\(\int \frac {A+B x^3}{\sqrt {e x} (a+b x^3)^{3/2}} \, dx\) [277]

Optimal result

Integrand size = 26, antiderivative size = 258 \[ \int \frac {A+B x^3}{\sqrt {e x} \left (a+b x^3\right )^{3/2}} \, dx=\frac {2 (A b-a B) \sqrt {e x}}{3 a b e \sqrt {a+b x^3}}+\frac {(2 A b+a B) \sqrt {e x} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}+\left (1-\sqrt {3}\right ) \sqrt [3]{b} x}{\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{3 \sqrt [4]{3} a^{4/3} b e \sqrt {\frac {\sqrt [3]{b} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\sqrt [3]{a}+\left (1+\sqrt {3}\right ) \sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

2/3*(A*b-B*a)*(e*x)^(1/2)/a/b/e/(b*x^3+a)^(1/2)+1/9*(2*A*b+B*a)*(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^(4/3)/b/e/(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 = 10.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.31 \[ \int \frac {A+B x^3}{\sqrt {e x} \left (a+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (A b-a B+(2 A b+a B) \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},-\frac {b x^3}{a}\right )\right )}{3 a b \sqrt {e x} \sqrt {a+b x^3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {957, 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.

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

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 766

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

Input:

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

Output:

(2*(A*b - a*B)*Sqrt[e*x])/(3*a*b*e*Sqrt[a + b*x^3]) + ((2*A*b + a*B)*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[Arc 
Cos[(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^(4/3)*b*e^2*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])
 

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 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 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.73 (sec) , antiderivative size = 754, normalized size of antiderivative = 2.92

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}\, \left (\frac {2 x \left (A b -B a \right )}{3 b a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b e x}}+\frac {2 \left (\frac {B}{b}+\frac {\frac {2 A b}{3}-\frac {2 B a}{3}}{a b}\right ) \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, {\left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}^{2} \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, \sqrt {\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{b \left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}\, b \operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) x}{\left (-\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}}, \sqrt {\frac {\left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}{\left (\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {b e x \left (x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(754\)
default	\(\text {Expression too large to display}\)	\(3565\)

Input:

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

Output:

((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^3+a)^(1/2)*(2/3/b*x/a*(A*b-B*a)/((x 
^3+a/b)*b*e*x)^(1/2)+2*(B/b+2/3*(A*b-B*a)/a/b)*(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)*El 
lipticF(((-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 [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.36 \[ \int \frac {A+B x^3}{\sqrt {e x} \left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (B a b + 2 \, A b^{2}\right )} x^{3} + B a^{2} + 2 \, A a b\right )} \sqrt {a e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, b}{a}, \frac {1}{x}\right ) + \sqrt {b x^{3} + a} {\left (B a^{2} - A a b\right )} \sqrt {e x}\right )}}{3 \, {\left (a^{2} b^{2} e x^{3} + a^{3} b e\right )}} \] Input:

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

Output:

-2/3*(((B*a*b + 2*A*b^2)*x^3 + B*a^2 + 2*A*a*b)*sqrt(a*e)*weierstrassPInve 
rse(0, -4*b/a, 1/x) + sqrt(b*x^3 + a)*(B*a^2 - A*a*b)*sqrt(e*x))/(a^2*b^2* 
e*x^3 + a^3*b*e)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 27.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.36 \[ \int \frac {A+B x^3}{\sqrt {e x} \left (a+b x^3\right )^{3/2}} \, dx=\frac {A \sqrt {x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{6}, \frac {3}{2} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \sqrt {e} \Gamma \left (\frac {7}{6}\right )} + \frac {B x^{\frac {7}{2}} \Gamma \left (\frac {7}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{6}, \frac {3}{2} \\ \frac {13}{6} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \sqrt {e} \Gamma \left (\frac {13}{6}\right )} \] Input:

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

Output:

A*sqrt(x)*gamma(1/6)*hyper((1/6, 3/2), (7/6,), b*x**3*exp_polar(I*pi)/a)/( 
3*a**(3/2)*sqrt(e)*gamma(7/6)) + B*x**(7/2)*gamma(7/6)*hyper((7/6, 3/2), ( 
13/6,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*sqrt(e)*gamma(13/6))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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