\(\int \frac {(a+b x^3)^{3/2} (A+B x^3)}{(e x)^{3/2}} \, dx\) [254]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [F]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 614 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{3/2}} \, dx=\frac {9 (14 A b+a B) (e x)^{5/2} \sqrt {a+b x^3}}{56 e^4}+\frac {27 \left (1+\sqrt {3}\right ) a (14 A b+a B) \sqrt {e x} \sqrt {a+b x^3}}{112 b^{2/3} e^2 \left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )}+\frac {(14 A b+a B) (e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}-\frac {27 \sqrt [4]{3} a^{4/3} (14 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{112 b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {9\ 3^{3/4} \left (1-\sqrt {3}\right ) a^{4/3} (14 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{224 b^{2/3} e^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

9/56*(14*A*b+B*a)*(e*x)^(5/2)*(b*x^3+a)^(1/2)/e^4+27/112*(1+3^(1/2))*a*(14 
*A*b+B*a)*(e*x)^(1/2)*(b*x^3+a)^(1/2)/b^(2/3)/e^2/(a^(1/3)+(1+3^(1/2))*b^( 
1/3)*x)+1/7*(14*A*b+B*a)*(e*x)^(5/2)*(b*x^3+a)^(3/2)/a/e^4-2*A*(b*x^3+a)^( 
5/2)/a/e/(e*x)^(1/2)-27/112*3^(1/4)*a^(4/3)*(14*A*b+B*a)*(e*x)^(1/2)*(a^(1 
/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+(1+3^(1/2 
))*b^(1/3)*x)^2)^(1/2)*EllipticE((1-(a^(1/3)+(1-3^(1/2))*b^(1/3)*x)^2/(a^( 
1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))/b^(2/3)/e^2/ 
(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)^2)^(1/2)/(b 
*x^3+a)^(1/2)-9/224*3^(3/4)*(1-3^(1/2))*a^(4/3)*(14*A*b+B*a)*(e*x)^(1/2)*( 
a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+(1+3^ 
(1/2))*b^(1/3)*x)^2)^(1/2)*InverseJacobiAM(arccos((a^(1/3)+(1-3^(1/2))*b^( 
1/3)*x)/(a^(1/3)+(1+3^(1/2))*b^(1/3)*x)),1/4*6^(1/2)+1/4*2^(1/2))/b^(2/3)/ 
e^2/(b^(1/3)*x*(a^(1/3)+b^(1/3)*x)/(a^(1/3)+(1+3^(1/2))*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 = 10.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.14 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{3/2}} \, dx=\frac {2 x \sqrt {a+b x^3} \left (-\frac {5 A \left (a+b x^3\right )^2}{a}+\frac {(14 A b+a B) x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{6},\frac {11}{6},-\frac {b x^3}{a}\right )}{\sqrt {1+\frac {b x^3}{a}}}\right )}{5 (e x)^{3/2}} \] Input:

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

Output:

(2*x*Sqrt[a + b*x^3]*((-5*A*(a + b*x^3)^2)/a + ((14*A*b + a*B)*x^3*Hyperge 
ometric2F1[-3/2, 5/6, 11/6, -((b*x^3)/a)])/Sqrt[1 + (b*x^3)/a]))/(5*(e*x)^ 
(3/2))
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 653, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {955, 811, 811, 851, 837, 25, 766, 2420}

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.

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

\(\Big \downarrow \) 955

\(\displaystyle \frac {(a B+14 A b) \int (e x)^{3/2} \left (b x^3+a\right )^{3/2}dx}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \int (e x)^{3/2} \sqrt {b x^3+a}dx+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3}{8} a \int \frac {(e x)^{3/2}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3 a \int \frac {e^2 x^2}{\sqrt {b x^3+a}}d\sqrt {e x}}{4 e}+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 837

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3 a \left (-\frac {\left (1-\sqrt {3}\right ) a^{2/3} e^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b^{2/3}}-\frac {\int -\frac {2 b^{2/3} x^2 e^2+\left (1-\sqrt {3}\right ) a^{2/3} e^2}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b^{2/3}}\right )}{4 e}+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3 a \left (\frac {\int \frac {2 b^{2/3} x^2 e^2+\left (1-\sqrt {3}\right ) a^{2/3} e^2}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) a^{2/3} e^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b^{2/3}}\right )}{4 e}+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3 a \left (\frac {\int \frac {2 b^{2/3} x^2 e^2+\left (1-\sqrt {3}\right ) a^{2/3} e^2}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}\right )}{4 e}+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

