3.302 \(\int \frac {x^4}{\sqrt {\frac {e (a+b x^2)}{c+d x^2}}} \, dx\)

Optimal. Leaf size=403 \[ -\frac {x \left (a+b x^2\right ) \left (-8 a^2 d^2+3 a b c d+2 b^2 c^2\right )}{15 b^3 d \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {\sqrt {c} \left (a+b x^2\right ) \left (-8 a^2 d^2+3 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^3 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {c^{3/2} \left (a+b x^2\right ) (b c-4 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x \left (a+b x^2\right ) (b c-4 a d)}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

1/15*(-4*a*d+b*c)*x*(b*x^2+a)/b^2/d/(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/5*x^3*(b*x^2+a)/b/(e*(b*x^2+a)/(d*x^2+c))^
(1/2)-1/15*(-8*a^2*d^2+3*a*b*c*d+2*b^2*c^2)*x*(b*x^2+a)/b^3/d/(d*x^2+c)/(e*(b*x^2+a)/(d*x^2+c))^(1/2)-1/15*c^(
3/2)*(-4*a*d+b*c)*(b*x^2+a)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticF(x*d^(1/2)/c^(1/2)/(1+d*x^2/c)^(1
/2),(1-b*c/a/d)^(1/2))/b^2/d^(3/2)/(d*x^2+c)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(e*(b*x^2+a)/(d*x^2+c))^(1/2)+1/1
5*(-8*a^2*d^2+3*a*b*c*d+2*b^2*c^2)*(b*x^2+a)*(1/(1+d*x^2/c))^(1/2)*(1+d*x^2/c)^(1/2)*EllipticE(x*d^(1/2)/c^(1/
2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))*c^(1/2)/b^3/d^(3/2)/(d*x^2+c)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(e*(b*x^
2+a)/(d*x^2+c))^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 403, normalized size of antiderivative = 1.00, 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, 478, 582, 531, 418, 492, 411} \[ -\frac {x \left (a+b x^2\right ) \left (-8 a^2 d^2+3 a b c d+2 b^2 c^2\right )}{15 b^3 d \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {\sqrt {c} \left (a+b x^2\right ) \left (-8 a^2 d^2+3 a b c d+2 b^2 c^2\right ) E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^3 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {c^{3/2} \left (a+b x^2\right ) (b c-4 a d) F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{3/2} \left (c+d x^2\right ) \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x \left (a+b x^2\right ) (b c-4 a d)}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 4*a*d)*x*(a + b*x^2))/(15*b^2*d*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (x^3*(a + b*x^2))/(5*b*Sqrt[(e*(a
+ b*x^2))/(c + d*x^2)]) - ((2*b^2*c^2 + 3*a*b*c*d - 8*a^2*d^2)*x*(a + b*x^2))/(15*b^3*d*Sqrt[(e*(a + b*x^2))/
(c + d*x^2)]*(c + d*x^2)) + (Sqrt[c]*(2*b^2*c^2 + 3*a*b*c*d - 8*a^2*d^2)*(a + b*x^2)*EllipticE[ArcTan[(Sqrt[d]
*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b^3*d^(3/2)*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)*(b*c - 4*a*d)*(a + b*x^2)*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(
a*d)])/(15*b^2*d^(3/2)*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 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]

