\(\int \sqrt {e x} (a+b x^3)^{5/2} (A+B x^3) \, dx\) [260]

Optimal result

Integrand size = 26, antiderivative size = 201 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {5 a^3 (8 A b-a B) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}} \] Output:

5/192*a^2*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b/e+5/288*a*(8*A*b-B*a)* 
(e*x)^(3/2)*(b*x^3+a)^(3/2)/b/e+1/72*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(5/ 
2)/b/e+1/12*B*(e*x)^(3/2)*(b*x^3+a)^(7/2)/b/e+5/192*a^3*(8*A*b-B*a)*e^(1/2 
)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.70 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {\sqrt {e x} \left (\sqrt {b} x^{3/2} \sqrt {a+b x^3} \left (15 a^3 B+16 b^3 x^6 \left (4 A+3 B x^3\right )+8 a b^2 x^3 \left (26 A+17 B x^3\right )+2 a^2 b \left (132 A+59 B x^3\right )\right )-15 a^3 (-8 A b+a B) \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{576 b^{3/2} \sqrt {x}} \] Input:

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

Output:

(Sqrt[e*x]*(Sqrt[b]*x^(3/2)*Sqrt[a + b*x^3]*(15*a^3*B + 16*b^3*x^6*(4*A + 
3*B*x^3) + 8*a*b^2*x^3*(26*A + 17*B*x^3) + 2*a^2*b*(132*A + 59*B*x^3)) - 1 
5*a^3*(-8*A*b + a*B)*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(576*b^(3/2) 
*Sqrt[x])
 

Rubi [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {959, 811, 811, 811, 851, 807, 224, 219}

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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(8 A b-a B) \int \sqrt {e x} \left (b x^3+a\right )^{5/2}dx}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \int \sqrt {e x} \left (b x^3+a\right )^{3/2}dx+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {e x} \sqrt {b x^3+a}dx+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 \sqrt {b}}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\)

Input:

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

Output:

(B*(e*x)^(3/2)*(a + b*x^3)^(7/2))/(12*b*e) + ((8*A*b - a*B)*(((e*x)^(3/2)* 
(a + b*x^3)^(5/2))/(9*e) + (5*a*(((e*x)^(3/2)*(a + b*x^3)^(3/2))/(6*e) + ( 
3*a*(((e*x)^(3/2)*Sqrt[a + b*x^3])/(3*e) + (a*Sqrt[e]*ArcTanh[(Sqrt[b]*(e* 
x)^(3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*Sqrt[b])))/4))/6))/(8*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 807

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

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 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 959

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

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {x^{2} \left (48 b^{3} B \,x^{9}+64 A \,b^{3} x^{6}+136 B a \,b^{2} x^{6}+208 a A \,b^{2} x^{3}+118 B \,a^{2} b \,x^{3}+264 a^{2} b A +15 a^{3} B \right ) \sqrt {b \,x^{3}+a}\, e}{576 b \sqrt {e x}}+\frac {5 a^{3} \left (8 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e \sqrt {\left (b \,x^{3}+a \right ) e x}}{192 b \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(158\)
default	\(\frac {\sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (48 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{3} x^{10}+64 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{3} x^{7}+136 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \,b^{2} x^{7}+208 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \,b^{2} x^{4}+118 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} b \,x^{4}+120 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{3} b e +264 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} b x -15 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{4} e +15 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{3} x \right )}{576 \sqrt {\left (b \,x^{3}+a \right ) e x}\, b \sqrt {b e}}\)	\(279\)
elliptic	\(\text {Expression too large to display}\)	\(1304\)

Input:

int((e*x)^(1/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
 

Output:

1/576/b*x^2*(48*B*b^3*x^9+64*A*b^3*x^6+136*B*a*b^2*x^6+208*A*a*b^2*x^3+118 
*B*a^2*b*x^3+264*A*a^2*b+15*B*a^3)*(b*x^3+a)^(1/2)*e/(e*x)^(1/2)+5/192*a^3 
/b*(8*A*b-B*a)/(b*e)^(1/2)*arctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))* 
e*((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^3+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.61 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{2304 \, b}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) + 2 \, {\left (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{1152 \, b}\right ] \] Input:

integrate((e*x)^(1/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="fricas")
 

Output:

