3.895 \(\int \frac{1}{(e x)^{5/2} (a-b x^2) (c-d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=397 \[ \frac{d^{3/4} \sqrt{1-\frac{d x^2}{c}} (2 b c-5 a d) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),-1\right )}{3 a c^{7/4} e^{5/2} \sqrt{c-d x^2} (b c-a d)}+\frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (-\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt{c-d x^2} (b c-a d)}+\frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt{c-d x^2} (b c-a d)}-\frac{\sqrt{c-d x^2} (2 b c-5 a d)}{3 a c^2 e (e x)^{3/2} (b c-a d)}-\frac{d}{c e (e x)^{3/2} \sqrt{c-d x^2} (b c-a d)} \]

[Out]

-(d/(c*(b*c - a*d)*e*(e*x)^(3/2)*Sqrt[c - d*x^2])) - ((2*b*c - 5*a*d)*Sqrt[c - d*x^2])/(3*a*c^2*(b*c - a*d)*e*
(e*x)^(3/2)) + (d^(3/4)*(2*b*c - 5*a*d)*Sqrt[1 - (d*x^2)/c]*EllipticF[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt
[e])], -1])/(3*a*c^(7/4)*(b*c - a*d)*e^(5/2)*Sqrt[c - d*x^2]) + (b^2*c^(1/4)*Sqrt[1 - (d*x^2)/c]*EllipticPi[-(
(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d])), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(a^2*d^(1/4)*(b*c -
a*d)*e^(5/2)*Sqrt[c - d*x^2]) + (b^2*c^(1/4)*Sqrt[1 - (d*x^2)/c]*EllipticPi[(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d]
), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(a^2*d^(1/4)*(b*c - a*d)*e^(5/2)*Sqrt[c - d*x^2])

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Rubi [A]  time = 0.789251, antiderivative size = 397, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {466, 472, 583, 523, 224, 221, 409, 1219, 1218} \[ \frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (-\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt{c-d x^2} (b c-a d)}+\frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} e^{5/2} \sqrt{c-d x^2} (b c-a d)}+\frac{d^{3/4} \sqrt{1-\frac{d x^2}{c}} (2 b c-5 a d) F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{3 a c^{7/4} e^{5/2} \sqrt{c-d x^2} (b c-a d)}-\frac{\sqrt{c-d x^2} (2 b c-5 a d)}{3 a c^2 e (e x)^{3/2} (b c-a d)}-\frac{d}{c e (e x)^{3/2} \sqrt{c-d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*x)^(5/2)*(a - b*x^2)*(c - d*x^2)^(3/2)),x]

[Out]

-(d/(c*(b*c - a*d)*e*(e*x)^(3/2)*Sqrt[c - d*x^2])) - ((2*b*c - 5*a*d)*Sqrt[c - d*x^2])/(3*a*c^2*(b*c - a*d)*e*
(e*x)^(3/2)) + (d^(3/4)*(2*b*c - 5*a*d)*Sqrt[1 - (d*x^2)/c]*EllipticF[ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt
[e])], -1])/(3*a*c^(7/4)*(b*c - a*d)*e^(5/2)*Sqrt[c - d*x^2]) + (b^2*c^(1/4)*Sqrt[1 - (d*x^2)/c]*EllipticPi[-(
(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d])), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(a^2*d^(1/4)*(b*c -
a*d)*e^(5/2)*Sqrt[c - d*x^2]) + (b^2*c^(1/4)*Sqrt[1 - (d*x^2)/c]*EllipticPi[(Sqrt[b]*Sqrt[c])/(Sqrt[a]*Sqrt[d]
), ArcSin[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], -1])/(a^2*d^(1/4)*(b*c - a*d)*e^(5/2)*Sqrt[c - d*x^2])

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -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 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (b*x^4)/a]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + (b*x^4)
/a], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 409

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-(d/c), 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-(d/c), 2]*x^2)), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1219

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4]
, Int[1/((d + e*x^2)*Sqrt[1 + (c*x^4)/a]), x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] &&  !GtQ[a, 0]