\(\Big \downarrow \) 2420

\(\displaystyle \frac {(a B+14 A b) \left (\frac {9}{14} a \left (\frac {3 a \left (\frac {\frac {\left (1+\sqrt {3}\right ) e^3 \sqrt {e x} \sqrt {a+b x^3}}{\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x}-\frac {\sqrt [4]{3} \sqrt [3]{a} e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} E\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}}{2 b^{2/3}}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a} e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b^{2/3} \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}\right )}{4 e}+\frac {(e x)^{5/2} \sqrt {a+b x^3}}{4 e}\right )+\frac {(e x)^{5/2} \left (a+b x^3\right )^{3/2}}{7 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{a e \sqrt {e x}}\)

Input:

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

Output:

(-2*A*(a + b*x^3)^(5/2))/(a*e*Sqrt[e*x]) + ((14*A*b + a*B)*(((e*x)^(5/2)*( 
a + b*x^3)^(3/2))/(7*e) + (9*a*(((e*x)^(5/2)*Sqrt[a + b*x^3])/(4*e) + (3*a 
*((((1 + Sqrt[3])*e^3*Sqrt[e*x]*Sqrt[a + b*x^3])/(a^(1/3)*e + (1 + Sqrt[3] 
)*b^(1/3)*e*x) - (3^(1/4)*a^(1/3)*e*Sqrt[e*x]*(a^(1/3)*e + b^(1/3)*e*x)*Sq 
rt[(a^(2/3)*e^2 - a^(1/3)*b^(1/3)*e^2*x + b^(2/3)*e^2*x^2)/(a^(1/3)*e + (1 
 + Sqrt[3])*b^(1/3)*e*x)^2]*EllipticE[ArcCos[(a^(1/3)*e + (1 - Sqrt[3])*b^ 
(1/3)*e*x)/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)], (2 + Sqrt[3])/4])/(Sq 
rt[(b^(1/3)*e*x*(a^(1/3)*e + b^(1/3)*e*x))/(a^(1/3)*e + (1 + Sqrt[3])*b^(1 
/3)*e*x)^2]*Sqrt[a + b*x^3]))/(2*b^(2/3)) - ((1 - Sqrt[3])*a^(1/3)*e*Sqrt[ 
e*x]*(a^(1/3)*e + b^(1/3)*e*x)*Sqrt[(a^(2/3)*e^2 - a^(1/3)*b^(1/3)*e^2*x + 
 b^(2/3)*e^2*x^2)/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)^2]*EllipticF[Arc 
Cos[(a^(1/3)*e + (1 - Sqrt[3])*b^(1/3)*e*x)/(a^(1/3)*e + (1 + Sqrt[3])*b^( 
1/3)*e*x)], (2 + Sqrt[3])/4])/(4*3^(1/4)*b^(2/3)*Sqrt[(b^(1/3)*e*x*(a^(1/3 
)*e + b^(1/3)*e*x))/(a^(1/3)*e + (1 + Sqrt[3])*b^(1/3)*e*x)^2]*Sqrt[a + b* 
x^3])))/(4*e)))/14))/(a*e^3)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 766
Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 811
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 837
Int[(x_)^4/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 
3]], s = Denom[Rt[b/a, 3]]}, Simp[(Sqrt[3] - 1)*(s^2/(2*r^2))   Int[1/Sqrt[ 
a + b*x^6], x], x] - Simp[1/(2*r^2)   Int[((Sqrt[3] - 1)*s^2 - 2*r^2*x^4)/S 
qrt[a + b*x^6], x], x]] /; FreeQ[{a, b}, x]
 

rule 851
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 955
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

rule 2420
Int[((c_) + (d_.)*(x_)^4)/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = 
 Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(1 + Sqrt[3])*d*s^3*x*(Sqr 
t[a + b*x^6]/(2*a*r^2*(s + (1 + Sqrt[3])*r*x^2))), x] - Simp[3^(1/4)*d*s*x* 
(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2 
*r^2*Sqrt[(r*x^2*(s + r*x^2))/(s + (1 + Sqrt[3])*r*x^2)^2]*Sqrt[a + b*x^6]) 
)*EllipticE[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 
 + Sqrt[3])/4], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[2*Rt[b/a, 3]^2*c - (1 
- Sqrt[3])*d, 0]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.98 (sec) , antiderivative size = 1140, normalized size of antiderivative = 1.86

method result size
risch \(\text {Expression too large to display}\) \(1140\)
elliptic \(\text {Expression too large to display}\) \(1233\)
default \(\text {Expression too large to display}\) \(6142\)

