3.437 \(\int \frac{x^5 (c+d x+e x^2)}{(a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=594 \[ -\frac{16 \sqrt{2+\sqrt{3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \left (14 \sqrt [3]{b} d-25 \left (1-\sqrt{3}\right ) \sqrt [3]{a} e\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{105 \sqrt [4]{3} b^{8/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{40 \sqrt{2-\sqrt{3}} a^{4/3} e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{7\ 3^{3/4} b^{8/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{4 c \sqrt{a+b x^3}}{3 b^2}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{80 a e \sqrt{a+b x^3}}{21 b^{8/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )} \]

[Out]

(2*x*(a*d + a*e*x - b*c*x^2))/(3*b^2*Sqrt[a + b*x^3]) + (4*c*Sqrt[a + b*x^3])/(3*b^2) + (2*d*x*Sqrt[a + b*x^3]
)/(5*b^2) + (2*e*x^2*Sqrt[a + b*x^3])/(7*b^2) - (80*a*e*Sqrt[a + b*x^3])/(21*b^(8/3)*((1 + Sqrt[3])*a^(1/3) +
b^(1/3)*x)) + (40*Sqrt[2 - Sqrt[3]]*a^(4/3)*e*(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]])/(7*3^(3/4)*b^(8/3)*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]) - (16*Sqrt[2 + Sqrt[3]]*a*(14*b^(1/3)*d - 25*(1 - Sqrt[3])*a^(1/
3)*e)*(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
]])/(105*3^(1/4)*b^(8/3)*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 = 0.640867, antiderivative size = 594, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {1828, 1888, 1886, 261, 1878, 218, 1877} \[ -\frac{16 \sqrt{2+\sqrt{3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \left (14 \sqrt [3]{b} d-25 \left (1-\sqrt{3}\right ) \sqrt [3]{a} e\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{105 \sqrt [4]{3} b^{8/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{40 \sqrt{2-\sqrt{3}} a^{4/3} e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{7\ 3^{3/4} b^{8/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{4 c \sqrt{a+b x^3}}{3 b^2}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{80 a e \sqrt{a+b x^3}}{21 b^{8/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(c + d*x + e*x^2))/(a + b*x^3)^(3/2),x]

[Out]

(2*x*(a*d + a*e*x - b*c*x^2))/(3*b^2*Sqrt[a + b*x^3]) + (4*c*Sqrt[a + b*x^3])/(3*b^2) + (2*d*x*Sqrt[a + b*x^3]
)/(5*b^2) + (2*e*x^2*Sqrt[a + b*x^3])/(7*b^2) - (80*a*e*Sqrt[a + b*x^3])/(21*b^(8/3)*((1 + Sqrt[3])*a^(1/3) +
b^(1/3)*x)) + (40*Sqrt[2 - Sqrt[3]]*a^(4/3)*e*(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]])/(7*3^(3/4)*b^(8/3)*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]) - (16*Sqrt[2 + Sqrt[3]]*a*(14*b^(1/3)*d - 25*(1 - Sqrt[3])*a^(1/
3)*e)*(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
]])/(105*3^(1/4)*b^(8/3)*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 1888

