a+b x^3}} \, dx}{3 a b^2}\\ &=-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}}-\frac{2 \int \frac{-a b c-2 a b d x}{\sqrt{a+b x^3}} \, dx}{3 a b^2}+\frac{(2 e) \int \frac{x^2}{\sqrt{a+b x^3}} \, dx}{b}\\ &=-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}}+\frac{4 e \sqrt{a+b x^3}}{3 b^2}+\frac{(4 d) \int \frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt{a+b x^3}} \, dx}{3 b^{4/3}}+\frac{\left (2 \left (\sqrt [3]{b} c-2 \left (1-\sqrt{3}\right ) \sqrt [3]{a} d\right )\right ) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{3 b^{4/3}}\\ &=-\frac{2 x \left (c+d x+e x^2\right )}{3 b \sqrt{a+b x^3}}+\frac{4 e \sqrt{a+b x^3}}{3 b^2}+\frac{8 d \sqrt{a+b x^3}}{3 b^{5/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac{4 \sqrt{2-\sqrt{3}} \sqrt [3]{a} d \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt{3}\right )}{3^{3/4} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{4 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{b} c-2 \left (1-\sqrt{3}\right ) \sqrt [3]{a} d\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt{3}\right )}{3 \sqrt [4]{3} b^{5/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0950857, size = 118, normalized size = 0.22 \[ \frac{2 \left (b c x \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )-3 b d x^2 \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{2}{3},\frac{3}{2};\frac{5}{3};-\frac{b x^3}{a}\right )+2 a e-b c x+3 b d x^2+b e x^3\right )}{3 b^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*a*e - b*c*x + 3*b*d*x^2 + b*e*x^3 + b*c*x*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)
/a)] - 3*b*d*x^2*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[2/3, 3/2, 5/3, -((b*x^3)/a)]))/(3*b^2*Sqrt[a + b*x^3])

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Maple [A]  time = 0.007, size = 800, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*(2/3/b^2*a/((x^3+1/b*a)*b)^(1/2)+2/3/b^2*(b*x^3+a)^(1/2))+d*(-2/3/b*x^2/((x^3+1/b*a)*b)^(1/2)-8/9*I/b^2*3^(1
/2)*(-b^2*a)^(1/3)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)*
((x-1/b*(-b^2*a)^(1/3))/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-b^2*a)^(1
/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*((-3/2/b*(-b^2*a)^(1/3)+1/
2*I*3^(1/2)/b*(-b^2*a)^(1/3))*EllipticE(1/3*3^(1/2)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))
*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2),(I*3^(1/2)/b*(-b^2*a)^(1/3)/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(
1/3)))^(1/2))+1/b*(-b^2*a)^(1/3)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/
3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2),(I*3^(1/2)/b*(-b^2*a)^(1/3)/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a
)^(1/3)))^(1/2))))+c*(-2/3/b*x/((x^3+1/b*a)*b)^(1/2)-4/9*I/b^2*3^(1/2)*(-b^2*a)^(1/3)*(I*(x+1/2/b*(-b^2*a)^(1/
3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)*((x-1/b*(-b^2*a)^(1/3))/(-3/2/b*(-b^2*a)^(1
/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)
*b/(-b^2*a)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^
2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2),(I*3^(1/2)/b*(-b^2*a)^(1/3)/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b
*(-b^2*a)^(1/3)))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x^{5} + d x^{4} + c x^{3}\right )} \sqrt{b x^{3} + a}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x^5 + d*x^4 + c*x^3)*sqrt(b*x^3 + a)/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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Sympy [A]  time = 13.5041, size = 129, normalized size = 0.24 \begin{align*} e \left (\begin{cases} \frac{4 a}{3 b^{2} \sqrt{a + b x^{3}}} + \frac{2 x^{3}}{3 b \sqrt{a + b x^{3}}} & \text{for}\: b \neq 0 \\\frac{x^{6}}{6 a^{\frac{3}{2}}} & \text{otherwise} \end{cases}\right ) + \frac{c x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{4}{3}, \frac{3}{2} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} \Gamma \left (\frac{7}{3}\right )} + \frac{d x^{5} \Gamma \left (\frac{5}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{5}{3} \\ \frac{8}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} \Gamma \left (\frac{8}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*Piecewise((4*a/(3*b**2*sqrt(a + b*x**3)) + 2*x**3/(3*b*sqrt(a + b*x**3)), Ne(b, 0)), (x**6/(6*a**(3/2)), Tru
e)) + c*x**4*gamma(4/3)*hyper((4/3, 3/2), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(7/3)) + d*x**5*g
amma(5/3)*hyper((3/2, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(8/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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