Rule 1218

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

Rubi steps

\begin{align*} \int \frac{1}{(e x)^{5/2} \left (a-b x^2\right ) \left (c-d x^2\right )^{3/2}} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^4 \left (a-\frac{b x^4}{e^2}\right ) \left (c-\frac{d x^4}{e^2}\right )^{3/2}} \, dx,x,\sqrt{e x}\right )}{e}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{e \operatorname{Subst}\left (\int \frac{-\frac{2 b c-5 a d}{e^2}-\frac{5 b d x^4}{e^4}}{x^4 \left (a-\frac{b x^4}{e^2}\right ) \sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{c (b c-a d)}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{(2 b c-5 a d) \sqrt{c-d x^2}}{3 a c^2 (b c-a d) e (e x)^{3/2}}+\frac{e \operatorname{Subst}\left (\int \frac{\frac{6 b^2 c^2+2 a b c d-5 a^2 d^2}{e^4}-\frac{b d (2 b c-5 a d) x^4}{e^6}}{\left (a-\frac{b x^4}{e^2}\right ) \sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 a c^2 (b c-a d)}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{(2 b c-5 a d) \sqrt{c-d x^2}}{3 a c^2 (b c-a d) e (e x)^{3/2}}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a-\frac{b x^4}{e^2}\right ) \sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{a (b c-a d) e^3}+\frac{(d (2 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 a c^2 (b c-a d) e^3}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{(2 b c-5 a d) \sqrt{c-d x^2}}{3 a c^2 (b c-a d) e (e x)^{3/2}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{a} e}\right ) \sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{a^2 (b c-a d) e^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{a} e}\right ) \sqrt{c-\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{a^2 (b c-a d) e^3}+\frac{\left (d (2 b c-5 a d) \sqrt{1-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{d x^4}{c e^2}}} \, dx,x,\sqrt{e x}\right )}{3 a c^2 (b c-a d) e^3 \sqrt{c-d x^2}}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{(2 b c-5 a d) \sqrt{c-d x^2}}{3 a c^2 (b c-a d) e (e x)^{3/2}}+\frac{d^{3/4} (2 b c-5 a d) \sqrt{1-\frac{d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{3 a c^{7/4} (b c-a d) e^{5/2} \sqrt{c-d x^2}}+\frac{\left (b^2 \sqrt{1-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{\sqrt{b} x^2}{\sqrt{a} e}\right ) \sqrt{1-\frac{d x^4}{c e^2}}} \, dx,x,\sqrt{e x}\right )}{a^2 (b c-a d) e^3 \sqrt{c-d x^2}}+\frac{\left (b^2 \sqrt{1-\frac{d x^2}{c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{\sqrt{b} x^2}{\sqrt{a} e}\right ) \sqrt{1-\frac{d x^4}{c e^2}}} \, dx,x,\sqrt{e x}\right )}{a^2 (b c-a d) e^3 \sqrt{c-d x^2}}\\ &=-\frac{d}{c (b c-a d) e (e x)^{3/2} \sqrt{c-d x^2}}-\frac{(2 b c-5 a d) \sqrt{c-d x^2}}{3 a c^2 (b c-a d) e (e x)^{3/2}}+\frac{d^{3/4} (2 b c-5 a d) \sqrt{1-\frac{d x^2}{c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{3 a c^{7/4} (b c-a d) e^{5/2} \sqrt{c-d x^2}}+\frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (-\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} (b c-a d) e^{5/2} \sqrt{c-d x^2}}+\frac{b^2 \sqrt [4]{c} \sqrt{1-\frac{d x^2}{c}} \Pi \left (\frac{\sqrt{b} \sqrt{c}}{\sqrt{a} \sqrt{d}};\left .\sin ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )\right |-1\right )}{a^2 \sqrt [4]{d} (b c-a d) e^{5/2} \sqrt{c-d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.247626, size = 197, normalized size = 0.5 \[ \frac{x \left (5 x^2 \sqrt{1-\frac{d x^2}{c}} \left (-5 a^2 d^2+2 a b c d+6 b^2 c^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\frac{d x^2}{c},\frac{b x^2}{a}\right )+b d x^4 \sqrt{1-\frac{d x^2}{c}} (5 a d-2 b c) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\frac{d x^2}{c},\frac{b x^2}{a}\right )+5 a \left (a d \left (2 c-5 d x^2\right )-2 b c \left (c-d x^2\right )\right )\right )}{15 a^2 c^2 (e x)^{5/2} \sqrt{c-d x^2} (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((e*x)^(5/2)*(a - b*x^2)*(c - d*x^2)^(3/2)),x]

