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]

-((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))])

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Rubi [A]  time = 0.337844, antiderivative size = 266, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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 = 0.288099, 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))

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Maple [A]  time = 0.01, size = 356, normalized size = 1.3 \begin{align*}{\frac{d{x}^{2}+c}{3\,{d}^{2}}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( \sqrt{-{\frac{b}{a}}}{x}^{5}b{d}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{3}a{d}^{2}+\sqrt{-{\frac{b}{a}}}{x}^{3}bcd-2\,ac\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) d+2\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) b{c}^{2}+\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) acd-2\,\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) b{c}^{2}+\sqrt{-{\frac{b}{a}}}xacd \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}} x^{2}\,{d x} \end{align*}

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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}, x\right ) \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}} x^{2}\,{d x} \end{align*}

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)