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

Optimal. Leaf size=187 \[ \frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+5 A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {\sqrt {e x} (a B+5 A b)}{6 a^2 b e \sqrt {a+b x^2}}+\frac {\sqrt {e x} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

[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)

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Rubi [A]  time = 0.11, antiderivative size = 187, 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 = {457, 290, 329, 220} \[ \frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+5 A b) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 a^{9/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {\sqrt {e x} (a B+5 A b)}{6 a^2 b e \sqrt {a+b x^2}}+\frac {\sqrt {e x} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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 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 457

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*} \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 {\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) \operatorname {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*}

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Mathematica [C]  time = 0.09, size = 108, normalized size = 0.58 \[ \frac {-a^2 B x+x \left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1} (a B+5 A b) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )+a b x \left (7 A+B x^2\right )+5 A b^2 x^3}{6 a^2 b \sqrt {e x} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[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))

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b^{3} e x^{7} + 3 \, a b^{2} e x^{5} + 3 \, a^{2} b e x^{3} + a^{3} e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.03, size = 425, normalized size = 2.27 \[ \frac {10 A \,b^{3} x^{3}+2 B a \,b^{2} x^{3}+5 \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A \,b^{2} x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+14 A a \,b^{2} x -2 B \,a^{2} b x +5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A a b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B \,a^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{12 \sqrt {e x}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(5*A*(-a*b)^(1/2)*((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))*x^2*b^2+B*(-a*b)^
(1/2)*((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))*x^2*a*b+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)*a*b+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)*a^2+10*A*b^3*x^3+2*B*a*b^2*x^3+14*A*a*b^2*x-2*B*a^2*b*x)/(e*x)^(1/2)/a
^2/b^2/(b*x^2+a)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [C]  time = 119.02, size = 94, normalized size = 0.50 \[ \frac {A \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \sqrt {e} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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