[-1/2304*(15*(B*a^4 - 8*A*a^3*b)*sqrt(e/b)*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 
- a^2*e - 4*(2*b^2*x^4 + a*b*x)*sqrt(b*x^3 + a)*sqrt(e*x)*sqrt(e/b)) - 4*( 
48*B*b^3*x^10 + 8*(17*B*a*b^2 + 8*A*b^3)*x^7 + 2*(59*B*a^2*b + 104*A*a*b^2 
)*x^4 + 3*(5*B*a^3 + 88*A*a^2*b)*x)*sqrt(b*x^3 + a)*sqrt(e*x))/b, 1/1152*( 
15*(B*a^4 - 8*A*a^3*b)*sqrt(-e/b)*arctan(2*sqrt(b*x^3 + a)*sqrt(e*x)*b*x*s 
qrt(-e/b)/(2*b*e*x^3 + a*e)) + 2*(48*B*b^3*x^10 + 8*(17*B*a*b^2 + 8*A*b^3) 
*x^7 + 2*(59*B*a^2*b + 104*A*a*b^2)*x^4 + 3*(5*B*a^3 + 88*A*a^2*b)*x)*sqrt 
(b*x^3 + a)*sqrt(e*x))/b]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 896 vs. \(2 (177) = 354\).

Time = 8.37 (sec) , antiderivative size = 896, normalized size of antiderivative = 4.46 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\text {Too large to display} \] Input:

integrate((e*x)**(1/2)*(b*x**3+a)**(5/2)*(B*x**3+A),x)
 

Output:

Piecewise((2*Piecewise((nan, Eq(e**3, 0)), ((A*a**2*e**3*Piecewise((a*Piec 
ewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt(b 
/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True))/2 
+ (e*x)**(3/2)*sqrt(a + b*x**3)/2, Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(3/2), 
True)) + 2*A*a*b*Piecewise((-a**2*e**3*Piecewise((log(2*b*(e*x)**(3/2)/e** 
3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2 
)*log((e*x)**(3/2))/sqrt(b*x**3), True))/(8*b) + sqrt(a + b*x**3)*(a*e**3* 
(e*x)**(3/2)/(8*b) + (e*x)**(9/2)/4), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(9/2 
)/3, True)) + A*b**2*Piecewise((a**3*e**6*Piecewise((log(2*b*(e*x)**(3/2)/ 
e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**( 
3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True))/(16*b**2) + sqrt(a + b*x**3)*( 
-a**2*e**6*(e*x)**(3/2)/(16*b**2) + a*e**3*(e*x)**(9/2)/(24*b) + (e*x)**(1 
5/2)/6), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(15/2)/5, True))/e**3 + B*a**2*Pi 
ecewise((-a**2*e**3*Piecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)* 
sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2)) 
/sqrt(b*x**3), True))/(8*b) + sqrt(a + b*x**3)*(a*e**3*(e*x)**(3/2)/(8*b) 
+ (e*x)**(9/2)/4), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(9/2)/3, True)) + 2*B*a 
*b*Piecewise((a**3*e**6*Piecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e* 
*3)*sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3 
/2))/sqrt(b*x**3), True))/(16*b**2) + sqrt(a + b*x**3)*(-a**2*e**6*(e*x...
 

Maxima [F]

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

integrate((e*x)^(1/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (159) = 318\).

