\(\int (e x)^{7/2} \sqrt {a+b x^3} (A+B x^3) \, dx\) [238]

Optimal result

Integrand size = 26, antiderivative size = 161 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {a (2 A b-a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{24 b^2}+\frac {(2 A b-a B) (e x)^{9/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}-\frac {a^2 (2 A b-a B) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{24 b^{5/2}} \] Output:

1/24*a*(2*A*b-B*a)*e^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/12*(2*A*b-B*a)*(e 
*x)^(9/2)*(b*x^3+a)^(1/2)/b/e+1/9*B*(e*x)^(9/2)*(b*x^3+a)^(3/2)/b/e-1/24*a 
^2*(2*A*b-B*a)*e^(7/2)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a)^(1/2) 
)/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.76 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {(e x)^{7/2} \sqrt {a+b x^3} \left (6 a A b-3 a^2 B+12 A b^2 x^3+2 a b B x^3+8 b^2 B x^6\right )}{72 b^2 x^2}+\frac {a^2 (-2 A b+a B) (e x)^{7/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{24 b^{5/2} x^{7/2}} \] Input:

Integrate[(e*x)^(7/2)*Sqrt[a + b*x^3]*(A + B*x^3),x]
 

Output:

((e*x)^(7/2)*Sqrt[a + b*x^3]*(6*a*A*b - 3*a^2*B + 12*A*b^2*x^3 + 2*a*b*B*x 
^3 + 8*b^2*B*x^6))/(72*b^2*x^2) + (a^2*(-2*A*b + a*B)*(e*x)^(7/2)*Log[Sqrt 
[b]*x^(3/2) + Sqrt[a + b*x^3]])/(24*b^(5/2)*x^(7/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {959, 811, 843, 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.

\(\displaystyle \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx\)

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {(2 A b-a B) \left (\frac {1}{4} a \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 b^{3/2}}\right )+\frac {(e x)^{9/2} \sqrt {a+b x^3}}{6 e}\right )}{2 b}+\frac {B (e x)^{9/2} \left (a+b x^3\right )^{3/2}}{9 b e}\)

Input:

Int[(e*x)^(7/2)*Sqrt[a + b*x^3]*(A + B*x^3),x]
 

Output:

(B*(e*x)^(9/2)*(a + b*x^3)^(3/2))/(9*b*e) + ((2*A*b - a*B)*(((e*x)^(9/2)*S 
qrt[a + b*x^3])/(6*e) + (a*((e^2*(e*x)^(3/2)*Sqrt[a + b*x^3])/(3*b) - (a*e 
^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*b^ 
(3/2))))/4))/(2*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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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 = 3.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {x^{2} \left (8 b^{2} B \,x^{6}+12 A \,b^{2} x^{3}+2 B a b \,x^{3}+6 a b A -3 a^{2} B \right ) \sqrt {b \,x^{3}+a}\, e^{4}}{72 b^{2} \sqrt {e x}}-\frac {a^{2} \left (2 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e^{4} \sqrt {\left (b \,x^{3}+a \right ) e x}}{24 b^{2} \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(138\)
default	\(-\frac {e^{3} \sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (-8 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{2} x^{7}-12 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b^{2} x^{4}-2 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a b \,x^{4}+6 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} b e -6 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a b x -3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{3} e +3 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a^{2} x \right )}{72 \sqrt {\left (b \,x^{3}+a \right ) e x}\, b^{2} \sqrt {b e}}\)	\(224\)
elliptic	\(\text {Expression too large to display}\)	\(1165\)

Input:

int((e*x)^(7/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
 

Output:

1/72*x^2*(8*B*b^2*x^6+12*A*b^2*x^3+2*B*a*b*x^3+6*A*a*b-3*B*a^2)*(b*x^3+a)^ 
(1/2)/b^2*e^4/(e*x)^(1/2)-1/24*a^2*(2*A*b-B*a)/b^2/(b*e)^(1/2)*arctanh(((b 
*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))*e^4*((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/ 
(b*x^3+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.83 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (8 \, B b^{2} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{288 \, b^{2}}, -\frac {3 \, {\left (B a^{3} - 2 \, A a^{2} b\right )} e^{3} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (8 \, B b^{2} e^{3} x^{7} + 2 \, {\left (B a b + 6 \, A b^{2}\right )} e^{3} x^{4} - 3 \, {\left (B a^{2} - 2 \, A a b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{144 \, b^{2}}\right ] \] Input:

integrate((e*x)^(7/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="fricas")
 

Output:

[-1/288*(3*(B*a^3 - 2*A*a^2*b)*e^3*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 
*(8*B*b^2*e^3*x^7 + 2*(B*a*b + 6*A*b^2)*e^3*x^4 - 3*(B*a^2 - 2*A*a*b)*e^3* 
x)*sqrt(b*x^3 + a)*sqrt(e*x))/b^2, -1/144*(3*(B*a^3 - 2*A*a^2*b)*e^3*sqrt( 
-e/b)*arctan(2*sqrt(b*x^3 + a)*sqrt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)) 
 - 2*(8*B*b^2*e^3*x^7 + 2*(B*a*b + 6*A*b^2)*e^3*x^4 - 3*(B*a^2 - 2*A*a*b)* 
e^3*x)*sqrt(b*x^3 + a)*sqrt(e*x))/b^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (141) = 282\).

