\(\int \frac {A+B x^2}{\sqrt {e x} (a+b x^2)^{5/2}} \, dx\) [819]

Optimal result

Integrand size = 26, antiderivative size = 187 \[ \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {(A b-a B) \sqrt {e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {(5 A b+a B) \sqrt {e x}}{6 a^2 b e \sqrt {a+b x^2}}+\frac {(5 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}} \]

[Out]

1/3*(A*b-B*a)*(e*x)^(1/2)/a/b/e/(b*x^2+a)^(3/2)+1/6*(5*A*b+B*a)*(e*x)^(1/2)/a^2/b/e/(b*x^2+a)^(1/2)+1/12*(5*A*
b+B*a)*(cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e
^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2
+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(9/4)/b^(5/4)/e^(1/2)/(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {468, 296, 335, 226} \[ \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+5 A b) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {\sqrt {e x} (a B+5 A b)}{6 a^2 b e \sqrt {a+b x^2}}+\frac {\sqrt {e x} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[In]

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

[Out]

((A*b - a*B)*Sqrt[e*x])/(3*a*b*e*(a + b*x^2)^(3/2)) + ((5*A*b + a*B)*Sqrt[e*x])/(6*a^2*b*e*Sqrt[a + b*x^2]) +
((5*A*b + a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqr
t[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(12*a^(9/4)*b^(5/4)*Sqrt[e]*Sqrt[a + b*x^2])

Rule 226

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]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[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 468

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] - Dist[(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] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) \sqrt {e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {\left (\frac {5 A b}{2}+\frac {a B}{2}\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/2}} \, dx}{3 a b} \\ & = \frac {(A b-a B) \sqrt {e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {(5 A b+a B) \sqrt {e x}}{6 a^2 b e \sqrt {a+b x^2}}+\frac {(5 A b+a B) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{12 a^2 b} \\ & = \frac {(A b-a B) \sqrt {e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {(5 A b+a B) \sqrt {e x}}{6 a^2 b e \sqrt {a+b x^2}}+\frac {(5 A b+a B) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{6 a^2 b e} \\ & = \frac {(A b-a B) \sqrt {e x}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {(5 A b+a B) \sqrt {e x}}{6 a^2 b e \sqrt {a+b x^2}}+\frac {(5 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {-a^2 B x+5 A b^2 x^3+a b x \left (7 A+B x^2\right )+(5 A b+a B) x \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )}{6 a^2 b \sqrt {e x} \left (a+b x^2\right )^{3/2}} \]

[In]

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

[Out]

(-(a^2*B*x) + 5*A*b^2*x^3 + a*b*x*(7*A + B*x^2) + (5*A*b + a*B)*x*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometr
ic2F1[1/4, 1/2, 5/4, -((b*x^2)/a)])/(6*a^2*b*Sqrt[e*x]*(a + b*x^2)^(3/2))

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.20

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {\left (A b -B a \right ) \sqrt {b e \,x^{3}+a e x}}{3 a e \,b^{3} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {x \left (5 A b +B a \right )}{6 b \,a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {\left (5 A b +B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{12 b^{2} a^{2} \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(225\)
default	\(\frac {5 A \sqrt {-a b}\, \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b^{2} x^{2}+B \sqrt {-a b}\, \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}+5 A \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a b +B \sqrt {2}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a^{2}+10 A \,b^{3} x^{3}+2 B a \,b^{2} x^{3}+14 a \,b^{2} A x -2 a^{2} b B x}{12 \sqrt {e x}\, a^{2} b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\)	\(425\)

[In]

int((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

((b*x^2+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^2+a)^(1/2)*(1/3/a/e/b^3*(A*b-B*a)*(b*e*x^3+a*e*x)^(1/2)/(x^2+a/b)^2+1/6
/b*x/a^2*(5*A*b+B*a)/((x^2+a/b)*b*e*x)^(1/2)+1/12/b^2/a^2*(5*A*b+B*a)*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*b)^
(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*e*x^3+a*e*x)^(1/2)*El
lipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (B a b^{2} + 5 \, A b^{3}\right )} x^{4} + B a^{3} + 5 \, A a^{2} b + 2 \, {\left (B a^{2} b + 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b e} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (B a^{2} b - 7 \, A a b^{2} - {\left (B a b^{2} + 5 \, A b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{6 \, {\left (a^{2} b^{4} e x^{4} + 2 \, a^{3} b^{3} e x^{2} + a^{4} b^{2} e\right )}} \]

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="fricas")

[Out]

1/6*(((B*a*b^2 + 5*A*b^3)*x^4 + B*a^3 + 5*A*a^2*b + 2*(B*a^2*b + 5*A*a*b^2)*x^2)*sqrt(b*e)*weierstrassPInverse
(-4*a/b, 0, x) - (B*a^2*b - 7*A*a*b^2 - (B*a*b^2 + 5*A*b^3)*x^2)*sqrt(b*x^2 + a)*sqrt(e*x))/(a^2*b^4*e*x^4 + 2
*a^3*b^3*e*x^2 + a^4*b^2*e)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 43.73 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.50 \[ \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/2}} \, dx=\frac {A \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate((B*x**2+A)/(b*x**2+a)**(5/2)/(e*x)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((B*x^2+A)/(b*x^2+a)^(5/2)/(e*x)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((A + B*x^2)/((e*x)^(1/2)*(a + b*x^2)^(5/2)),x)

[Out]

int((A + B*x^2)/((e*x)^(1/2)*(a + b*x^2)^(5/2)), x)