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

Optimal. Leaf size=357 \[ \frac {x \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b^2 d^2}-\frac {\sqrt {c} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {c^{3/2} (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {x \left (c+d x^2\right ) (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b d^2}+\frac {x^3 \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 357, 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} \[ \frac {x \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b^2 d^2}-\frac {\sqrt {c} \left (-2 a^2 d^2-3 a b c d+8 b^2 c^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {c^{3/2} (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {x \left (c+d x^2\right ) (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b d^2}+\frac {x^3 \left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b^2*c^2 - 3*a*b*c*d - 2*a^2*d^2)*x*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(15*b^2*d^2) - ((4*b*c - a*d)*x*Sqrt
[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(15*b*d^2) + (x^3*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(5
*d) - (Sqrt[c]*(8*b^2*c^2 - 3*a*b*c*d - 2*a^2*d^2)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]
*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b^2*d^(5/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]) + (c^(3/2)*(4*b*c - a*d
)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b*d^(5/2)*Sqr
t[(c*(a + b*x^2))/(a*(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 x^4 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \, dx &=\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^4 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx}{\sqrt {a+b x^2}}\\ &=\frac {x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d}-\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2 \left (3 a c+(4 b c-a d) x^2\right )}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{5 d \sqrt {a+b x^2}}\\ &=-\frac {(4 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{15 b d^2}+\frac {x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d}+\frac {\left (\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {a c (4 b c-a d)+\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2 \sqrt {a+b x^2}}\\ &=-\frac {(4 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{15 b d^2}+\frac {x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d}+\frac {\left (a c (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2 \sqrt {a+b x^2}}+\frac {\left (\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {x^2}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx}{15 b d^2 \sqrt {a+b x^2}}\\ &=\frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b^2 d^2}-\frac {(4 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{15 b d^2}+\frac {x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d}+\frac {c^{3/2} (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac {\left (c \left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2}\right ) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 b^2 d^2 \sqrt {a+b x^2}}\\ &=\frac {\left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{15 b^2 d^2}-\frac {(4 b c-a d) x \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{15 b d^2}+\frac {x^3 \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{5 d}-\frac {\sqrt {c} \left (8 b^2 c^2-3 a b c d-2 a^2 d^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} E\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b^2 d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac {c^{3/2} (4 b c-a d) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} F\left (\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{5/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.48, size = 255, normalized size = 0.71 \[ \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \left (-i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (a^2 d^2+7 a b c d-8 b^2 c^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+i c \sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (2 a^2 d^2+3 a b c d-8 b^2 c^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+d x \sqrt {\frac {b}{a}} \left (a+b x^2\right ) \left (c+d x^2\right ) \left (a d-4 b c+3 b d x^2\right )\right )}{15 b d^3 \sqrt {\frac {b}{a}} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(Sqrt[b/a]*d*x*(a + b*x^2)*(c + d*x^2)*(-4*b*c + a*d + 3*b*d*x^2) + I*c*(-8
*b^2*c^2 + 3*a*b*c*d + 2*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*(-8*b^2*c^2 + 7*a*b*c*d + a^2*d^2)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSin
h[Sqrt[b/a]*x], (a*d)/(b*c)]))/(15*b*Sqrt[b/a]*d^3*(a + b*x^2))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}, x\right ) \]

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

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

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

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maple [A]  time = 0.05, size = 552, normalized size = 1.55 \[ \frac {\sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (3 \sqrt {-\frac {b}{a}}\, b^{2} d^{3} x^{7}+4 \sqrt {-\frac {b}{a}}\, a b \,d^{3} x^{5}-\sqrt {-\frac {b}{a}}\, b^{2} c \,d^{2} x^{5}+\sqrt {-\frac {b}{a}}\, a^{2} d^{3} x^{3}-4 \sqrt {-\frac {b}{a}}\, b^{2} c^{2} d \,x^{3}+\sqrt {-\frac {b}{a}}\, a^{2} c \,d^{2} x -2 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+\sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-4 \sqrt {-\frac {b}{a}}\, a b \,c^{2} d x -3 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+7 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, a b \,c^{2} d \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )+8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticE \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )-8 \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {d \,x^{2}+c}{c}}\, b^{2} c^{3} \EllipticF \left (\sqrt {-\frac {b}{a}}\, x , \sqrt {\frac {a d}{b c}}\right )\right )}{15 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, b \,d^{3}} \]

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

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

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

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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