\(\int \frac {(e x)^{5/2} (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [274]

Optimal result

Integrand size = 26, antiderivative size = 285 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 (A b-a B) e^2 \sqrt {e x}}{3 b^2 \sqrt {a+b x^3}}+\frac {B e^2 \sqrt {e x} \sqrt {a+b x^3}}{2 b^2}+\frac {(4 A b-7 a B) e^2 \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{12 \sqrt [4]{3} \sqrt [3]{a} b^2 \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

-2/3*(A*b-B*a)*e^2*(e*x)^(1/2)/b^2/(b*x^3+a)^(1/2)+1/2*B*e^2*(e*x)^(1/2)*( 
b*x^3+a)^(1/2)/b^2+1/36*(4*A*b-7*B*a)*e^2*(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))*3^(3/4)/a^(1/3)/b^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 = 10.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.31 \[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {e^2 \sqrt {e x} \left (-4 A b+7 a B+3 b B x^3+(4 A b-7 a B) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{6 b^2 \sqrt {a+b x^3}} \] Input:

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

Output:

(e^2*Sqrt[e*x]*(-4*A*b + 7*a*B + 3*b*B*x^3 + (4*A*b - 7*a*B)*Sqrt[1 + (b*x 
^3)/a]*Hypergeometric2F1[1/6, 1/2, 7/6, -((b*x^3)/a)]))/(6*b^2*Sqrt[a + b* 
x^3])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {959, 817, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 766

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

Input:

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

Output:

(B*(e*x)^(7/2))/(2*b*e*Sqrt[a + b*x^3]) + ((4*A*b - 7*a*B)*((-2*e^2*Sqrt[e 
*x])/(3*b*Sqrt[a + b*x^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 + S 
qrt[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])/(3*3^(1 
/4)*a^(1/3)*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])))/(4*b)
 

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 817

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*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 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 [C] (verified)

Result contains complex when optimal does not.

Time = 5.74 (sec) , antiderivative size = 787, normalized size of antiderivative = 2.76

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(787\)
risch	\(\text {Expression too large to display}\)	\(2115\)
default	\(\text {Expression too large to display}\)	\(3760\)

Input:

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

Output:

1/e/x*(e*x)^(1/2)/(b*x^3+a)^(1/2)*((b*x^3+a)*e*x)^(1/2)*(-2/3/b^2*e^3*x*(A 
*b-B*a)/((x^3+a/b)*b*e*x)^(1/2)+1/2*B*e^2/b^2*(b*e*x^4+a*e*x)^(1/2)+2*(1/3 
*(A*b-B*a)*e^3/b^2-1/4*B/b^2*e^3*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))*b/(-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)))
 

Fricas [F]

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

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

Output:

integral((B*e^2*x^5 + A*e^2*x^2)*sqrt(b*x^3 + a)*sqrt(e*x)/(b^2*x^6 + 2*a* 
b*x^3 + a^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

Reduce [F]

\[ \int \frac {(e x)^{5/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {e}\, e^{2} \left (2 \sqrt {x}\, \sqrt {b \,x^{3}+a}-\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{3}+a}}{b \,x^{4}+a x}d x \right ) a \right )}{4 b} \] Input:

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

Output:

(sqrt(e)*e**2*(2*sqrt(x)*sqrt(a + b*x**3) - int((sqrt(x)*sqrt(a + b*x**3)) 
/(a*x + b*x**4),x)*a))/(4*b)