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 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 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 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 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 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]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \, dx &=\frac {\sqrt {a+b x^2} \int \frac {x^4 \sqrt {c+d x^2}}{\sqrt {a+b x^2}} \, dx}{\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}\\ &=\frac {x^3 \left (a+b x^2\right )}{5 b \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 a c+(-b c+4 a d) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}\\ &=\frac {(b c-4 a d) x \left (a+b x^2\right )}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {\sqrt {a+b x^2} \int \frac {-a c (b c-4 a d)+\left (-2 b^2 c^2-3 a b c d+8 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}\\ &=\frac {(b c-4 a d) x \left (a+b x^2\right )}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {\left (a c (b c-4 a d) \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}+\frac {\left (\left (-2 b^2 c^2-3 a b c d+8 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}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}\\ &=\frac {(b c-4 a d) x \left (a+b x^2\right )}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {\left (2 b^2 c^2+3 a b c d-8 a^2 d^2\right ) x \left (a+b x^2\right )}{15 b^3 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}-\frac {c^{3/2} (b c-4 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 )}{15 b^2 d^{3/2} \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 (-2 b^2 c^2-3 a b c d+8 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}{15 b^3 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}}\\ &=\frac {(b c-4 a d) x \left (a+b x^2\right )}{15 b^2 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}+\frac {x^3 \left (a+b x^2\right )}{5 b \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}-\frac {\left (2 b^2 c^2+3 a b c d-8 a^2 d^2\right ) x \left (a+b x^2\right )}{15 b^3 d \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}+\frac {\sqrt {c} \left (2 b^2 c^2+3 a b c d-8 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 )}{15 b^3 d^{3/2} \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} (b c-4 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 )}{15 b^2 d^{3/2} \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*}

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Mathematica [C]  time = 0.48, size = 258, normalized size = 0.64 \[ \frac {2 i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (2 a^2 d^2-a b c d-b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (8 a^2 d^2-3 a b c d-2 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+d x \left (-\sqrt {\frac {b}{a}}\right ) \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 a d-b \left (c+3 d x^2\right )\right )}{15 a^2 d^2 \left (\frac {b}{a}\right )^{5/2} \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(4*a*d - b*(c + 3*d*x^2))) - I*c*(-2*b^2*c^2 - 3*a*b*c*d + 8*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)] + (2*I)*c*(-(b^2*c^2)
- a*b*c*d + 2*a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])
/(15*a^2*(b/a)^(5/2)*d^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x^{6} + c x^{4}\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{b e x^{2} + a e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

integral((d*x^6 + c*x^4)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b*e*x^2 + a*e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 553, normalized size = 1.37 \[ \frac {\left (b \,x^{2}+a \right ) \left (3 \sqrt {-\frac {b}{a}}\, b^{2} d^{3} x^{7}-\sqrt {-\frac {b}{a}}\, a b \,d^{3} x^{5}+4 \sqrt {-\frac {b}{a}}\, b^{2} c \,d^{2} x^{5}-4 \sqrt {-\frac {b}{a}}\, a^{2} d^{3} x^{3}+\sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}-4 \sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x +8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-4 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {-\frac {b}{a}}\, a b \,c^{2} d x -3 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{15 \sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((b*x^2+a)/(d*x^2+c)*e)^(1/2),x)

[Out]

1/15*(b*x^2+a)*(3*(-1/a*b)^(1/2)*b^2*d^3*x^7-(-1/a*b)^(1/2)*a*b*d^3*x^5+4*(-1/a*b)^(1/2)*b^2*c*d^2*x^5-4*(-1/a
*b)^(1/2)*a^2*d^3*x^3+(-1/a*b)^(1/2)*b^2*c^2*d*x^3-4*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b
)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*c*d^2+2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b
/c*d)^(1/2))*a*b*c^2*d+2*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b
^2*c^3+8*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a^2*c*d^2-3*((b*x
^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*a*b*c^2*d-2*((b*x^2+a)/a)^(1/2)
*((d*x^2+c)/c)^(1/2)*EllipticE((-1/a*b)^(1/2)*x,(a/b/c*d)^(1/2))*b^2*c^3-4*(-1/a*b)^(1/2)*a^2*c*d^2*x+(-1/a*b)
^(1/2)*a*b*c^2*d*x)/b^2/((b*x^2+a)/(d*x^2+c)*e)^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^2/(-1/a*b)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{\sqrt {\frac {e\,\left (b\,x^2+a\right )}{d\,x^2+c}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((e*(a + b*x^2))/(c + d*x^2))^(1/2),x)

[Out]

int(x^4/((e*(a + b*x^2))/(c + d*x^2))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(e*(b*x**2+a)/(d*x**2+c))**(1/2),x)

[Out]

Timed out

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