\(\int \frac {(e x)^{3/2} (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [307]

Optimal result

Integrand size = 26, antiderivative size = 174 \[ \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 (7 A b-5 a B) e \sqrt {e x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 B (e x)^{5/2} \sqrt {a+b x^2}}{7 b e}-\frac {a^{3/4} (7 A b-5 a B) e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a+b x^2}} \] Output:

2/21*(7*A*b-5*B*a)*e*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b^2+2/7*B*(e*x)^(5/2)*(b* 
x^2+a)^(1/2)/b/e-1/21*a^(3/4)*(7*A*b-5*B*a)*e^(3/2)*(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))/b^(9/4)/(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 = 10.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55 \[ \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 e \sqrt {e x} \left (-\left (\left (a+b x^2\right ) \left (-7 A b+5 a B-3 b B x^2\right )\right )+a (-7 A b+5 a B) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{21 b^2 \sqrt {a+b x^2}} \] Input:

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

Output:

(2*e*Sqrt[e*x]*(-((a + b*x^2)*(-7*A*b + 5*a*B - 3*b*B*x^2)) + a*(-7*A*b + 
5*a*B)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]) 
)/(21*b^2*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 186, 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 = {363, 262, 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.

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

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {(7 A b-5 a B) \left (\frac {2 e \sqrt {e x} \sqrt {a+b x^2}}{3 b}-\frac {a e^2 \int \frac {1}{\sqrt {e x} \sqrt {b x^2+a}}dx}{3 b}\right )}{7 b}+\frac {2 B (e x)^{5/2} \sqrt {a+b x^2}}{7 b e}\)

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

\(\displaystyle \frac {(7 A b-5 a B) \left (\frac {2 e \sqrt {e x} \sqrt {a+b x^2}}{3 b}-\frac {a^{3/4} \sqrt {e} \left (\sqrt {a} e+\sqrt {b} e x\right ) \sqrt {\frac {a e^2+b e^2 x^2}{\left (\sqrt {a} e+\sqrt {b} e x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a+b x^2}}\right )}{7 b}+\frac {2 B (e x)^{5/2} \sqrt {a+b x^2}}{7 b e}\)

Input:

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

Output:

(2*B*(e*x)^(5/2)*Sqrt[a + b*x^2])/(7*b*e) + ((7*A*b - 5*a*B)*((2*e*Sqrt[e* 
x]*Sqrt[a + b*x^2])/(3*b) - (a^(3/4)*Sqrt[e]*(Sqrt[a]*e + Sqrt[b]*e*x)*Sqr 
t[(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])/(3*b^(5/4)*Sqrt[a + b*x^2])))/(7 
*b)
 

Defintions of rubi rules used

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 363

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

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 = 0.61 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {2 \left (3 b B \,x^{2}+7 A b -5 B a \right ) x \sqrt {b \,x^{2}+a}\, e^{2}}{21 b^{2} \sqrt {e x}}-\frac {a \left (7 A b -5 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) e^{2} \sqrt {\left (b \,x^{2}+a \right ) e x}}{21 b^{3} \sqrt {b e \,x^{3}+a e x}\, \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(190\)
elliptic	\(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {2 B e \,x^{2} \sqrt {b e \,x^{3}+a e x}}{7 b}+\frac {2 \left (A \,e^{2}-\frac {5 B \,e^{2} a}{7 b}\right ) \sqrt {b e \,x^{3}+a e x}}{3 b e}-\frac {\left (A \,e^{2}-\frac {5 B \,e^{2} a}{7 b}\right ) a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\)	\(222\)
default	\(-\frac {e \sqrt {e x}\, \left (7 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, \sqrt {-a b}\, a b -5 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, \sqrt {-a b}\, a^{2}-6 b^{3} B \,x^{5}-14 A \,b^{3} x^{3}+4 B a \,b^{2} x^{3}-14 A x a \,b^{2}+10 B x \,a^{2} b \right )}{21 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(250\)

Input:

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

Output:

2/21*(3*B*b*x^2+7*A*b-5*B*a)*x*(b*x^2+a)^(1/2)/b^2*e^2/(e*x)^(1/2)-1/21*a* 
(7*A*b-5*B*a)/b^3*(-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))*e^2*((b*x^2+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^2+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.43 \[ \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 \, {\left ({\left (5 \, B a^{2} - 7 \, A a b\right )} \sqrt {b e} e {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (3 \, B b^{2} e x^{2} - {\left (5 \, B a b - 7 \, A b^{2}\right )} e\right )} \sqrt {b x^{2} + a} \sqrt {e x}\right )}}{21 \, b^{3}} \] Input:

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

Output:

2/21*((5*B*a^2 - 7*A*a*b)*sqrt(b*e)*e*weierstrassPInverse(-4*a/b, 0, x) + 
(3*B*b^2*e*x^2 - (5*B*a*b - 7*A*b^2)*e)*sqrt(b*x^2 + a)*sqrt(e*x))/b^3
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.48 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.54 \[ \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {A e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \] Input:

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

Output:

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**2*exp_polar( 
I*pi)/a)/(2*sqrt(a)*gamma(9/4)) + B*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((1/ 
2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(13/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {2 \sqrt {e}\, e \left (2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a +3 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, b \,x^{2}-\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{3}+a x}d x \right ) a^{2}\right )}{21 b} \] Input:

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

Output:

(2*sqrt(e)*e*(2*sqrt(x)*sqrt(a + b*x**2)*a + 3*sqrt(x)*sqrt(a + b*x**2)*b* 
x**2 - int((sqrt(x)*sqrt(a + b*x**2))/(a*x + b*x**3),x)*a**2))/(21*b)