3.313 \(\int \frac{x^4}{(\frac{e (a+b x^2)}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=453 \[ \frac{x \left (a+b x^2\right ) \left (16 a^2 d^2-16 a b c d+b^2 c^2\right )}{5 b^4 e \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{c} \left (a+b x^2\right ) \left (16 a^2 d^2-16 a b c d+b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{5 b^4 \sqrt{d} e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{c^{3/2} \left (a+b x^2\right ) (7 b c-8 a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{5 b^3 \sqrt{d} e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{x \left (a+b x^2\right ) (7 b c-8 a d)}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

((7*b*c - 8*a*d)*x*(a + b*x^2))/(5*b^3*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (6*d*x^3*(a + b*x^2))/(5*b^2*e*S
qrt[(e*(a + b*x^2))/(c + d*x^2)]) + ((b^2*c^2 - 16*a*b*c*d + 16*a^2*d^2)*x*(a + b*x^2))/(5*b^4*e*Sqrt[(e*(a +
b*x^2))/(c + d*x^2)]*(c + d*x^2)) - (x^3*(c + d*x^2))/(b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - (Sqrt[c]*(b^2*
c^2 - 16*a*b*c*d + 16*a^2*d^2)*(a + b*x^2)*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*b^4*Sqr
t[d]*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)*(7*b*c
- 8*a*d)*(a + b*x^2)*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*b^3*Sqrt[d]*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))

________________________________________________________________________________________

Rubi [A]  time = 0.675178, antiderivative size = 453, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6719, 467, 581, 582, 531, 418, 492, 411} \[ \frac{x \left (a+b x^2\right ) \left (16 a^2 d^2-16 a b c d+b^2 c^2\right )}{5 b^4 e \left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{c} \left (a+b x^2\right ) \left (16 a^2 d^2-16 a b c d+b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{5 b^4 \sqrt{d} e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{c^{3/2} \left (a+b x^2\right ) (7 b c-8 a d) F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{5 b^3 \sqrt{d} e \left (c+d x^2\right ) \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{x \left (a+b x^2\right ) (7 b c-8 a d)}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((e*(a + b*x^2))/(c + d*x^2))^(3/2),x]

[Out]

((7*b*c - 8*a*d)*x*(a + b*x^2))/(5*b^3*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (6*d*x^3*(a + b*x^2))/(5*b^2*e*S
qrt[(e*(a + b*x^2))/(c + d*x^2)]) + ((b^2*c^2 - 16*a*b*c*d + 16*a^2*d^2)*x*(a + b*x^2))/(5*b^4*e*Sqrt[(e*(a +
b*x^2))/(c + d*x^2)]*(c + d*x^2)) - (x^3*(c + d*x^2))/(b*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) - (Sqrt[c]*(b^2*
c^2 - 16*a*b*c*d + 16*a^2*d^2)*(a + b*x^2)*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*b^4*Sqr
t[d]*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)*(7*b*c
- 8*a*d)*(a + b*x^2)*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(5*b^3*Sqrt[d]*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 467

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*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 581