[Out]

(x*(5*a*(a*d*(2*c - 5*d*x^2) - 2*b*c*(c - d*x^2)) + 5*(6*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*x^2*Sqrt[1 - (d*x^2)
/c]*AppellF1[1/4, 1/2, 1, 5/4, (d*x^2)/c, (b*x^2)/a] + b*d*(-2*b*c + 5*a*d)*x^4*Sqrt[1 - (d*x^2)/c]*AppellF1[5
/4, 1/2, 1, 9/4, (d*x^2)/c, (b*x^2)/a]))/(15*a^2*c^2*(b*c - a*d)*(e*x)^(5/2)*Sqrt[c - d*x^2])

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Maple [B]  time = 0.033, size = 896, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*d*(3*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*d),1/2*2
^(1/2))*2^(1/2)*x*b^3*c^3*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-x*d/(
c*d)^(1/2))^(1/2)+3*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((c*d)^(1/2)*b-(a*b)^(1/2)*
d),1/2*2^(1/2))*2^(1/2)*x*b^2*c^2*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)
*(-x*d/(c*d)^(1/2))^(1/2)*(a*b)^(1/2)*(c*d)^(1/2)-5*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2
))*2^(1/2)*x*a^2*d^2*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^
(1/2))^(1/2)*(a*b)^(1/2)*(c*d)^(1/2)+7*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x*
a*b*c*d*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)*
(a*b)^(1/2)*(c*d)^(1/2)-2*EllipticF(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),1/2*2^(1/2))*2^(1/2)*x*b^2*c^2*((d*x
+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-x*d/(c*d)^(1/2))^(1/2)*(a*b)^(1/2)*(
c*d)^(1/2)-3*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((a*b)^(1/2)*d+(c*d)^(1/2)*b),1/2*
2^(1/2))*2^(1/2)*x*b^3*c^3*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*(-x*d/
(c*d)^(1/2))^(1/2)+3*EllipticPi(((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2),(c*d)^(1/2)*b/((a*b)^(1/2)*d+(c*d)^(1/2)
*b),1/2*2^(1/2))*2^(1/2)*x*b^2*c^2*((d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2)*((-d*x+(c*d)^(1/2))/(c*d)^(1/2))^(1/2
)*(-x*d/(c*d)^(1/2))^(1/2)*(a*b)^(1/2)*(c*d)^(1/2)-10*x^2*a^2*d^3*(a*b)^(1/2)+14*x^2*a*b*c*d^2*(a*b)^(1/2)-4*x
^2*b^2*c^2*d*(a*b)^(1/2)+4*a^2*c*d^2*(a*b)^(1/2)-8*a*b*c^2*d*(a*b)^(1/2)+4*b^2*c^3*(a*b)^(1/2))*(-d*x^2+c)^(1/
2)/x/c^2/a/((c*d)^(1/2)*b-(a*b)^(1/2)*d)/((a*b)^(1/2)*d+(c*d)^(1/2)*b)/(a*b)^(1/2)/(a*d-b*c)/(d*x^2-c)/e^2/(e*
x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (b x^{2} - a\right )}{\left (-d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(5/2)/(-b*x^2+a)/(-d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(5/2)/(-b*x^2+a)/(-d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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/(e*x)**(5/2)/(-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 (b x^{2} - a\right )}{\left (-d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(5/2)/(-b*x^2+a)/(-d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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