3.271 \(\int x^2 \sqrt {\frac {e (a+b x^2)}{c+d x^2}} \, dx\)

Optimal. Leaf size=266 \[ -\frac {c^{3/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\sqrt {c} (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 d}-\frac {x (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 b d} \]

[Out]

-1/3*(-a*d+2*b*c)*x*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/b/d+1/3*x*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(1/2)/d-1/3*c^(3
/2)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*(
e*(b*x^2+a)/(d*x^2+c))^(1/2)/d^(3/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)+1/3*(-a*d+2*b*c)*(1/(1+d*x^2/c))^(1/2)*(1
+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)*(e*(b*x^2+a)/(d*x^2+c
))^(1/2)/b/d^(3/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 266, 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 = {6719, 478, 531, 418, 492, 411} \[ -\frac {c^{3/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\sqrt {c} (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {x \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 d}-\frac {x (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]

[Out]

-((2*b*c - a*d)*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(3*b*d) + (x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2
))/(3*d) + (Sqrt[c]*(2*b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 -
(b*c)/(a*d)])/(3*b*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) - (c^(3/2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)
]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))])

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 478

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int x^2 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx &=\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx}{\sqrt {a+b x^2}}\\ &=\frac {x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 d}-\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {a c+(2 b c-a d) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 d \sqrt {a+b x^2}}\\ &=\frac {x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 d}-\frac {\left (a c \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 d \sqrt {a+b x^2}}-\frac {\left ((2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{3 d \sqrt {a+b x^2}}\\ &=-\frac {(2 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 b d}+\frac {x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 d}-\frac {c^{3/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {\left (c (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{3 b d \sqrt {a+b x^2}}\\ &=-\frac {(2 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{3 b d}+\frac {x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{3 d}+\frac {\sqrt {c} (2 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 b d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {c^{3/2} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{3/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.31, size = 208, normalized size = 0.78 \[ \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (d x \sqrt {\frac {b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right )+2 i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )\right )}{3 d^2 \sqrt {\frac {b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)],x]

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2) - I*c*(-2*b*c + a*d)*Sqrt[1 + (b*x^2
)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + (2*I)*c*(-(b*c) + a*d)*Sqrt[1 + (b*x
^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(3*Sqrt[b/a]*d^2*(a + b*x^2))

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))*x^2, x)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 356, normalized size = 1.34 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (\sqrt {-\frac {b}{a}}\, b \,d^{2} x^{5}+\sqrt {-\frac {b}{a}}\, a \,d^{2} x^{3}+\sqrt {-\frac {b}{a}}\, b c d \,x^{3}+\sqrt {-\frac {b}{a}}\, a c d x +\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a c d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b \,c^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{3 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((b*x^2+a)/(d*x^2+c)*e)^(1/2),x)

[Out]

1/3*((b*x^2+a)/(d*x^2+c)*e)^(1/2)*(d*x^2+c)*((-1/a*b)^(1/2)*x^5*b*d^2+(-1/a*b)^(1/2)*x^3*a*d^2+(-1/a*b)^(1/2)*
x^3*b*c*d-2*a*c*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*d+2*((b*x^
2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b*c^2+((b*x^2+a)/a)^(1/2)*((d*x^
2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*c*d-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Elli
pticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b*c^2+(-1/a*b)^(1/2)*x*a*c*d)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^2/(-1/a*b)
^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))*x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((e*(a + b*x^2))/(c + d*x^2))^(1/2),x)

[Out]

int(x^2*((e*(a + b*x^2))/(c + d*x^2))^(1/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________