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, D
ist[1/(b*(q + n*p + 1)), Int[ExpandToSum[b*(q + n*p + 1)*(Pq - Pqq*x^q) - a*Pqq*(q - n + 1)*x^(q - n), x]*(a +
b*x^n)^p, x], x] + Simp[(Pqq*x^(q - n + 1)*(a + b*x^n)^(p + 1))/(b*(q + n*p + 1)), x]] /; NeQ[q + n*p + 1, 0]
&& q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IG
tQ[n, 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^5 \left (c+d x+e x^2\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}-\frac{2 \int \frac{a^2 b d+2 a^2 b e x-3 a b^2 c x^2-\frac{3}{2} a b^2 d x^3-\frac{3}{2} a b^2 e x^4}{\sqrt{a+b x^3}} \, dx}{3 a b^3}\\ &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{4 \int \frac{\frac{7}{2} a^2 b^2 d+10 a^2 b^2 e x-\frac{21}{2} a b^3 c x^2-\frac{21}{4} a b^3 d x^3}{\sqrt{a+b x^3}} \, dx}{21 a b^4}\\ &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{8 \int \frac{14 a^2 b^3 d+25 a^2 b^3 e x-\frac{105}{4} a b^4 c x^2}{\sqrt{a+b x^3}} \, dx}{105 a b^5}\\ &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{8 \int \frac{14 a^2 b^3 d+25 a^2 b^3 e x}{\sqrt{a+b x^3}} \, dx}{105 a b^5}+\frac{(2 c) \int \frac{x^2}{\sqrt{a+b x^3}} \, dx}{b}\\ &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{4 c \sqrt{a+b x^3}}{3 b^2}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{(40 a e) \int \frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt{a+b x^3}} \, dx}{21 b^{7/3}}-\frac{\left (8 a \left (14 \sqrt [3]{b} d-25 \left (1-\sqrt{3}\right ) \sqrt [3]{a} e\right )\right ) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{105 b^{7/3}}\\ &=\frac{2 x \left (a d+a e x-b c x^2\right )}{3 b^2 \sqrt{a+b x^3}}+\frac{4 c \sqrt{a+b x^3}}{3 b^2}+\frac{2 d x \sqrt{a+b x^3}}{5 b^2}+\frac{2 e x^2 \sqrt{a+b x^3}}{7 b^2}-\frac{80 a e \sqrt{a+b x^3}}{21 b^{8/3} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac{40 \sqrt{2-\sqrt{3}} a^{4/3} e \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 )}{7\ 3^{3/4} b^{8/3} \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{16 \sqrt{2+\sqrt{3}} a \left (14 \sqrt [3]{b} d-25 \left (1-\sqrt{3}\right ) \sqrt [3]{a} e\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 )}{105 \sqrt [4]{3} b^{8/3} \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 = 0.111454, size = 134, normalized size = 0.23 \[ \frac{2 \left (-56 a d x \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )+150 a e x^2 \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{2}{3},\frac{3}{2};\frac{5}{3};-\frac{b x^3}{a}\right )+70 a c+56 a d x-150 a e x^2+35 b c x^3+21 b d x^4+15 b e x^5\right )}{105 b^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(c + d*x + e*x^2))/(a + b*x^3)^(3/2),x]

[Out]

(2*(70*a*c + 56*a*d*x - 150*a*e*x^2 + 35*b*c*x^3 + 21*b*d*x^4 + 15*b*e*x^5 - 56*a*d*x*Sqrt[1 + (b*x^3)/a]*Hype
rgeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)] + 150*a*e*x^2*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[2/3, 3/2, 5/3,
-((b*x^3)/a)]))/(105*b^2*Sqrt[a + b*x^3])

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Maple [A]  time = 0.02, size = 836, normalized size = 1.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(e*x^2+d*x+c)/(b*x^3+a)^(3/2),x)

[Out]

e*(2/3/b^2*x^2*a/((x^3+1/b*a)*b)^(1/2)+2/7/b^2*x^2*(b*x^3+a)^(1/2)+80/63*I/b^3*a*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))))+d*(2/3
/b^2*x*a/((x^3+1/b*a)*b)^(1/2)+2/5/b^2*x*(b*x^3+a)^(1/2)+32/45*I/b^3*a*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)))+c*(2/3/b^2*a/((x^3+1/b*a)*b)^(1/2)+2/3/b^2*(b*x^3+a)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2}{3} \, c{\left (\frac{\sqrt{b x^{3} + a}}{b^{2}} + \frac{a}{\sqrt{b x^{3} + a} b^{2}}\right )} + \int \frac{{\left (e x^{7} + d x^{6}\right )} \sqrt{b x^{3} + a}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(e*x^2+d*x+c)/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

2/3*c*(sqrt(b*x^3 + a)/b^2 + a/(sqrt(b*x^3 + a)*b^2)) + integrate((e*x^7 + d*x^6)*sqrt(b*x^3 + a)/(b^2*x^6 + 2
*a*b*x^3 + a^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x^{7} + d x^{6} + c x^{5}\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^5*(e*x^2+d*x+c)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

integral((e*x^7 + d*x^6 + c*x^5)*sqrt(b*x^3 + a)/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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Sympy [A]  time = 27.0308, size = 129, normalized size = 0.22 \begin{align*} c \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{d x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} \Gamma \left (\frac{10}{3}\right )} + \frac{e x^{8} \Gamma \left (\frac{8}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{8}{3} \\ \frac{11}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{3}{2}} \Gamma \left (\frac{11}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(e*x**2+d*x+c)/(b*x**3+a)**(3/2),x)

[Out]

c*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)) + d*x**7*gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(10/3)) + e*x**8
*gamma(8/3)*hyper((3/2, 8/3), (11/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(11/3))

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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^{5}}{{\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^5*(e*x^2+d*x+c)/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d*x + c)*x^5/(b*x^3 + a)^(3/2), x)