\(\int \frac {c+d x+e x^2+f x^3+g x^4}{(a+b x^3)^{5/2}} \, dx\) [68]

Optimal result

Integrand size = 32, antiderivative size = 628 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 x \left (b c-a f+(b d-a g) x+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {2 (5 b d+4 a g) \sqrt {a+b x^3}}{27 a^2 b^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {2 (3 a e-x (7 b c+2 a f+(5 b d+4 a g) x))}{27 a^2 b \sqrt {a+b x^3}}+\frac {\sqrt {2-\sqrt {3}} (5 b d+4 a g) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} a^{5/3} b^{5/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{b} (7 b c+2 a f)+\left (1-\sqrt {3}\right ) \sqrt [3]{a} (5 b d+4 a g)\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} a^2 b^{5/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

2/9*x*(b*c-a*f+(-a*g+b*d)*x+b*e*x^2)/a/b/(b*x^3+a)^(3/2)-2/27*(3*a*e-x*(7*b*c+2*a*f+(4*a*g+5*b*d)*x))/a^2/b/(b
*x^3+a)^(1/2)-2/27*(4*a*g+5*b*d)*(b*x^3+a)^(1/2)/a^2/b^(5/3)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))+1/27*(4*a*g+5*b*d
)*(a^(1/3)+b^(1/3)*x)*EllipticE((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)
*(1/2*6^(1/2)-1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3
^(1/4)/a^(5/3)/b^(5/3)/(b*x^3+a)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)+2
/81*(a^(1/3)+b^(1/3)*x)*EllipticF((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*
I)*(b^(1/3)*(2*a*f+7*b*c)+a^(1/3)*(4*a*g+5*b*d)*(1-3^(1/2)))*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1/
3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/a^2/b^(5/3)/(b*x^3+a)^(1/2)/(a^(1/3)*(a^(1/
3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 628, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1872, 1868, 1892, 224, 1891} \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (4 a g+5 b d) E\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} a^{5/3} b^{5/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {2 \sqrt {a+b x^3} (4 a g+5 b d)}{27 a^2 b^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {2 (3 a e-x (x (4 a g+5 b d)+2 a f+7 b c))}{27 a^2 b \sqrt {a+b x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right ) \left (\sqrt [3]{b} (2 a f+7 b c)+\left (1-\sqrt {3}\right ) \sqrt [3]{a} (4 a g+5 b d)\right )}{27 \sqrt [4]{3} a^2 b^{5/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {2 x \left (x (b d-a g)-a f+b c+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}} \]

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^3)^(5/2),x]

[Out]

(2*x*(b*c - a*f + (b*d - a*g)*x + b*e*x^2))/(9*a*b*(a + b*x^3)^(3/2)) - (2*(5*b*d + 4*a*g)*Sqrt[a + b*x^3])/(2
7*a^2*b^(5/3)*((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)) - (2*(3*a*e - x*(7*b*c + 2*a*f + (5*b*d + 4*a*g)*x)))/(27*a
^2*b*Sqrt[a + b*x^3]) + (Sqrt[2 - Sqrt[3]]*(5*b*d + 4*a*g)*(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]])/(9*3^(3/4)*a^(5/3)*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]) + (2*Sqrt[2 + Sqrt[3]]*(b^(1/3)*(7*b*c + 2*
a*f) + (1 - Sqrt[3])*a^(1/3)*(5*b*d + 4*a*g))*(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]])/(27*3^(1/4)*a^2*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 224

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

Rule 1868

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

Rule 1872

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*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]

Rule 1891

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]/(r^2
*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1
+ Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rule 1892

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x \left (b c-a f+(b d-a g) x+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {2 \int \frac {-\frac {1}{2} b (7 b c+2 a f)-\frac {1}{2} b (5 b d+4 a g) x-\frac {3}{2} b^2 e x^2}{\left (a+b x^3\right )^{3/2}} \, dx}{9 a b^2} \\ & = \frac {2 x \left (b c-a f+(b d-a g) x+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {2 (3 a e-x (7 b c+2 a f+(5 b d+4 a g) x))}{27 a^2 b \sqrt {a+b x^3}}+\frac {4 \int \frac {\frac {1}{4} b (7 b c+2 a f)-\frac {1}{4} b (5 b d+4 a g) x}{\sqrt {a+b x^3}} \, dx}{27 a^2 b^2} \\ & = \frac {2 x \left (b c-a f+(b d-a g) x+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {2 (3 a e-x (7 b c+2 a f+(5 b d+4 a g) x))}{27 a^2 b \sqrt {a+b x^3}}-\frac {(5 b d+4 a g) \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx}{27 a^2 b^{4/3}}+\frac {\left (\sqrt [3]{b} (7 b c+2 a f)+\left (1-\sqrt {3}\right ) \sqrt [3]{a} (5 b d+4 a g)\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{27 a^2 b^{4/3}} \\ & = \frac {2 x \left (b c-a f+(b d-a g) x+b e x^2\right )}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {2 (5 b d+4 a g) \sqrt {a+b x^3}}{27 a^2 b^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {2 (3 a e-x (7 b c+2 a f+(5 b d+4 a g) x))}{27 a^2 b \sqrt {a+b x^3}}+\frac {\sqrt {2-\sqrt {3}} (5 b d+4 a g) \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 )}{9\ 3^{3/4} a^{5/3} 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 {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{b} (7 b c+2 a f)+\left (1-\sqrt {3}\right ) \sqrt [3]{a} (5 b d+4 a g)\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 )}{27 \sqrt [4]{3} a^2 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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.27 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {140 b^2 c x^4+40 a b x \left (5 c+f x^3\right )-4 a^2 (15 e+x (5 f+27 g x))+10 (7 b c+2 a f) x \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )+27 (5 b d+4 a g) x^2 \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{2},\frac {5}{3},-\frac {b x^3}{a}\right )}{270 a^2 b \left (a+b x^3\right )^{3/2}} \]