Time = 0.30 (sec) , antiderivative size = 590, normalized size of antiderivative = 2.94 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} B a b x {\left | e \right |}^{2}}{36 \, e^{3}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} A b^{2} x {\left | e \right |}^{2}}{72 \, e^{3}} - \frac {{\left (25 \, B^{2} a^{8} + 240 \, A B a^{7} b + 576 \, A^{2} a^{6} b^{2}\right )} e^{4} \log \left ({\left | {\left (5 \, \sqrt {e x} B a^{4} x + 24 \, \sqrt {e x} A a^{3} b x\right )} \sqrt {b e} + \sqrt {25 \, B^{2} a^{9} e^{2} + 240 \, A B a^{8} b e^{2} + 576 \, A^{2} a^{7} b^{2} e^{2} + {\left (5 \, \sqrt {e x} B a^{4} x + 24 \, \sqrt {e x} A a^{3} b x\right )}^{2} b e} \right |}\right )}{192 \, \sqrt {b e} b {\left | 5 \, B a^{4} e + 24 \, A a^{3} b e \right |} {\left | e \right |}^{2}} - \frac {{\left (\frac {a e^{4} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{\sqrt {b e}} - \sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} e x\right )} A a^{2} {\left | e \right |}^{2}}{3 \, e^{5}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} + \frac {a e^{3}}{b}\right )} \sqrt {e x} B a^{2} x {\left | e \right |}^{2}}{12 \, e^{7}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} + \frac {a e^{3}}{b}\right )} \sqrt {e x} A a b x {\left | e \right |}^{2}}{6 \, e^{7}} + \frac {{\left (\frac {15 \, a^{3} e^{9}}{b^{3}} + 2 \, {\left (4 \, {\left (6 \, e^{3} x^{3} + \frac {a e^{3}}{b}\right )} e^{3} x^{3} - \frac {5 \, a^{2} e^{6}}{b^{2}}\right )} e^{3} x^{3}\right )} \sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} B b^{2} x {\left | e \right |}^{2}}{576 \, e^{13}} \] Input:

integrate((e*x)^(1/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x, algorithm="giac")
 

Output:

1/36*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3*(4*x^3/e^4 + a/(b*e^4)) - 3*a^2/(b 
^2*e))*sqrt(e*x)*B*a*b*x*abs(e)^2/e^3 + 1/72*sqrt(b*e^4*x^3 + a*e^4)*(2*e^ 
3*x^3*(4*x^3/e^4 + a/(b*e^4)) - 3*a^2/(b^2*e))*sqrt(e*x)*A*b^2*x*abs(e)^2/ 
e^3 - 1/192*(25*B^2*a^8 + 240*A*B*a^7*b + 576*A^2*a^6*b^2)*e^4*log(abs((5* 
sqrt(e*x)*B*a^4*x + 24*sqrt(e*x)*A*a^3*b*x)*sqrt(b*e) + sqrt(25*B^2*a^9*e^ 
2 + 240*A*B*a^8*b*e^2 + 576*A^2*a^7*b^2*e^2 + (5*sqrt(e*x)*B*a^4*x + 24*sq 
rt(e*x)*A*a^3*b*x)^2*b*e)))/(sqrt(b*e)*b*abs(5*B*a^4*e + 24*A*a^3*b*e)*abs 
(e)^2) - 1/3*(a*e^4*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a* 
e^4)))/sqrt(b*e) - sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*e*x)*A*a^2*abs(e)^2/e 
^5 + 1/12*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3 + a*e^3/b)*sqrt(e*x)*B*a^2*x* 
abs(e)^2/e^7 + 1/6*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3 + a*e^3/b)*sqrt(e*x) 
*A*a*b*x*abs(e)^2/e^7 + 1/576*(15*a^3*e^9/b^3 + 2*(4*(6*e^3*x^3 + a*e^3/b) 
*e^3*x^3 - 5*a^2*e^6/b^2)*e^3*x^3)*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*B*b^2 
*x*abs(e)^2/e^13
 

Mupad [F(-1)]

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

int((A + B*x^3)*(e*x)^(1/2)*(a + b*x^3)^(5/2),x)
 

Output:

int((A + B*x^3)*(e*x)^(1/2)*(a + b*x^3)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.65 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {\sqrt {e}\, \left (558 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a^{3} b x +652 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a^{2} b^{2} x^{4}+400 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a \,b^{3} x^{7}+96 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{4} x^{10}-105 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{4}+105 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{4}\right )}{1152 b} \] Input:

int((e*x)^(1/2)*(b*x^3+a)^(5/2)*(B*x^3+A),x)
 

Output:

(sqrt(e)*(558*sqrt(x)*sqrt(a + b*x**3)*a**3*b*x + 652*sqrt(x)*sqrt(a + b*x 
**3)*a**2*b**2*x**4 + 400*sqrt(x)*sqrt(a + b*x**3)*a*b**3*x**7 + 96*sqrt(x 
)*sqrt(a + b*x**3)*b**4*x**10 - 105*sqrt(b)*log(sqrt(a + b*x**3) - sqrt(x) 
*sqrt(b)*x)*a**4 + 105*sqrt(b)*log(sqrt(a + b*x**3) + sqrt(x)*sqrt(b)*x)*a 
**4))/(1152*b)