3.6.28 \(\int (e x)^{5/2} (a+b x^3)^{3/2} (A+B x^3) \, dx\) [528]

3.6.28.1 Optimal result
3.6.28.2 Mathematica [C] (verified)
3.6.28.3 Rubi [A] (verified)
3.6.28.4 Maple [C] (verified)
3.6.28.5 Fricas [F]
3.6.28.6 Sympy [C] (verification not implemented)
3.6.28.7 Maxima [F]
3.6.28.8 Giac [F]
3.6.28.9 Mupad [F(-1)]

3.6.28.1 Optimal result

Integrand size = 26, antiderivative size = 364 \[ \int (e x)^{5/2} \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {27 a^2 (22 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^3}}{7040 b^2}+\frac {9 a (22 A b-7 a B) (e x)^{7/2} \sqrt {a+b x^3}}{1760 b e}+\frac {(22 A b-7 a B) (e x)^{7/2} \left (a+b x^3\right )^{3/2}}{176 b e}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}-\frac {9\ 3^{3/4} a^{8/3} (22 A b-7 a B) e^2 \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{14080 b^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

output
1/176*(22*A*b-7*B*a)*(e*x)^(7/2)*(b*x^3+a)^(3/2)/b/e+1/11*B*(e*x)^(7/2)*(b 
*x^3+a)^(5/2)/b/e+9/1760*a*(22*A*b-7*B*a)*(e*x)^(7/2)*(b*x^3+a)^(1/2)/b/e+ 
27/7040*a^2*(22*A*b-7*B*a)*e^2*(e*x)^(1/2)*(b*x^3+a)^(1/2)/b^2-9/14080*3^( 
3/4)*a^(8/3)*(22*A*b-7*B*a)*e^2*(a^(1/3)+b^(1/3)*x)*((a^(1/3)+b^(1/3)*x*(1 
-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)/(a^(1/3)+b^(1/3)*x*( 
1-3^(1/2)))*(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))*EllipticF((1-(a^(1/3)+b^(1/3)* 
x*(1-3^(1/2)))^2/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4* 
2^(1/2))*(e*x)^(1/2)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(a^(1/3)+b^( 
1/3)*x*(1+3^(1/2)))^2)^(1/2)/b^2/(b*x^3+a)^(1/2)/(b^(1/3)*x*(a^(1/3)+b^(1/ 
3)*x)/(a^(1/3)+b^(1/3)*x*(1+3^(1/2)))^2)^(1/2)
 
3.6.28.2 Mathematica [C] (verified)

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

Time = 10.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.32 \[ \int (e x)^{5/2} \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {e^2 \sqrt {e x} \sqrt {a+b x^3} \left (-\left (a+b x^3\right )^2 \sqrt {1+\frac {b x^3}{a}} \left (-22 A b+7 a B-16 b B x^3\right )+a^2 (-22 A b+7 a B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{6},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{176 b^2 \sqrt {1+\frac {b x^3}{a}}} \]

input
Integrate[(e*x)^(5/2)*(a + b*x^3)^(3/2)*(A + B*x^3),x]
 
output
(e^2*Sqrt[e*x]*Sqrt[a + b*x^3]*(-((a + b*x^3)^2*Sqrt[1 + (b*x^3)/a]*(-22*A 
*b + 7*a*B - 16*b*B*x^3)) + a^2*(-22*A*b + 7*a*B)*Hypergeometric2F1[-3/2, 
1/6, 7/6, -((b*x^3)/a)]))/(176*b^2*Sqrt[1 + (b*x^3)/a])
 
3.6.28.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {959, 811, 811, 843, 851, 766}

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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(22 A b-7 a B) \int (e x)^{5/2} \left (b x^3+a\right )^{3/2}dx}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(22 A b-7 a B) \left (\frac {9}{16} a \int (e x)^{5/2} \sqrt {b x^3+a}dx+\frac {(e x)^{7/2} \left (a+b x^3\right )^{3/2}}{8 e}\right )}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(22 A b-7 a B) \left (\frac {9}{16} a \left (\frac {3}{10} a \int \frac {(e x)^{5/2}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{7/2} \sqrt {a+b x^3}}{5 e}\right )+\frac {(e x)^{7/2} \left (a+b x^3\right )^{3/2}}{8 e}\right )}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(22 A b-7 a B) \left (\frac {9}{16} a \left (\frac {3}{10} a \left (\frac {e^2 \sqrt {e x} \sqrt {a+b x^3}}{2 b}-\frac {a e^3 \int \frac {1}{\sqrt {e x} \sqrt {b x^3+a}}dx}{4 b}\right )+\frac {(e x)^{7/2} \sqrt {a+b x^3}}{5 e}\right )+\frac {(e x)^{7/2} \left (a+b x^3\right )^{3/2}}{8 e}\right )}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(22 A b-7 a B) \left (\frac {9}{16} a \left (\frac {3}{10} a \left (\frac {e^2 \sqrt {e x} \sqrt {a+b x^3}}{2 b}-\frac {a e^2 \int \frac {1}{\sqrt {b x^3+a}}d\sqrt {e x}}{2 b}\right )+\frac {(e x)^{7/2} \sqrt {a+b x^3}}{5 e}\right )+\frac {(e x)^{7/2} \left (a+b x^3\right )^{3/2}}{8 e}\right )}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {(22 A b-7 a B) \left (\frac {9}{16} a \left (\frac {3}{10} a \left (\frac {e^2 \sqrt {e x} \sqrt {a+b x^3}}{2 b}-\frac {a^{2/3} e \sqrt {e x} \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right ) \sqrt {\frac {a^{2/3} e^2-\sqrt [3]{a} \sqrt [3]{b} e^2 x+b^{2/3} e^2 x^2}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}{\left (1+\sqrt {3}\right ) \sqrt [3]{b} x e+\sqrt [3]{a} e}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} b \sqrt {a+b x^3} \sqrt {\frac {\sqrt [3]{b} e x \left (\sqrt [3]{a} e+\sqrt [3]{b} e x\right )}{\left (\sqrt [3]{a} e+\left (1+\sqrt {3}\right ) \sqrt [3]{b} e x\right )^2}}}\right )+\frac {(e x)^{7/2} \sqrt {a+b x^3}}{5 e}\right )+\frac {(e x)^{7/2} \left (a+b x^3\right )^{3/2}}{8 e}\right )}{22 b}+\frac {B (e x)^{7/2} \left (a+b x^3\right )^{5/2}}{11 b e}\)

