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]

((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))

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Rubi [A]  time = 0.522041, antiderivative size = 403, 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, 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 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 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 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}{\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*}

Mathematica [C]  time = 0.4799, 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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Maple [A]  time = 0.018, size = 553, normalized size = 1.4 \begin{align*}{\frac{b{x}^{2}+a}{15\,{b}^{2}{d}^{2}} \left ( 3\,\sqrt{-{\frac{b}{a}}}{x}^{7}{b}^{2}{d}^{3}-\sqrt{-{\frac{b}{a}}}{x}^{5}ab{d}^{3}+4\,\sqrt{-{\frac{b}{a}}}{x}^{5}{b}^{2}c{d}^{2}-4\,\sqrt{-{\frac{b}{a}}}{x}^{3}{a}^{2}{d}^{3}+\sqrt{-{\frac{b}{a}}}{x}^{3}{b}^{2}{c}^{2}d-4\,\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 ){a}^{2}c{d}^{2}+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 ) ab{c}^{2}d+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 ){b}^{2}{c}^{3}+8\,\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 ){a}^{2}c{d}^{2}-3\,\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 ) ab{c}^{2}d-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 ){b}^{2}{c}^{3}-4\,\sqrt{-{\frac{b}{a}}}x{a}^{2}c{d}^{2}+\sqrt{-{\frac{b}{a}}}xab{c}^{2}d \right ){\frac{1}{\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \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(x^4/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x)

[Out]

1/15*(b*x^2+a)*(3*(-b/a)^(1/2)*x^7*b^2*d^3-(-b/a)^(1/2)*x^5*a*b*d^3+4*(-b/a)^(1/2)*x^5*b^2*c*d^2-4*(-b/a)^(1/2
)*x^3*a^2*d^3+(-b/a)^(1/2)*x^3*b^2*c^2*d-4*((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+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+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+8*((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^2-3*((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-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))*b^2*c^3-4*(-b/a)^(1/2)*x*a^2*c*d^2+(-b/a)^(1/2)*x*a*b*c^2*d)/b^2/(e
*(b*x^2+a)/(d*x^2+c))^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^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}}{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \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))^(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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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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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