Input:

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

Output:

-1/56*(b*x^3+a)^(1/2)*(-8*B*b*x^6-14*A*b*x^3-17*B*a*x^3+112*A*a)/e/(e*x)^( 
1/2)+27/112*a*(14*A*b+B*a)*(x*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a* 
b^2)^(1/3))*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))+(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))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a 
*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2)*(x-1/b*(-a*b^2)^(1/3))^2*(1/b*( 
-a*b^2)^(1/3)*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-1/ 
2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3))) 
^(1/2)*(1/b*(-a*b^2)^(1/3)*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2 
)^(1/3))/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(x-1/b*(-a 
*b^2)^(1/3)))^(1/2)*(((-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3 
))/b*(-a*b^2)^(1/3)+1/b^2*(-a*b^2)^(2/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^( 
1/2)/b*(-a*b^2)^(1/3))*b/(-a*b^2)^(1/3)*EllipticF(((-3/2/b*(-a*b^2)^(1/3)+ 
1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*( 
-a*b^2)^(1/3))/(x-1/b*(-a*b^2)^(1/3)))^(1/2),((3/2/b*(-a*b^2)^(1/3)+1/2*I* 
3^(1/2)/b*(-a*b^2)^(1/3))*(1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^( 
1/3))/(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))+(1/2/b*(-a*b^2)^(1/3)+1/2* 
I*3^(1/2)/b*(-a*b^2)^(1/3))*EllipticE(((-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2 
)/b*(-a*b^2)^(1/3))*x/(-1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(...
 

Fricas [F]

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

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

Output:

integral((B*b*x^6 + (B*a + A*b)*x^3 + A*a)*sqrt(b*x^3 + a)*sqrt(e*x)/(e^2* 
x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.72 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{3/2}} \, dx=\frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{6} \\ \frac {5}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {5}{6}\right )} + \frac {A \sqrt {a} b x^{\frac {5}{2}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {3}{2}} \Gamma \left (\frac {11}{6}\right )} + \frac {B a^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {3}{2}} \Gamma \left (\frac {11}{6}\right )} + \frac {B \sqrt {a} b x^{\frac {11}{2}} \Gamma \left (\frac {11}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {11}{6} \\ \frac {17}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 e^{\frac {3}{2}} \Gamma \left (\frac {17}{6}\right )} \] Input:

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

Output:

A*a**(3/2)*gamma(-1/6)*hyper((-1/2, -1/6), (5/6,), b*x**3*exp_polar(I*pi)/ 
a)/(3*e**(3/2)*sqrt(x)*gamma(5/6)) + A*sqrt(a)*b*x**(5/2)*gamma(5/6)*hyper 
((-1/2, 5/6), (11/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*gamma(11/6)) 
+ B*a**(3/2)*x**(5/2)*gamma(5/6)*hyper((-1/2, 5/6), (11/6,), b*x**3*exp_po 
lar(I*pi)/a)/(3*e**(3/2)*gamma(11/6)) + B*sqrt(a)*b*x**(11/2)*gamma(11/6)* 
hyper((-1/2, 11/6), (17/6,), b*x**3*exp_polar(I*pi)/a)/(3*e**(3/2)*gamma(1 
7/6))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (362 \sqrt {b \,x^{3}+a}\, a^{2}+124 \sqrt {b \,x^{3}+a}\, a b \,x^{3}+32 \sqrt {b \,x^{3}+a}\, b^{2} x^{6}+405 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}}{b \,x^{5}+a \,x^{2}}d x \right ) a^{3}\right )}{224 \sqrt {x}\, e^{2}} \] Input:

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

Output:

(sqrt(e)*(362*sqrt(a + b*x**3)*a**2 + 124*sqrt(a + b*x**3)*a*b*x**3 + 32*s 
qrt(a + b*x**3)*b**2*x**6 + 405*sqrt(x)*int((sqrt(x)*sqrt(a + b*x**3))/(a* 
x**2 + b*x**5),x)*a**3))/(224*sqrt(x)*e**2)