\(\int \frac {A+B x^2}{(e x)^{5/2} (a+b x^2)^{3/2}} \, dx\) [318]

Optimal result

Integrand size = 26, antiderivative size = 176 \[ \int \frac {A+B x^2}{(e x)^{5/2} \left (a+b x^2\right )^{3/2}} \, dx=-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}}-\frac {(5 A b-3 a B) \sqrt {e x}}{3 a^2 e^3 \sqrt {a+b x^2}}-\frac {(5 A b-3 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{6 a^{9/4} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}} \] Output:

-2/3*A/a/e/(e*x)^(3/2)/(b*x^2+a)^(1/2)-1/3*(5*A*b-3*B*a)*(e*x)^(1/2)/a^2/e 
^3/(b*x^2+a)^(1/2)-1/6*(5*A*b-3*B*a)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/ 
2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4 
)/e^(1/2)),1/2*2^(1/2))/a^(9/4)/b^(1/4)/e^(5/2)/(b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.52 \[ \int \frac {A+B x^2}{(e x)^{5/2} \left (a+b x^2\right )^{3/2}} \, dx=\frac {x \left (-2 a A-5 A b x^2+3 a B x^2+(-5 A b+3 a B) x^2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 a^2 (e x)^{5/2} \sqrt {a+b x^2}} \] Input:

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

Output:

(x*(-2*a*A - 5*A*b*x^2 + 3*a*B*x^2 + (-5*A*b + 3*a*B)*x^2*Sqrt[1 + (b*x^2) 
/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(3*a^2*(e*x)^(5/2)*Sq 
rt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {359, 253, 266, 761}

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.

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

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(5 A b-3 a B) \int \frac {1}{\sqrt {e x} \left (b x^2+a\right )^{3/2}}dx}{3 a e^2}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(5 A b-3 a B) \left (\frac {\int \frac {1}{\sqrt {e x} \sqrt {b x^2+a}}dx}{2 a}+\frac {\sqrt {e x}}{a e \sqrt {a+b x^2}}\right )}{3 a e^2}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(5 A b-3 a B) \left (\frac {\int \frac {1}{\sqrt {b x^2+a}}d\sqrt {e x}}{a e}+\frac {\sqrt {e x}}{a e \sqrt {a+b x^2}}\right )}{3 a e^2}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {(5 A b-3 a B) \left (\frac {\left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} e^{3/2} \sqrt {a+b x^2}}+\frac {\sqrt {e x}}{a e \sqrt {a+b x^2}}\right )}{3 a e^2}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}}\)

Input:

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

Output:

(-2*A)/(3*a*e*(e*x)^(3/2)*Sqrt[a + b*x^2]) - ((5*A*b - 3*a*B)*(Sqrt[e*x]/( 
a*e*Sqrt[a + b*x^2]) + ((Sqrt[a]*e + Sqrt[b]*e*x)*Sqrt[(a*e^2 + b*e^2*x^2) 
/(Sqrt[a]*e + Sqrt[b]*e*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1 
/4)*Sqrt[e])], 1/2])/(2*a^(5/4)*b^(1/4)*e^(3/2)*Sqrt[a + b*x^2])))/(3*a*e^ 
2)
 

Defintions of rubi rules used

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.26

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {x \left (A b -B a \right )}{e^{2} a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}-\frac {2 A \sqrt {b e \,x^{3}+a e x}}{3 a^{2} e^{3} x^{2}}+\frac {\left (-\frac {A b -B a}{2 a^{2} e^{2}}-\frac {b A}{3 a^{2} e^{2}}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(222\)
default	\(-\frac {5 A \sqrt {2}\, \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b x -3 B \sqrt {2}\, \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a x +10 A \,b^{2} x^{2}-6 B a b \,x^{2}+4 a b A}{6 x \sqrt {b \,x^{2}+a}\, b \,a^{2} e^{2} \sqrt {e x}}\)	\(232\)
risch	\(-\frac {2 A \sqrt {b \,x^{2}+a}}{3 a^{2} x \,e^{2} \sqrt {e x}}-\frac {\left (\frac {A \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b e \,x^{3}+a e x}}+3 a \left (A b -B a \right ) \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b e \,x^{3}+a e x}}\right )\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{3 a^{2} e^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(315\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^2}{(e x)^{5/2} \left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{4} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \sqrt {b e} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (2 \, A a b - {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{3 \, {\left (a^{2} b^{2} e^{3} x^{4} + a^{3} b e^{3} x^{2}\right )}} \] Input:

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

Output:

1/3*(((3*B*a*b - 5*A*b^2)*x^4 + (3*B*a^2 - 5*A*a*b)*x^2)*sqrt(b*e)*weierst 
rassPInverse(-4*a/b, 0, x) - (2*A*a*b - (3*B*a*b - 5*A*b^2)*x^2)*sqrt(b*x^ 
2 + a)*sqrt(e*x))/(a^2*b^2*e^3*x^4 + a^3*b*e^3*x^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 22.78 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.55 \[ \int \frac {A+B x^2}{(e x)^{5/2} \left (a+b x^2\right )^{3/2}} \, dx=\frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

A*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**( 
3/2)*e**(5/2)*x**(3/2)*gamma(1/4)) + B*sqrt(x)*gamma(1/4)*hyper((1/4, 3/2) 
, (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*e**(5/2)*gamma(5/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*int((sqrt(x)*sqrt(a + b*x**2))/(a*x**3 + b*x**5),x))/e**3