\(\int \frac {x^4 (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\) [234]

Optimal result

Integrand size = 22, antiderivative size = 557 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 (A b-a B) x^2}{9 b^2 \left (a+b x^3\right )^{3/2}}+\frac {2 (4 A b-13 a B) x^2}{27 a b^2 \sqrt {a+b x^3}}-\frac {8 (A b-10 a B) \sqrt {a+b x^3}}{27 a b^{8/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac {4 \sqrt {2-\sqrt {3}} (A b-10 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} a^{2/3} b^{8/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {8 \sqrt {2} (A b-10 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} a^{2/3} b^{8/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

-2/9*(A*b-B*a)*x^2/b^2/(b*x^3+a)^(3/2)+2/27*(4*A*b-13*B*a)*x^2/a/b^2/(b*x^ 
3+a)^(1/2)-8/27*(A*b-10*B*a)*(b*x^3+a)^(1/2)/a/b^(8/3)/((1+3^(1/2))*a^(1/3 
)+b^(1/3)*x)+4/27*(1/2*6^(1/2)-1/2*2^(1/2))*(A*b-10*B*a)*(a^(1/3)+b^(1/3)* 
x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x 
)^2)^(1/2)*EllipticE(((1-3^(1/2))*a^(1/3)+b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+ 
b^(1/3)*x),I*3^(1/2)+2*I)*3^(1/4)/a^(2/3)/b^(8/3)/(a^(1/3)*(a^(1/3)+b^(1/3 
)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)/(b*x^3+a)^(1/2)-8/81*2^(1/2) 
*(A*b-10*B*a)*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2) 
/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*a^(1/3)+b 
^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/a^(2/3)/b 
^(8/3)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/ 
2)/(b*x^3+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.17 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 x^2 \left (-a A b+5 a B \left (2 a+b x^3\right )+(A b-10 a B) \left (a+b x^3\right ) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{2},\frac {5}{3},-\frac {b x^3}{a}\right )\right )}{5 a b^2 \left (a+b x^3\right )^{3/2}} \] Input:

Integrate[(x^4*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(2*x^2*(-(a*A*b) + 5*a*B*(2*a + b*x^3) + (A*b - 10*a*B)*(a + b*x^3)*Sqrt[1 
 + (b*x^3)/a]*Hypergeometric2F1[2/3, 5/2, 5/3, -((b*x^3)/a)]))/(5*a*b^2*(a 
 + b*x^3)^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 1.04 (sec) , antiderivative size = 564, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {957, 817, 832, 759, 2416}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {2 x^5 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {(A b-10 a B) \int \frac {x^4}{\left (b x^3+a\right )^{3/2}}dx}{9 a b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {2 x^5 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {(A b-10 a B) \left (\frac {4 \int \frac {x}{\sqrt {b x^3+a}}dx}{3 b}-\frac {2 x^2}{3 b \sqrt {a+b x^3}}\right )}{9 a b}\)

\(\Big \downarrow \) 832

\(\displaystyle \frac {2 x^5 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {(A b-10 a B) \left (\frac {4 \left (\frac {\int \frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt {b x^3+a}}dx}{\sqrt [3]{b}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} \int \frac {1}{\sqrt {b x^3+a}}dx}{\sqrt [3]{b}}\right )}{3 b}-\frac {2 x^2}{3 b \sqrt {a+b x^3}}\right )}{9 a b}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {2 x^5 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {(A b-10 a B) \left (\frac {4 \left (\frac {\int \frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt {b x^3+a}}dx}{\sqrt [3]{b}}-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{3 b}-\frac {2 x^2}{3 b \sqrt {a+b x^3}}\right )}{9 a b}\)

\(\Big \downarrow \) 2416

\(\displaystyle \frac {2 x^5 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}}-\frac {(A b-10 a B) \left (\frac {4 \left (\frac {\frac {2 \sqrt {a+b x^3}}{\sqrt [3]{b} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}}{\sqrt [3]{b}}-\frac {2 \left (1-\sqrt {3}\right ) \sqrt {2+\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\right )}{3 b}-\frac {2 x^2}{3 b \sqrt {a+b x^3}}\right )}{9 a b}\)

Input:

