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

Optimal. Leaf size=405 \[ \frac{x \left (a^2 c^2-14 a b c+b^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{5 a d^2}-\frac{\sqrt{c} \left (a^2 c^2-14 a b c+b^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{5 a d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}-\frac{c^{3/2} (7 b-a c) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{5 d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}}}+\frac{x (7 b-a c) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{5 d^2}+\frac{6 a x^3 \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{5 d}-\frac{x^3 \left (a c+a d x^2+b\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{d} \]

[Out]

((b^2 - 14*a*b*c + a^2*c^2)*x*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(5*a*d^2) + ((7*b - a*c)*x*(c + d*x^2)*Sq
rt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(5*d^2) + (6*a*x^3*(c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/(5*
d) - (x^3*(b + a*c + a*d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/d - (Sqrt[c]*(b^2 - 14*a*b*c + a^2*c^2)*S
qrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(5*a*d^(5/2)*Sqrt[(c
*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*x^2))]) - (c^(3/2)*(7*b - a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*E
llipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(5*d^(5/2)*Sqrt[(c*(b + a*c + a*d*x^2))/((b + a*c)*(c + d*
x^2))])

________________________________________________________________________________________

Rubi [A]  time = 0.876041, antiderivative size = 526, normalized size of antiderivative = 1.3, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6722, 1975, 467, 581, 582, 531, 418, 492, 411} \[ \frac{x \left (a^2 c^2-14 a b c+b^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{5 a d^2 \sqrt{a \left (c+d x^2\right )+b}}-\frac{\sqrt{c} \left (a^2 c^2-14 a b c+b^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{5 a d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}-\frac{c^{3/2} (7 b-a c) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|\frac{b}{b+a c}\right )}{5 d^{5/2} \sqrt{\frac{c \left (a c+a d x^2+b\right )}{(a c+b) \left (c+d x^2\right )}} \sqrt{a \left (c+d x^2\right )+b}}+\frac{x (7 b-a c) \left (c+d x^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{5 d^2 \sqrt{a \left (c+d x^2\right )+b}}-\frac{x^3 \left (a c+a d x^2+b\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{d \sqrt{a \left (c+d x^2\right )+b}}+\frac{6 a x^3 \left (c+d x^2\right ) \sqrt{a c+a d x^2+b} \sqrt{a+\frac{b}{c+d x^2}}}{5 d \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b/(c + d*x^2))^(3/2),x]

[Out]