input
Int[(e*x)^(5/2)*(a + b*x^3)^(3/2)*(A + B*x^3),x]
 
output
(B*(e*x)^(7/2)*(a + b*x^3)^(5/2))/(11*b*e) + ((22*A*b - 7*a*B)*(((e*x)^(7/ 
2)*(a + b*x^3)^(3/2))/(8*e) + (9*a*(((e*x)^(7/2)*Sqrt[a + b*x^3])/(5*e) + 
(3*a*((e^2*Sqrt[e*x]*Sqrt[a + b*x^3])/(2*b) - (a^(2/3)*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[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])/(4*3^(1/4)*b*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])))/10))/16))/ 
(22*b)
 

3.6.28.3.1 Defintions of rubi rules used

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 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]
 
3.6.28.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.53 (sec) , antiderivative size = 801, normalized size of antiderivative = 2.20

method result size
risch \(\text {Expression too large to display}\) \(801\)
elliptic \(\text {Expression too large to display}\) \(976\)
default \(\text {Expression too large to display}\) \(4619\)

input
int((e*x)^(5/2)*(b*x^3+a)^(3/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
 
output
1/7040/b^2*(640*B*b^3*x^9+880*A*b^3*x^6+1000*B*a*b^2*x^6+1672*A*a*b^2*x^3+ 
108*B*a^2*b*x^3+594*A*a^2*b-189*B*a^3)*x*(b*x^3+a)^(1/2)*e^3/(e*x)^(1/2)-2 
7/7040*a^3/b*(22*A*b-7*B*a)*(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)/(-3/2/b*(-a*b^2)^(1/ 
3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))/(-a*b^2)^(1/3)/(b*e*x*(x-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))*(x+1/2/b*(-a 
*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*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))*e^3*((b*x^3+a)* 
e*x)^(1/2)/(e*x)^(1/2)/(b*x^3+a)^(1/2)
 
3.6.28.5 Fricas [F]

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

input
integrate((e*x)^(5/2)*(b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="fricas")
 
output
integral((B*b*e^2*x^8 + (B*a + A*b)*e^2*x^5 + A*a*e^2*x^2)*sqrt(b*x^3 + a) 
*sqrt(e*x), x)
 
3.6.28.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 63.41 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.55 \[ \int (e x)^{5/2} \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {A a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{6} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {13}{6}\right )} + \frac {A \sqrt {a} b e^{\frac {5}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{6} \\ \frac {19}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {19}{6}\right )} + \frac {B a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {13}{6} \\ \frac {19}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {19}{6}\right )} + \frac {B \sqrt {a} b e^{\frac {5}{2}} x^{\frac {19}{2}} \Gamma \left (\frac {19}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {19}{6} \\ \frac {25}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {25}{6}\right )} \]

input
integrate((e*x)**(5/2)*(b*x**3+a)**(3/2)*(B*x**3+A),x)
 
output
A*a**(3/2)*e**(5/2)*x**(7/2)*gamma(7/6)*hyper((-1/2, 7/6), (13/6,), b*x**3 
*exp_polar(I*pi)/a)/(3*gamma(13/6)) + A*sqrt(a)*b*e**(5/2)*x**(13/2)*gamma 
(13/6)*hyper((-1/2, 13/6), (19/6,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(19/ 
6)) + B*a**(3/2)*e**(5/2)*x**(13/2)*gamma(13/6)*hyper((-1/2, 13/6), (19/6, 
), b*x**3*exp_polar(I*pi)/a)/(3*gamma(19/6)) + B*sqrt(a)*b*e**(5/2)*x**(19 
/2)*gamma(19/6)*hyper((-1/2, 19/6), (25/6,), b*x**3*exp_polar(I*pi)/a)/(3* 
gamma(25/6))
 
3.6.28.7 Maxima [F]

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

input
integrate((e*x)^(5/2)*(b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="maxima")
 
output
integrate((B*x^3 + A)*(b*x^3 + a)^(3/2)*(e*x)^(5/2), x)
 
3.6.28.8 Giac [F]

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

input
integrate((e*x)^(5/2)*(b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="giac")
 
output
integrate((B*x^3 + A)*(b*x^3 + a)^(3/2)*(e*x)^(5/2), x)
 
3.6.28.9 Mupad [F(-1)]

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

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