Int[(x^4*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(2*(A*b - a*B)*x^5)/(9*a*b*(a + b*x^3)^(3/2)) - ((A*b - 10*a*B)*((-2*x^2)/ 
(3*b*Sqrt[a + b*x^3]) + (4*(((2*Sqrt[a + b*x^3])/(b^(1/3)*((1 + Sqrt[3])*a 
^(1/3) + b^(1/3)*x)) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*a^(1/3)*(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]])/(b^(1/3)*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]))/b^(1/3) - (2*(1 - Sqrt[3])*Sqrt[2 + Sqrt[3]]*a^(1/3)*(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^(1/4)*b^(2/3) 
*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])))/(3*b)))/(9*a*b)
 

Defintions of rubi rules used

rule 759

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]*Sqrt[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 817

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 832

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

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[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] - S 
imp[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] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 528, normalized size of antiderivative = 0.95

method	result	size
elliptic	\(-\frac {2 x^{2} \left (A b -B a \right ) \sqrt {b \,x^{3}+a}}{9 b^{4} \left (x^{3}+\frac {a}{b}\right )^{2}}+\frac {2 x^{2} \left (4 A b -13 B a \right )}{27 b^{2} a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 i \left (\frac {B}{b^{2}}-\frac {4 A b -13 B a}{27 a \,b^{2}}\right ) \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \operatorname {EllipticE}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )+\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{b}\right )}{3 b \sqrt {b \,x^{3}+a}}\)	\(528\)
default	\(\text {Expression too large to display}\)	\(981\)

Input:

int(x^4*(B*x^3+A)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/9*x^2/b^4*(A*b-B*a)*(b*x^3+a)^(1/2)/(x^3+a/b)^2+2/27/b^2*x^2/a*(4*A*b-1 
3*B*a)/((x^3+a/b)*b)^(1/2)-2/3*I*(B/b^2-1/27*(4*A*b-13*B*a)/a/b^2)*3^(1/2) 
/b*(-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 [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.28 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 \, {\left (4 \, {\left ({\left (10 \, B a b^{2} - A b^{3}\right )} x^{6} + 10 \, B a^{3} - A a^{2} b + 2 \, {\left (10 \, B a^{2} b - A a b^{2}\right )} x^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right ) + {\left ({\left (13 \, B a b^{2} - 4 \, A b^{3}\right )} x^{5} + {\left (10 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{3} + a}\right )}}{27 \, {\left (a b^{5} x^{6} + 2 \, a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}} \] Input:

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")
 

Output:

-2/27*(4*((10*B*a*b^2 - A*b^3)*x^6 + 10*B*a^3 - A*a^2*b + 2*(10*B*a^2*b - 
A*a*b^2)*x^3)*sqrt(b)*weierstrassZeta(0, -4*a/b, weierstrassPInverse(0, -4 
*a/b, x)) + ((13*B*a*b^2 - 4*A*b^3)*x^5 + (10*B*a^2*b - A*a*b^2)*x^2)*sqrt 
(b*x^3 + a))/(a*b^5*x^6 + 2*a^2*b^4*x^3 + a^3*b^3)
 

Sympy [A] (verification not implemented)

Time = 41.92 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.14 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {A x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{3}, \frac {5}{2} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {8}{3}\right )} + \frac {B x^{8} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{2}} \Gamma \left (\frac {11}{3}\right )} \] Input:

integrate(x**4*(B*x**3+A)/(b*x**3+a)**(5/2),x)
 

Output:

A*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)) + B*x**8*gamma(8/3)*hyper((5/2, 8/3), (11/3,), b*x**3* 
exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(11/3))
 

Maxima [F]

\[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*x^4/(b*x^3 + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^4*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*x^4/(b*x^3 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int \frac {x^4\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{5/2}} \,d x \] Input:

int((x^4*(A + B*x^3))/(a + b*x^3)^(5/2),x)
 

Output:

int((x^4*(A + B*x^3))/(a + b*x^3)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {b \,x^{3}+a}\, x^{2}-4 \left (\int \frac {\sqrt {b \,x^{3}+a}\, x}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a^{2}-4 \left (\int \frac {\sqrt {b \,x^{3}+a}\, x}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a b \,x^{3}}{b \left (b \,x^{3}+a \right )} \] Input:

int(x^4*(B*x^3+A)/(b*x^3+a)^(5/2),x)
 

Output:

(2*(sqrt(a + b*x**3)*x**2 - 2*int((sqrt(a + b*x**3)*x)/(a**2 + 2*a*b*x**3 
+ b**2*x**6),x)*a**2 - 2*int((sqrt(a + b*x**3)*x)/(a**2 + 2*a*b*x**3 + b** 
2*x**6),x)*a*b*x**3))/(b*(a + b*x**3))