3.316 \(\int \frac{1}{x^2 (\frac{e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=380 \[ \frac{c^{3/2} \sqrt{d} \left (a+b x^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a^2 e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\left (a+b x^2\right ) (2 b c-a d)}{a^2 b e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{d x \left (a+b x^2\right ) (2 b c-a d)}{a^2 b e \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{c} \sqrt{d} \left (a+b x^2\right ) (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a^2 b e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{a b e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.470159, antiderivative size = 380, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {6719, 468, 583, 531, 418, 492, 411} \[ \frac{c^{3/2} \sqrt{d} \left (a+b x^2\right ) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a^2 e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\left (a+b x^2\right ) (2 b c-a d)}{a^2 b e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{d x \left (a+b x^2\right ) (2 b c-a d)}{a^2 b e \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{c} \sqrt{d} \left (a+b x^2\right ) (2 b c-a d) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a^2 b e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{b c-a d}{a b e x \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x]

[Out]

(b*c - a*d)/(a*b*e*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - ((2*b*c - a*d)*(a + b*x^2))/(a^2*b*e*x*Sqrt[(e*(a +
b*x^2))/(c + d*x^2)]) + (d*(2*b*c - a*d)*x*(a + b*x^2))/(a^2*b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)
) - (Sqrt[c]*Sqrt[d]*(2*b*c - a*d)*(a + b*x^2)*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a^2*b
*e*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2)) + (c^(3/2)*Sqrt[d]*(a
+ b*x^2)*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a^2*e*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]
*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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

Mathematica [C]  time = 0.365372, size = 223, normalized size = 0.59 \[ \frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (-\sqrt{\frac{b}{a}} \left (c+d x^2\right ) \left (a c-a d x^2+2 b c x^2\right )-2 i c x \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )+i c x \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} (a d-2 b c) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )\right )}{a^2 e^2 x \sqrt{\frac{b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*((e*(a + b*x^2))/(c + d*x^2))^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.016, size = 654, normalized size = 1.7 \begin{align*}{\frac{b{x}^{2}+a}{ \left ( d{x}^{2}+c \right ) ^{2}{a}^{2}x} \left ( \sqrt{-{\frac{b}{a}}}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{4}a{d}^{2}-\sqrt{-{\frac{b}{a}}}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{4}bcd-\sqrt{-{\frac{b}{a}}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{4}bcd+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 ) }xacd-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 ) }xb{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 ) }xacd+2\,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 ) \sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }x+\sqrt{-{\frac{b}{a}}}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}acd-\sqrt{-{\frac{b}{a}}}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{2}b{c}^{2}-\sqrt{-{\frac{b}{a}}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}acd-\sqrt{-{\frac{b}{a}}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}b{c}^{2}-\sqrt{-{\frac{b}{a}}}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }a{c}^{2} \right ) \left ({\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}} \right ) ^{-{\frac{3}{2}}}{\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(1/x^2/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{b^{2} e^{2} x^{6} + 2 \, a b e^{2} x^{4} + a^{2} e^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*e^2*x^6 + 2*a*b*e^2*x^4 + a^2*e^2*
x^2), 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(1/x**2/(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 \frac{1}{\left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

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