3.8.89 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{(e x)^{5/2}} \, dx\) [789]

Optimal. Leaf size=172 \[ \frac {2 (A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{3 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {2 (A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}} \]

[Out]

-2/3*A*(b*x^2+a)^(3/2)/a/e/(e*x)^(3/2)+2/3*(A*b+B*a)*(e*x)^(1/2)*(b*x^2+a)^(1/2)/a/e^3+2/3*(A*b+B*a)*(cos(2*ar
ctan(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)))*Ellipti
cF(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^(1/4)/b^(1/4)/e^(5/2)/(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 172, 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 = {464, 285, 335, 226} \begin {gather*} \frac {2 \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 )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{5/2} \sqrt {a+b x^2}}+\frac {2 \sqrt {e x} \sqrt {a+b x^2} (a B+A b)}{3 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(3*a*e^3) - (2*A*(a + b*x^2)^(3/2))/(3*a*e*(e*x)^(3/2)) + (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])/(3*a^(1/4)*b^(1/4)*e^(5/2)*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 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && 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 464

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 {\sqrt {a+b x^2} \left (A+B x^2\right )}{(e x)^{5/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(A b+a B) \int \frac {\sqrt {a+b x^2}}{\sqrt {e x}} \, dx}{a e^2}\\ &=\frac {2 (A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{3 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(2 (A b+a B)) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{3 e^2}\\ &=\frac {2 (A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{3 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {(4 (A b+a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 e^3}\\ &=\frac {2 (A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{3 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{3 a e (e x)^{3/2}}+\frac {2 (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 )}{3 \sqrt [4]{a} \sqrt [4]{b} e^{5/2} \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 = 9.82, size = 82, normalized size = 0.48 \begin {gather*} \frac {2 x \sqrt {a+b x^2} \left (-A \left (a+b x^2\right )+\frac {3 (A b+a B) x^2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )}{\sqrt {1+\frac {b x^2}{a}}}\right )}{3 a (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 234, normalized size = 1.36

method	result	size
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-B \,x^{2}+A \right )}{3 x \,e^{2} \sqrt {e x}}+\frac {\left (\frac {2 A b}{3}+\frac {2 B a}{3}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{b \sqrt {b e \,x^{3}+a e x}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(179\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {2 A \sqrt {b e \,x^{3}+a e x}}{3 e^{3} x^{2}}+\frac {2 B \sqrt {b e \,x^{3}+a e x}}{3 e^{3}}+\frac {\left (\frac {A b +B a}{e^{2}}-\frac {b A}{3 e^{2}}-\frac {B a}{3 e^{2}}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(205\)
default	\(\frac {\frac {2 A \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b x}{3}+\frac {2 B \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a x}{3}+\frac {2 b^{2} B \,x^{4}}{3}-\frac {2 A \,b^{2} x^{2}}{3}+\frac {2 B a b \,x^{2}}{3}-\frac {2 a b A}{3}}{\sqrt {b \,x^{2}+a}\, x \,e^{2} \sqrt {e x}\, b}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.37, size = 59, normalized size = 0.34 \begin {gather*} \frac {2 \, {\left (2 \, {\left (B a + A b\right )} \sqrt {b} x^{2} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (B b x^{2} - A b\right )} \sqrt {b x^{2} + a} \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{3 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 3.83, size = 100, normalized size = 0.58 \begin {gather*} \frac {A \sqrt {a} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {a} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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