((b^2 - 14*a*b*c + a^2*c^2)*x*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(5*a*d^2*Sqrt[b + a*(c + d*x^2)
]) + ((7*b - a*c)*x*(c + d*x^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)])/(5*d^2*Sqrt[b + a*(c + d*x^2)
]) + (6*a*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)]) - (x^
3*(b + a*c + a*d*x^2)^(3/2)*Sqrt[a + b/(c + d*x^2)])/(d*Sqrt[b + a*(c + d*x^2)]) - (Sqrt[c]*(b^2 - 14*a*b*c +
a^2*c^2)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/
(5*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)]) - (c^(3/2)*(7*b -
a*c)*Sqrt[b + a*c + a*d*x^2]*Sqrt[a + b/(c + d*x^2)]*EllipticF[ArcTan[(Sqrt[d]*x)/Sqrt[c]], b/(b + a*c)])/(5*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 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 x^4 \left (a+\frac{b}{c+d x^2}\right )^{3/2} \, dx &=\frac{\left (\sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}\right ) \int \frac{x^4 \left (b+a \left (c+d x^2\right )\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, 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 \left (b+a c+a d x^2\right )^{3/2}}{\left (c+d x^2\right )^{3/2}} \, dx}{\sqrt{b+a \left (c+d x^2\right )}}\\ &=-\frac{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{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 \sqrt{b+a c+a d x^2} \left (3 (b+a c)+6 a d x^2\right )}{\sqrt{c+d x^2}} \, dx}{d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{6 a 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{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{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 (5 b-a c) (b+a c) d+3 a (7 b-a c) d^2 x^2\right )}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{(7 b-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}}}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{6 a 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{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{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{3 a c (7 b-a c) (b+a c) d^2-3 a \left (b^2-14 a b c+a^2 c^2\right ) d^3 x^2}{\sqrt{c+d x^2} \sqrt{b+a c+a d x^2}} \, dx}{15 a d^4 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{(7 b-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}}}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{6 a 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{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{d \sqrt{b+a \left (c+d x^2\right )}}-\frac{\left (c (7 b-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}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{\left (\left (b^2-14 a b c+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}{5 d \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2-14 a b c+a^2 c^2\right ) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{(7 b-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}}}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{6 a 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{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{d \sqrt{b+a \left (c+d x^2\right )}}-\frac{c^{3/2} (7 b-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 )}{5 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 (b^2-14 a b c+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}{5 a d^2 \sqrt{b+a \left (c+d x^2\right )}}\\ &=\frac{\left (b^2-14 a b c+a^2 c^2\right ) x \sqrt{b+a c+a d x^2} \sqrt{a+\frac{b}{c+d x^2}}}{5 a d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{(7 b-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}}}{5 d^2 \sqrt{b+a \left (c+d x^2\right )}}+\frac{6 a 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{x^3 \left (b+a c+a d x^2\right )^{3/2} \sqrt{a+\frac{b}{c+d x^2}}}{d \sqrt{b+a \left (c+d x^2\right )}}-\frac{\sqrt{c} \left (b^2-14 a b c+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 )}{5 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{c^{3/2} (7 b-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 )}{5 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 = 0.848245, size = 308, normalized size = 0.76 \[ \frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (x \sqrt{\frac{a d}{a c+b}} \left (-a^2 \left (c-d x^2\right ) \left (c+d x^2\right )^2+3 a b \left (2 c^2+3 c d x^2+d^2 x^4\right )+b^2 \left (7 c+2 d x^2\right )\right )-i c \left (a^2 c^2-14 a b c+b^2\right ) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} E\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )+8 i b c (b-a c) \sqrt{\frac{d x^2}{c}+1} \sqrt{\frac{a c+a d x^2+b}{a c+b}} F\left (i \sinh ^{-1}\left (\sqrt{\frac{a d}{b+a c}} x\right )|\frac{b}{a c}+1\right )\right )}{5 d^2 \sqrt{\frac{a d}{a c+b}} \left (a \left (c+d x^2\right )+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(Sqrt[(a*d)/(b + a*c)]*x*(-(a^2*(c - d*x^2)*(c + d*x^2)^2) + b^2*(7*c +
2*d*x^2) + 3*a*b*(2*c^2 + 3*c*d*x^2 + d^2*x^4)) - I*c*(b^2 - 14*a*b*c + 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)] + (8*I)*b*c*(b - a*c)*S
qrt[(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)]))/(5*d^2*Sqrt[(a*d)/(b + a*c)]*(b + a*(c + d*x^2)))

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Maple [B]  time = 0.033, size = 1098, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x^7*a^2*d^3+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*
d/(a*c+b))^(1/2)*x^5*a^2*c*d^2+3*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x^5*a*b*d^2-((d*x^2+c)
*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x^3*a^2*c^2*d+4*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^
(1/2)*x^3*a*b*c*d+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((a*d*x^2+a*c+b)/(a*c+b))^(1/2)*((d*x^2+c)/c)^(1/2)*Ellipt
icE(x*(-a*d/(a*c+b))^(1/2),((a*c+b)/a/c)^(1/2))*a^2*c^3+5*(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)*x^3*a*b*c*d+2*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x^3*b^2*d-((d*x^2+c)*(a*
d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x*a^2*c^3+8*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((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))*a*b*c^2-14*((d*x^2+c)*(a*d*
x^2+a*c+b))^(1/2)*((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))*a*b*c^2+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x*a*b*c^2-8*((d*x^2+c)*(a*d*x^2+a
*c+b))^(1/2)*((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+((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*((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+5*(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)*x*a*b*c^2+2*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*(-a*d/(a*c+b))^(1/2)*x*b^2*c+5*(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)*x*b^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+a*c+b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a + \frac{b}{d x^{2} + c}\right )}^{\frac{3}{2}} 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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a d x^{6} +{\left (a c + b\right )} x^{4}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (\frac{a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*((a*c + a*d*x**2 + b)/(c + d*x**2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a + \frac{b}{d x^{2} + c}\right )}^{\frac{3}{2}} 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))^(3/2),x, algorithm="giac")

[Out]

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