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

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
 x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 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{x^4}{\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{x^4 \left (c+d x^2\right )^{3/2}}{\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{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\sqrt{a+b x^2} \int \frac{x^2 \sqrt{c+d x^2} \left (3 c+6 d x^2\right )}{\sqrt{a+b x^2}} \, dx}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}\\ &=\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\sqrt{a+b x^2} \int \frac{x^2 \left (3 c (5 b c-6 a d)+3 d (7 b c-8 a d) x^2\right )}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}\\ &=\frac{(7 b c-8 a d) x \left (a+b x^2\right )}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{a+b x^2} \int \frac{3 a c d (7 b c-8 a d)-3 d \left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{15 b^3 d e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}\\ &=\frac{(7 b c-8 a d) x \left (a+b x^2\right )}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\left (a c (7 b c-8 a d) \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}+\frac{\left (\left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) \sqrt{a+b x^2}\right ) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}\\ &=\frac{(7 b c-8 a d) x \left (a+b x^2\right )}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) x \left (a+b x^2\right )}{5 b^4 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{c^{3/2} (7 b c-8 a 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 )}{5 b^3 \sqrt{d} 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 \left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) \sqrt{a+b x^2}\right ) \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{5 b^4 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \sqrt{c+d x^2}}\\ &=\frac{(7 b c-8 a d) x \left (a+b x^2\right )}{5 b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{6 d x^3 \left (a+b x^2\right )}{5 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{\left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) x \left (a+b x^2\right )}{5 b^4 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}-\frac{x^3 \left (c+d x^2\right )}{b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}-\frac{\sqrt{c} \left (b^2 c^2-16 a b c d+16 a^2 d^2\right ) \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 )}{5 b^4 \sqrt{d} 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} (7 b c-8 a 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 )}{5 b^3 \sqrt{d} 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 = 0.504485, size = 271, normalized size = 0.6 \[ \frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (8 a^2 d^2-9 a b c d+b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )-i c \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (16 a^2 d^2-16 a b c d+b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )+d x \sqrt{\frac{b}{a}} \left (c+d x^2\right ) \left (-8 a^2 d+a b \left (7 c-2 d x^2\right )+b^2 x^2 \left (2 c+d x^2\right )\right )\right )}{5 b^3 d e^2 \sqrt{\frac{b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((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]*d*x*(c + d*x^2)*(-8*a^2*d + a*b*(7*c - 2*d*x^2) + b^2*x^2*(2*c +
 d*x^2)) - I*c*(b^2*c^2 - 16*a*b*c*d + 16*a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh
[Sqrt[b/a]*x], (a*d)/(b*c)] + I*c*(b^2*c^2 - 9*a*b*c*d + 8*a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*El
lipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(5*b^3*Sqrt[b/a]*d*e^2*(a + b*x^2))

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Maple [A]  time = 0.035, size = 935, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x)

[Out]

1/5*(b*x^2+a)*(((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)*x^7*b^2*d^3-2*((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)
*x^5*a*b*d^3+3*((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)*x^5*b^2*c*d^2-3*((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/
2)*x^3*a^2*d^3+2*((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)*x^3*b^2*c^2*d-5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(-b/a)^(1/2)*x^3*a^2*d^3+5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x^3*a*b*c*d^2-8*((d*x^2+c)*(b*x^2+
a))^(1/2)*((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))*a^2*c*d^2+9*((d*x^
2+c)*(b*x^2+a))^(1/2)*((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))*a*b*c^
2*d-((d*x^2+c)*(b*x^2+a))^(1/2)*((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^2*c^3+16*((d*x^2+c)*(b*x^2+a))^(1/2)*((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^2*c*d^2-16*((d*x^2+c)*(b*x^2+a))^(1/2)*((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*b*c^2*d+((d*x^2+c)*(b*x^2+a))^(1/2)*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*Elli
pticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))*b^2*c^3-3*((d*x^2+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)*x*a^2*c*d^2+2*((d*x^2
+c)*(b*x^2+a))^(1/2)*(-b/a)^(1/2)*x*a*b*c^2*d-5*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x*a^2*c*d^2+5
*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(-b/a)^(1/2)*x*a*b*c^2*d)/b^3/d/(e*(b*x^2+a)/(d*x^2+c))^(3/2)/(d*x^2+c)^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 \frac{x^{4}}{\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(x^4/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/((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 (d^{2} x^{8} + 2 \, c d x^{6} + c^{2} x^{4}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{b^{2} e^{2} x^{4} + 2 \, a b e^{2} x^{2} + a^{2} e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="fricas")

[Out]

integral((d^2*x^8 + 2*c*d*x^6 + c^2*x^4)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*e^2*x^4 + 2*a*b*e^2*x^2 + a^2*
e^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(x**4/(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{x^{4}}{\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(x^4/(e*(b*x^2+a)/(d*x^2+c))^(3/2),x, algorithm="giac")

[Out]

integrate(x^4/((b*x^2 + a)*e/(d*x^2 + c))^(3/2), x)