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

Optimal. Leaf size=368 \[ -\frac{x \left (-3 a^2 c^2+7 a b c+2 b^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{15 a^2 d^2}+\frac{\sqrt{c} \left (-3 a^2 c^2+7 a b c+2 b^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a^2 d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac{c^{3/2} (b-3 a c) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac{x (b-3 a c) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{15 a d^2}+\frac{x^3 \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{5 d} \]

[Out]

-((2*b^2 + 7*a*b*c - 3*a^2*c^2)*x*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(15*a^2*d^2) + ((b - 3*a*c)*x*(c + d*
x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(15*a*d^2) + (x^3*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)
])/(5*d) + (Sqrt[c]*(2*b^2 + 7*a*b*c - 3*a^2*c^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt
[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*a^2*d^(5/2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]) - (c^(3/
2)*(b - 3*a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*
a*d^(5/2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))])

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Rubi [A]  time = 0.718072, antiderivative size = 478, normalized size of antiderivative = 1.3, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {6722, 1975, 478, 582, 531, 418, 492, 411} \[ -\frac{x \left (-3 a^2 c^2+7 a b c+2 b^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{15 a^2 d^2 \sqrt{a \left (c+d x^2\right )+b}}+\frac{\sqrt{c} \left (-3 a^2 c^2+7 a b c+2 b^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a^2 d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}-\frac{c^{3/2} (b-3 a c) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}+\frac{x (b-3 a c) \left (c+d x^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{15 a d^2 \sqrt{a \left (c+d x^2\right )+b}}+\frac{x^3 \left (c+d x^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a + b/(c + d*x^2)],x]

[Out]

-((2*b^2 + 7*a*b*c - 3*a^2*c^2)*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(15*a^2*d^2*Sqrt[b + a*(c +
d*x^2)]) + ((b - 3*a*c)*x*(c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(15*a*d^2*Sqrt[b + a*(
c + d*x^2)]) + (x^3*(c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(5*d*Sqrt[b + a*(c + d*x^2)])
+ (Sqrt[c]*(2*b^2 + 7*a*b*c - 3*a^2*c^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticE[ArcTan[(Sq
rt[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*a^2*d^(5/2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b
+ a*(c + d*x^2)]) - (c^(3/2)*(b - 3*a*c)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticF[ArcTan[(Sqr
t[d]*x)/Sqrt[c]], b/(b + a*c)])/(15*a*d^(5/2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]*Sqrt[b + a
*(c + d*x^2)])

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, 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 x^4 \sqrt{a+\frac{b}{c+d x^2}} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^4 \sqrt{b+a \left (c+d x^2\right )}}{\sqrt{c+d x^2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^4 \sqrt{b+a c+a d x^2}}{\sqrt{c+d x^2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{x^3 \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^2 \left (3 c (b+a c)-(b-3 a c) d x^2\right )}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{5 d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{(b-3 a c) x \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{x^3 \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{-c (b-3 a c) (b+a c) d-\left (2 b^2+7 a b c-3 a^2 c^2\right ) d^2 x^2}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{15 a d^3 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{(b-3 a c) x \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{x^3 \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left (c (b-3 a c) (b+a c) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{1}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{15 a d^2 \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left (\left (2 b^2+7 a b c-3 a^2 c^2\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^2}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{15 a d \sqrt{b+a \left (c+d x^2\right )}}\\ &=-\frac{\left (2 b^2+7 a b c-3 a^2 c^2\right ) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a^2 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{(b-3 a c) x \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{x^3 \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{b+a \left (c+d x^2\right )}}-\frac{c^{3/2} (b-3 a c) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a d^{5/2} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (c \left (2 b^2+7 a b c-3 a^2 c^2\right ) \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{\sqrt{b+a c+a d x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{15 a^2 d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=-\frac{\left (2 b^2+7 a b c-3 a^2 c^2\right ) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a^2 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{(b-3 a c) x \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{15 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{x^3 \left (c+d x^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{b+a \left (c+d x^2\right )}}+\frac{\sqrt{c} \left (2 b^2+7 a b c-3 a^2 c^2\right ) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a^2 d^{5/2} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}-\frac{c^{3/2} (b-3 a c) \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{15 a d^{5/2} \sqrt{\frac{c \left (b+a c+a d x^2\right )}{(b+a c) \left (c+d x^2\right )}} \sqrt{b+a \left (c+d x^2\right )}}\\ \end{align*}

