3.786 \(\int \sqrt {e x} \sqrt {a+b x^2} (A+B x^2) \, dx\)

Optimal. Leaf size=337 \[ \frac {2 a^{5/4} \sqrt {e} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^2}}-\frac {4 a^{5/4} \sqrt {e} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^2}}+\frac {4 a \sqrt {e x} \sqrt {a+b x^2} (3 A b-a B)}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 (e x)^{3/2} \sqrt {a+b x^2} (3 A b-a B)}{15 b e}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e} \]

[Out]

2/9*B*(e*x)^(3/2)*(b*x^2+a)^(3/2)/b/e+2/15*(3*A*b-B*a)*(e*x)^(3/2)*(b*x^2+a)^(1/2)/b/e+4/15*a*(3*A*b-B*a)*(e*x
)^(1/2)*(b*x^2+a)^(1/2)/b^(3/2)/(a^(1/2)+x*b^(1/2))-4/15*a^(5/4)*(3*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)))*EllipticE(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))*e^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)
/b^(7/4)/(b*x^2+a)^(1/2)+2/15*a^(5/4)*(3*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))*e^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/b^(7/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {459, 279, 329, 305, 220, 1196} \[ \frac {2 a^{5/4} \sqrt {e} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^2}}-\frac {4 a^{5/4} \sqrt {e} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-a B) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^2}}+\frac {4 a \sqrt {e x} \sqrt {a+b x^2} (3 A b-a B)}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 (e x)^{3/2} \sqrt {a+b x^2} (3 A b-a B)}{15 b e}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A*b - a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*b*e) + (4*a*(3*A*b - a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(
3/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*B*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*b*e) - (4*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)
*Sqrt[e])], 1/2])/(15*b^(7/4)*Sqrt[a + b*x^2]) + (2*a^(5/4)*(3*A*b - a*B)*Sqrt[e]*(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])/(15*b^(7/
4)*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 279

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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 459

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \sqrt {e x} \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx &=\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}-\frac {\left (2 \left (-\frac {9 A b}{2}+\frac {3 a B}{2}\right )\right ) \int \sqrt {e x} \sqrt {a+b x^2} \, dx}{9 b}\\ &=\frac {2 (3 A b-a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 b e}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac {(2 a (3 A b-a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{15 b}\\ &=\frac {2 (3 A b-a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 b e}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac {(4 a (3 A b-a B)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 b e}\\ &=\frac {2 (3 A b-a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 b e}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}+\frac {\left (4 a^{3/2} (3 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 b^{3/2}}-\frac {\left (4 a^{3/2} (3 A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a} e}}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{15 b^{3/2}}\\ &=\frac {2 (3 A b-a B) (e x)^{3/2} \sqrt {a+b x^2}}{15 b e}+\frac {4 a (3 A b-a B) \sqrt {e x} \sqrt {a+b x^2}}{15 b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 B (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 b e}-\frac {4 a^{5/4} (3 A b-a B) \sqrt {e} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {a+b x^2}}+\frac {2 a^{5/4} (3 A b-a B) \sqrt {e} \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 )}{15 b^{7/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 93, normalized size = 0.28 \[ \frac {2 x \sqrt {e x} \sqrt {a+b x^2} \left ((3 A b-a B) \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )+B \sqrt {\frac {b x^2}{a}+1} \left (a+b x^2\right )\right )}{9 b \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[e*x]*Sqrt[a + b*x^2]*(B*(a + b*x^2)*Sqrt[1 + (b*x^2)/a] + (3*A*b - a*B)*Hypergeometric2F1[-1/2, 3/4,
7/4, -((b*x^2)/a)]))/(9*b*Sqrt[1 + (b*x^2)/a])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x^{2} + A\right )} \sqrt {b x^{2} + a} \sqrt {e x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 414, normalized size = 1.23 \[ \frac {2 \sqrt {e x}\, \left (5 B \,b^{3} x^{6}+9 A \,b^{3} x^{4}+7 B a \,b^{2} x^{4}+9 A a \,b^{2} x^{2}+2 B \,a^{2} b \,x^{2}+18 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A \,a^{2} b \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-9 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, A \,a^{2} b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{3} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right )}{45 \sqrt {b \,x^{2}+a}\, b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(e*x)^(1/2)/(b*x^2+a)^(1/2)/b^2*(5*B*x^6*b^3+18*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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2
),1/2*2^(1/2))*a^2*b-9*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^2*b-6*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
)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^3+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^3+9*A*b^3*x^4+7*B*a*b^2*x^4+9*A*a*b^2*x^2+2*B*a^2*b*x^2)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.75, size = 95, normalized size = 0.28 \[ \frac {A \sqrt {a} \left (e x\right )^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e \Gamma \left (\frac {7}{4}\right )} + \frac {B \sqrt {a} \left (e x\right )^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{3} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*sqrt(a)*(e*x)**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e*gamma(7/4)) + B*sq
rt(a)*(e*x)**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**3*gamma(11/4))

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