Time = 18.55 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.85 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {2 \left (\begin {cases} \text {NaN} & \text {for}\: e^{3} = 0 \\\frac {A e^{3} \left (\begin {cases} - \frac {a^{2} e^{3} \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x^{3}} \left (\frac {a e^{3} \left (e x\right )^{\frac {3}{2}}}{8 b} + \frac {\left (e x\right )^{\frac {9}{2}}}{4}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\sqrt {a} \left (e x\right )^{\frac {9}{2}}}{3} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {a^{3} e^{6} \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x^{3}} \left (- \frac {a^{2} e^{6} \left (e x\right )^{\frac {3}{2}}}{16 b^{2}} + \frac {a e^{3} \left (e x\right )^{\frac {9}{2}}}{24 b} + \frac {\left (e x\right )^{\frac {15}{2}}}{6}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\sqrt {a} \left (e x\right )^{\frac {15}{2}}}{5} & \text {otherwise} \end {cases}\right )}{3 e^{3}} & \text {otherwise} \end {cases}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

integrate((e*x)**(7/2)*(b*x**3+a)**(1/2)*(B*x**3+A),x)
 

Output:

Piecewise((2*Piecewise((nan, Eq(e**3, 0)), ((A*e**3*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))/sq 
rt(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), N 
e(b/e**3, 0)), (sqrt(a)*(e*x)**(9/2)/3, True)) + B*Piecewise((a**3*e**6*Pi 
ecewise((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)**(15/2)/6), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(15 
/2)/5, True)))/(3*e**3), True))/e, Ne(e, 0)), (0, True))
 

Maxima [F]

\[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} \sqrt {b x^{3} + a} \left (e x\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((e*x)^(7/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*sqrt(b*x^3 + a)*(e*x)^(7/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (126) = 252\).

Time = 0.23 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.05 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {1}{72} \, \sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} {\left (\frac {4 \, x^{3}}{e^{4}} + \frac {a}{b e^{4}}\right )} - \frac {3 \, a^{2}}{b^{2} e}\right )} \sqrt {e x} B x {\left | e \right |}^{2} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} + \frac {a e^{3}}{b}\right )} \sqrt {e x} A x {\left | e \right |}^{2}}{12 \, e^{4}} - \frac {{\left (B^{2} a^{6} e - 4 \, A B a^{5} b e + 4 \, A^{2} a^{4} b^{2} e\right )}^{2} e^{5} \log \left ({\left | -{\left (\sqrt {e x} B a^{3} e^{2} x - 2 \, \sqrt {e x} A a^{2} b e^{2} x\right )} \sqrt {b e} + \sqrt {B^{2} a^{7} e^{6} - 4 \, A B a^{6} b e^{6} + 4 \, A^{2} a^{5} b^{2} e^{6} + {\left (\sqrt {e x} B a^{3} e^{2} x - 2 \, \sqrt {e x} A a^{2} b e^{2} x\right )}^{2} b e} \right |}\right )}{24 \, \sqrt {b e} b^{2} {\left | B^{2} a^{6} e - 4 \, A B a^{5} b e + 4 \, A^{2} a^{4} b^{2} e \right |} {\left | -B a^{3} + 2 \, A a^{2} b \right |} {\left | e \right |}^{2}} \] Input:

integrate((e*x)^(7/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="giac")
 

Output:

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)*B*x*abs(e)^2 + 1/12*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3 + 
a*e^3/b)*sqrt(e*x)*A*x*abs(e)^2/e^4 - 1/24*(B^2*a^6*e - 4*A*B*a^5*b*e + 4* 
A^2*a^4*b^2*e)^2*e^5*log(abs(-(sqrt(e*x)*B*a^3*e^2*x - 2*sqrt(e*x)*A*a^2*b 
*e^2*x)*sqrt(b*e) + sqrt(B^2*a^7*e^6 - 4*A*B*a^6*b*e^6 + 4*A^2*a^5*b^2*e^6 
 + (sqrt(e*x)*B*a^3*e^2*x - 2*sqrt(e*x)*A*a^2*b*e^2*x)^2*b*e)))/(sqrt(b*e) 
*b^2*abs(B^2*a^6*e - 4*A*B*a^5*b*e + 4*A^2*a^4*b^2*e)*abs(-B*a^3 + 2*A*a^2 
*b)*abs(e)^2)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^{7/2}\,\sqrt {b\,x^3+a} \,d x \] Input:

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((A + B*x^3)*(e*x)^(7/2)*(a + b*x^3)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.70 \[ \int (e x)^{7/2} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {\sqrt {e}\, e^{3} \left (6 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a^{2} b x +28 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a \,b^{2} x^{4}+16 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{3} x^{7}+3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{3}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{3}\right )}{144 b^{2}} \] Input:

int((e*x)^(7/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x)
 

Output:

(sqrt(e)*e**3*(6*sqrt(x)*sqrt(a + b*x**3)*a**2*b*x + 28*sqrt(x)*sqrt(a + b 
*x**3)*a*b**2*x**4 + 16*sqrt(x)*sqrt(a + b*x**3)*b**3*x**7 + 3*sqrt(b)*log 
(sqrt(a + b*x**3) - sqrt(x)*sqrt(b)*x)*a**3 - 3*sqrt(b)*log(sqrt(a + b*x** 
3) + sqrt(x)*sqrt(b)*x)*a**3))/(144*b**2)