3.285 \(\int (\frac{e (a+b x^2)}{c+d x^2})^{3/2} \, dx\)

Optimal. Leaf size=262 \[ \frac{b \sqrt{c} e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{e (2 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{e x (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{e x (2 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

[Out]

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

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Rubi [A]  time = 0.202139, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6719, 413, 531, 418, 492, 411} \[ \frac{b \sqrt{c} e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{e (2 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{e x (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{e x (2 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

Antiderivative was successfully verified.

[In]

Int[((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

-(((b*c - a*d)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(c*d)) + ((2*b*c - a*d)*e*x*Sqrt[(e*(a + b*x^2))/(c + d*
x^2)])/(c*d) - ((2*b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (
b*c)/(a*d)])/(Sqrt[c]*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) + (b*Sqrt[c]*e*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
 FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 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 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx &=\frac{\left (e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}\right ) \int \frac{\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{a+b x^2}}\\ &=-\frac{(b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{\left (e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}\right ) \int \frac{a b c+b (2 b c-a d) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{c d \sqrt{a+b x^2}}\\ &=-\frac{(b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{\left (a b e \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}{d \sqrt{a+b x^2}}+\frac{\left (b (2 b c-a d) e \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}{c d \sqrt{a+b x^2}}\\ &=-\frac{(b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{(2 b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{b \sqrt{c} e \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 )}{d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\left ((2 b c-a d) e \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}{d \sqrt{a+b x^2}}\\ &=-\frac{(b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{(2 b c-a d) e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}-\frac{(2 b c-a d) e \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 )}{\sqrt{c} d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{b \sqrt{c} e \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 )}{d^{3/2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\\ \end{align*}

Mathematica [C]  time = 0.317002, size = 206, normalized size = 0.79 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left ((a d-b c) \left (d x \sqrt{\frac{b}{a}} \left (a+b x^2\right )-2 i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} F\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )+i b c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )}{c d^2 \sqrt{\frac{b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

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

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Maple [A]  time = 0.012, size = 527, normalized size = 2. \begin{align*}{\frac{d{x}^{2}+c}{ \left ( b{x}^{2}+a \right ) ^{2}{d}^{2}c} \left ({\frac{ \left ( b{x}^{2}+a \right ) e}{d{x}^{2}+c}} \right ) ^{{\frac{3}{2}}} \left ( \sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{-{\frac{b}{a}}}{x}^{3}ab{d}^{2}-\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{-{\frac{b}{a}}}{x}^{3}{b}^{2}cd+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 ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }abcd-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 ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{b}^{2}{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 ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }abcd+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 ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{b}^{2}{c}^{2}+\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{-{\frac{b}{a}}}x{a}^{2}{d}^{2}-\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{-{\frac{b}{a}}}xabcd \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((e*(b*x^2+a)/(d*x^2+c))^(3/2),x)

[Out]

(e*(b*x^2+a)/(d*x^2+c))^(3/2)*(d*x^2+c)*((b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x^3*a*b*d^2-(b*d*x^4
+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x^3*b^2*c*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))*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*c*d-2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Ellip
ticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*((d*x^2+c)*(b*x^2+a))^(1/2)*b^2*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))*((d*x^2+c)*(b*x^2+a))^(1/2)*a*b*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))*((d*x^2+c)*(b*x^2+a))^(1/2)*b^2*c^2+(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x*a^2*d^2-(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x*a*b*c*d)/(b*x^2+a)^
2/d^2/c/(-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 \left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(((b*x^2 + a)*e/(d*x^2 + c))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*e*x^2 + a*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(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((e*(b*x**2+a)/(d*x**2+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

integrate(((b*x^2 + a)*e/(d*x^2 + c))^(3/2), x)