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^3)^(5/2),x]

[Out]

(140*b^2*c*x^4 + 40*a*b*x*(5*c + f*x^3) - 4*a^2*(15*e + x*(5*f + 27*g*x)) + 10*(7*b*c + 2*a*f)*x*(a + b*x^3)*S
qrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)] + 27*(5*b*d + 4*a*g)*x^2*(a + b*x^3)*Sqrt[1
+ (b*x^3)/a]*Hypergeometric2F1[2/3, 5/2, 5/3, -((b*x^3)/a)])/(270*a^2*b*(a + b*x^3)^(3/2))

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 861, normalized size of antiderivative = 1.37

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(861\)
default	\(\text {Expression too large to display}\)	\(1673\)

[In]

int((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-2/9/a/b^3*(a*g-b*d)*x^2-2/9/a/b^3*(a*f-b*c)*x-2/9/b^3*e)*(b*x^3+a)^(1/2)/(x^3+a/b)^2-2*b*(-1/27/a^2/b^2*(4*a
*g+5*b*d)*x^2-1/27/a^2/b^2*(2*a*f+7*b*c)*x)/((x^3+a/b)*b)^(1/2)-2/81*I/a^2/b^2*(2*a*f+7*b*c)*3^(1/2)*(-a*b^2)^
(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b
^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(
1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b
^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*
(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))+2/81*I/a^2/b^2*(4*a*g+5*b*d)*3^(1/2)*(-a*b^2)^(1/3)*(I*
(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3)
)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-
a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)
^(1/3))*EllipticE(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1
/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))+1/b*(-a*
b^2)^(1/3)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)
^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.15 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.42 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (7 \, b^{3} c + 2 \, a b^{2} f\right )} x^{6} + 7 \, a^{2} b c + 2 \, a^{3} f + 2 \, {\left (7 \, a b^{2} c + 2 \, a^{2} b f\right )} x^{3}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) + {\left ({\left (5 \, b^{3} d + 4 \, a b^{2} g\right )} x^{6} + 5 \, a^{2} b d + 4 \, a^{3} g + 2 \, {\left (5 \, a b^{2} d + 4 \, a^{2} b g\right )} x^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right ) + {\left ({\left (5 \, b^{3} d + 4 \, a b^{2} g\right )} x^{5} + {\left (7 \, b^{3} c + 2 \, a b^{2} f\right )} x^{4} - 3 \, a^{2} b e + {\left (8 \, a b^{2} d + a^{2} b g\right )} x^{2} + {\left (10 \, a b^{2} c - a^{2} b f\right )} x\right )} \sqrt {b x^{3} + a}\right )}}{27 \, {\left (a^{2} b^{4} x^{6} + 2 \, a^{3} b^{3} x^{3} + a^{4} b^{2}\right )}} \]

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

2/27*(((7*b^3*c + 2*a*b^2*f)*x^6 + 7*a^2*b*c + 2*a^3*f + 2*(7*a*b^2*c + 2*a^2*b*f)*x^3)*sqrt(b)*weierstrassPIn
verse(0, -4*a/b, x) + ((5*b^3*d + 4*a*b^2*g)*x^6 + 5*a^2*b*d + 4*a^3*g + 2*(5*a*b^2*d + 4*a^2*b*g)*x^3)*sqrt(b
)*weierstrassZeta(0, -4*a/b, weierstrassPInverse(0, -4*a/b, x)) + ((5*b^3*d + 4*a*b^2*g)*x^5 + (7*b^3*c + 2*a*
b^2*f)*x^4 - 3*a^2*b*e + (8*a*b^2*d + a^2*b*g)*x^2 + (10*a*b^2*c - a^2*b*f)*x)*sqrt(b*x^3 + a))/(a^2*b^4*x^6 +
2*a^3*b^3*x^3 + a^4*b^2)

Sympy [A] (verification not implemented)

Time = 47.42 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.33 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=e \left (\begin {cases} - \frac {2}{9 a b \sqrt {a + b x^{3}} + 9 b^{2} x^{3} \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{2} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {5}{3}\right )} + \frac {f x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {5}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {g x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{3}, \frac {5}{2} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {8}{3}\right )} \]

[In]

integrate((g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**3+a)**(5/2),x)

[Out]

e*Piecewise((-2/(9*a*b*sqrt(a + b*x**3) + 9*b**2*x**3*sqrt(a + b*x**3)), Ne(b, 0)), (x**3/(3*a**(5/2)), True))
+ c*x*gamma(1/3)*hyper((1/3, 5/2), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(4/3)) + d*x**2*gamma(2
/3)*hyper((2/3, 5/2), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(5/3)) + f*x**4*gamma(4/3)*hyper((4/3
, 5/2), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(7/3)) + g*x**5*gamma(5/3)*hyper((5/3, 5/2), (8/3,)
, b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(8/3))

Maxima [F]

\[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {g x^{4} + f x^{3} + e x^{2} + d x + c}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((g*x^4 + f*x^3 + e*x^2 + d*x + c)/(b*x^3 + a)^(5/2), x)

Giac [F]

\[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {g x^{4} + f x^{3} + e x^{2} + d x + c}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

integrate((g*x^4 + f*x^3 + e*x^2 + d*x + c)/(b*x^3 + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^3\right )^{5/2}} \, dx=\int \frac {g\,x^4+f\,x^3+e\,x^2+d\,x+c}{{\left (b\,x^3+a\right )}^{5/2}} \,d x \]

[In]

int((c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^3)^(5/2),x)

[Out]

int((c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^3)^(5/2), x)