Integrand size = 26, antiderivative size = 201 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {5 a^2 (8 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{192 b e}+\frac {5 a (8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{288 b e}+\frac {(8 A b-a B) (e x)^{3/2} \left (a+b x^3\right )^{5/2}}{72 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}+\frac {5 a^3 (8 A b-a B) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{192 b^{3/2}} \]
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)*arctanh((e*x)^(3/2)*b^(1/2)/e^(3/2)/(b*x^3+a)^(1/2))*e^(1/2) /b^(3/2)+5/192*a^2*(8*A*b-B*a)*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b/e
Time = 0.64 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.70 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {\sqrt {e x} \left (\sqrt {b} x^{3/2} \sqrt {a+b x^3} \left (15 a^3 B+16 b^3 x^6 \left (4 A+3 B x^3\right )+8 a b^2 x^3 \left (26 A+17 B x^3\right )+2 a^2 b \left (132 A+59 B x^3\right )\right )-15 a^3 (-8 A b+a B) \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{576 b^{3/2} \sqrt {x}} \]
(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])
Time = 0.33 (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 |
\(\displaystyle \frac {(8 A b-a B) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 \sqrt {b}}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{5/2}}{9 e}\right )}{8 b}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{7/2}}{12 b e}\) |
(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)
3.6.38.3.1 Defintions of rubi rules used
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])
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]
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]
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]
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]
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]
Time = 4.46 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x^{2} \left (48 b^{3} B \,x^{9}+64 x^{6} b^{3} A +136 B \,x^{6} a \,b^{2}+208 a A \,b^{2} x^{3}+118 B \,a^{2} b \,x^{3}+264 a^{2} b A +15 a^{3} B \right ) \sqrt {b \,x^{3}+a}\, e}{576 b \sqrt {e x}}+\frac {5 a^{3} \left (8 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e \sqrt {\left (b \,x^{3}+a \right ) e x}}{192 b \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\) | \(158\) |
default | \(\frac {\sqrt {b \,x^{3}+a}\, \sqrt {e x}\, \left (48 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{3} x^{10}+64 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{3} x^{7}+136 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \,b^{2} x^{7}+208 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \,b^{2} x^{4}+118 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} b \,x^{4}+264 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} b x +120 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{3} b e +15 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{3} x -15 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{4} e \right )}{576 b \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}}\) | \(279\) |
elliptic | \(\text {Expression too large to display}\) | \(1304\) |
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)
Time = 0.57 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.61 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\left [-\frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{2304 \, b}, \frac {15 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) + 2 \, {\left (48 \, B b^{3} x^{10} + 8 \, {\left (17 \, B a b^{2} + 8 \, A b^{3}\right )} x^{7} + 2 \, {\left (59 \, B a^{2} b + 104 \, A a b^{2}\right )} x^{4} + 3 \, {\left (5 \, B a^{3} + 88 \, A a^{2} b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{1152 \, b}\right ] \]
[-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]
Leaf count of result is larger than twice the leaf count of optimal. 896 vs. \(2 (177) = 354\).
Time = 7.75 (sec) , antiderivative size = 896, normalized size of antiderivative = 4.46 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\text {Too large to display} \]
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...
\[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {5}{2}} \sqrt {e x} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (159) = 318\).
Time = 0.52 (sec) , antiderivative size = 590, normalized size of antiderivative = 2.94 \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\frac {\sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} {\left (\frac {2 \, x^{3}}{e} + \frac {a}{b e}\right )} B a^{2} x {\left | e \right |}^{2}}{12 \, e^{3}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} {\left (\frac {2 \, x^{3}}{e} + \frac {a}{b e}\right )} A a b x {\left | e \right |}^{2}}{6 \, e^{3}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} B a b x {\left | e \right |}^{2}}{36 \, e^{3}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} A b^{2} x {\left | e \right |}^{2}}{72 \, e^{3}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, {\left (4 \, e^{3} x^{3} {\left (\frac {6 \, x^{3}}{e^{7}} + \frac {a}{b e^{7}}\right )} - \frac {5 \, a^{2}}{b^{2} e^{4}}\right )} e^{3} x^{3} + \frac {15 \, a^{3}}{b^{3} e}\right )} \sqrt {e x} B b^{2} x {\left | e \right |}^{2}}{576 \, e^{3}} - \frac {{\left (25 \, B^{2} a^{8} + 240 \, A B a^{7} b + 576 \, A^{2} a^{6} b^{2}\right )} e^{4} \log \left ({\left | {\left (5 \, \sqrt {e x} B a^{4} x + 24 \, \sqrt {e x} A a^{3} b x\right )} \sqrt {b e} + \sqrt {25 \, B^{2} a^{9} e^{2} + 240 \, A B a^{8} b e^{2} + 576 \, A^{2} a^{7} b^{2} e^{2} + {\left (5 \, \sqrt {e x} B a^{4} x + 24 \, \sqrt {e x} A a^{3} b x\right )}^{2} b e} \right |}\right )}{192 \, \sqrt {b e} b {\left | 5 \, B a^{4} e + 24 \, A a^{3} b e \right |} {\left | e \right |}^{2}} - \frac {{\left (\frac {a e^{4} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{\sqrt {b e}} - \sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} e x\right )} A a^{2} {\left | e \right |}^{2}}{3 \, e^{5}} \]
1/12*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*(2*x^3/e + a/(b*e))*B*a^2*x*abs(e)^ 2/e^3 + 1/6*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*(2*x^3/e + a/(b*e))*A*a*b*x* abs(e)^2/e^3 + 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 + 1/576*sqrt(b*e^4*x^3 + a*e^4)*(2*(4*e^3*x^3*(6*x^3/e^ 7 + a/(b*e^7)) - 5*a^2/(b^2*e^4))*e^3*x^3 + 15*a^3/(b^3*e))*sqrt(e*x)*B*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*sqrt(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
Timed out. \[ \int \sqrt {e x} \left (a+b x^3\right )^{5/2} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,\sqrt {e\,x}\,{\left (b\,x^3+a\right )}^{5/2} \,d x \]