Mathematica [C]  time = 0.821535, size = 293, normalized size = 0.8 \[ \frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (x \left (c+d x^2\right ) \sqrt{\frac{a d}{a c+b}} \left (-3 a^2 \left (c^2-d^2 x^4\right )-2 a b \left (c-2 d x^2\right )+b^2\right )+i c \left (-3 a^2 c^2+7 a b c+2 b^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )-i b c (9 a c+b) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )\right )}{15 a d^2 \sqrt{\frac{a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*x*(c + d*x^2)*(b^2 - 2*a*b*(c - 2*d*x^2) - 3*a^2
*(c^2 - d^2*x^4)) + I*c*(2*b^2 + 7*a*b*c - 3*a^2*c^2)*Sqrt[(b + a*c + a*d*x^2)/(b + a*c)]*Sqrt[1 + (d*x^2)/c]*
EllipticE[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)] - I*b*c*(b + 9*a*c)*Sqrt[(b + a*c + a*d*x^2)/(b + a
*c)]*Sqrt[1 + (d*x^2)/c]*EllipticF[I*ArcSinh[Sqrt[(a*d)/(b + a*c)]*x], 1 + b/(a*c)]))/(15*a*d^2*Sqrt[(a*d)/(b
+ a*c)]*(b + a*(c + d*x^2)))

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Maple [A]  time = 0.034, size = 662, normalized size = 1.8 \begin{align*}{\frac{d{x}^{2}+c}{15\,a{d}^{2}} \left ( 3\,\sqrt{-{\frac{ad}{ac+b}}}{x}^{7}{a}^{2}{d}^{3}+3\,\sqrt{-{\frac{ad}{ac+b}}}{x}^{5}{a}^{2}c{d}^{2}+4\,\sqrt{-{\frac{ad}{ac+b}}}{x}^{5}ab{d}^{2}-3\,\sqrt{-{\frac{ad}{ac+b}}}{x}^{3}{a}^{2}{c}^{2}d+2\,\sqrt{-{\frac{ad}{ac+b}}}{x}^{3}abcd+3\,\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ){a}^{2}{c}^{3}+\sqrt{-{\frac{ad}{ac+b}}}{x}^{3}{b}^{2}d-3\,\sqrt{-{\frac{ad}{ac+b}}}x{a}^{2}{c}^{3}+9\,\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ) ab{c}^{2}-7\,\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ) ab{c}^{2}-2\,\sqrt{-{\frac{ad}{ac+b}}}xab{c}^{2}+\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ){b}^{2}c-2\,\sqrt{{\frac{ad{x}^{2}+ac+b}{ac+b}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}{\it EllipticE} \left ( x\sqrt{-{\frac{ad}{ac+b}}},\sqrt{{\frac{ac+b}{ac}}} \right ){b}^{2}c+\sqrt{-{\frac{ad}{ac+b}}}x{b}^{2}c \right ) \sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}}{\frac{1}{\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}}}{\frac{1}{\sqrt{-{\frac{ad}{ac+b}}}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*(-a*d/(a*c+b))^(1/2)*x^7*a^2*d^3+3*(-a*d/(a*c+b))^(1/2)*x^5*a^2*c*d^2+4*(-a*d/(a*c+b))^(1/2)*x^5*a*b*d
^2-3*(-a*d/(a*c+b))^(1/2)*x^3*a^2*c^2*d+2*(-a*d/(a*c+b))^(1/2)*x^3*a*b*c*d+3*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*(
(d*x^2+c)/c)^(1/2)*EllipticE(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*a^2*c^3+(-a*d/(a*c+b))^(1/2)*x^3*b^2*
d-3*(-a*d/(a*c+b))^(1/2)*x*a^2*c^3+9*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-a*d/(a*
c+b))^(1/2),((a*c+b)/a/c)^(1/2))*a*b*c^2-7*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-a
*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*a*b*c^2-2*(-a*d/(a*c+b))^(1/2)*x*a*b*c^2+((a*d*x^2+a*c+b)/(a*c+b))^(1/2
)*((d*x^2+c)/c)^(1/2)*EllipticF(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*b^2*c-2*((a*d*x^2+a*c+b)/(a*c+b))^
(1/2)*((d*x^2+c)/c)^(1/2)*EllipticE(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*b^2*c+(-a*d/(a*c+b))^(1/2)*x*b
^2*c)*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/d^2/(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)/(-a*d/(a
*c+b))^(1/2)/a/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + \frac{b}{d x^{2} + c}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{4} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \sqrt{\frac{a c + a d x^{2} + b}{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + \frac{b}{d x^{2} + c}} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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