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

Optimal. Leaf size=176 \[ -\frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-3 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{6 a^{9/4} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}}-\frac {\sqrt {e x} (5 A b-3 a B)}{3 a^2 e^3 \sqrt {a+b x^2}}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}} \]

[Out]

-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)
*(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^(1/4)/e^(5/2)/(b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {453, 290, 329, 220} \[ -\frac {\sqrt {e x} (5 A b-3 a B)}{3 a^2 e^3 \sqrt {a+b x^2}}-\frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-3 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{6 a^{9/4} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}}-\frac {2 A}{3 a e (e x)^{3/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps

\begin {align*} \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) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/2}} \, dx}{3 a e^2}\\ &=-\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) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{6 a^2 e^2}\\ &=-\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) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 a^2 e^3}\\ &=-\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}} F\left (2 \tan ^{-1}\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}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 91, normalized size = 0.52 \[ \frac {x \left (x^2 \sqrt {\frac {b x^2}{a}+1} (3 a B-5 A b) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )-2 a A+3 a B x^2-5 A b x^2\right )}{3 a^2 (e x)^{5/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)*Sqrt[a + b*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b^{2} e^{3} x^{7} + 2 \, a b e^{3} x^{5} + a^{2} e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(b^2*e^3*x^7 + 2*a*b*e^3*x^5 + a^2*e^3*x^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 232, normalized size = 1.32 \[ -\frac {10 A \,b^{2} x^{2}-6 B a b \,x^{2}+5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A b x \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B a x \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+4 A a b}{6 \sqrt {b \,x^{2}+a}\, \sqrt {e x}\, a^{2} b \,e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/x*(5*A*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b
)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x*b-3*B*((b*x+(
-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*
EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*x*a+10*A*b^2*x^2-6*B*a*b*x^2+4*A*a
*b)/(b*x^2+a)^(1/2)/b/a^2/e^2/(e*x)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {B\,x^2+A}{{\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 56.00, size = 97, normalized size = 0.55 \[ \frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

________________________________________________________________________________________