3.9.11 \(\int \frac {A+B x^2}{\sqrt {e x} (a+b x^2)^{3/2}} \, dx\) [811]

Optimal. Leaf size=144 \[ \frac {(A b-a B) \sqrt {e x}}{a b e \sqrt {a+b x^2}}+\frac {(A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{2 a^{5/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {468, 335, 226} \begin {gather*} \frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+A b) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{2 a^{5/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}}+\frac {\sqrt {e x} (A b-a B)}{a b e \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*Sqrt[e*x])/(a*b*e*Sqrt[a + b*x^2]) + ((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)*Sqrt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(2*a^(5/4)*b^(5/4)*Sqrt[e]*S
qrt[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 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*} \int \frac {A+B x^2}{\sqrt {e x} \left (a+b x^2\right )^{3/2}} \, dx &=\frac {(A b-a B) \sqrt {e x}}{a b e \sqrt {a+b x^2}}+\frac {(A b+a B) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{2 a b}\\ &=\frac {(A b-a B) \sqrt {e x}}{a b e \sqrt {a+b x^2}}+\frac {(A b+a B) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{a b e}\\ &=\frac {(A b-a B) \sqrt {e x}}{a b e \sqrt {a+b x^2}}+\frac {(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 )}{2 a^{5/4} b^{5/4} \sqrt {e} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 75, normalized size = 0.52 \begin {gather*} \frac {x \left (A b-a B+(A b+a B) \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )\right )}{a b \sqrt {e x} \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 213, normalized size = 1.48

method	result	size
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {x \left (A b -B a \right )}{b a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {\left (\frac {B}{b}+\frac {A b -B a}{2 a b}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(191\)
default	\(\frac {A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, b +B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a +2 A \,b^{2} x -2 B a b x}{2 \sqrt {b \,x^{2}+a}\, a \sqrt {e x}\, b^{2}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.27, size = 84, normalized size = 0.58 \begin {gather*} \frac {{\left ({\left (B a^{2} + A a b + {\left (B a b + A b^{2}\right )} x^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (B a b - A b^{2}\right )} \sqrt {b x^{2} + a} \sqrt {x}\right )} e^{\left (-\frac {1}{2}\right )}}{a b^{3} x^{2} + a^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 5.15, size = 94, normalized size = 0.65 \begin {gather*} \frac {A \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}} \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 {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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