3.217 \(\int \frac{x^{3/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=343 \[ \frac{11 c^{3/4} (7 b B-15 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}+\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{19/4}}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}+\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2} \]

[Out]

(11*(7*b*B - 15*A*c))/(112*b^3*c*x^(7/2)) - (11*(7*b*B - 15*A*c))/(48*b^4*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(7/2
)*(b + c*x^2)^2) - (7*b*B - 15*A*c)/(16*b^2*c*x^(7/2)*(b + c*x^2)) + (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 - (
Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) + (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)
*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4))

________________________________________________________________________________________

Rubi [A]  time = 0.277669, antiderivative size = 343, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {1584, 457, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{11 c^{3/4} (7 b B-15 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}+\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{19/4}}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}+\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(3/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(11*(7*b*B - 15*A*c))/(112*b^3*c*x^(7/2)) - (11*(7*b*B - 15*A*c))/(48*b^4*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(7/2
)*(b + c*x^2)^2) - (7*b*B - 15*A*c)/(16*b^2*c*x^(7/2)*(b + c*x^2)) + (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 - (
Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) + (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)
*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4))

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{A+B x^2}{x^{9/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}+\frac{\left (-\frac{7 b B}{2}+\frac{15 A c}{2}\right ) \int \frac{1}{x^{9/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{(11 (7 b B-15 A c)) \int \frac{1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac{(11 (7 b B-15 A c)) \int \frac{1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{(11 c (7 b B-15 A c)) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{(11 c (7 b B-15 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 b^4}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{(11 c (7 b B-15 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{9/2}}-\frac{(11 c (7 b B-15 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{9/2}}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac{\left (11 \sqrt{c} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{64 b^{9/2}}-\frac{\left (11 \sqrt{c} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{64 b^{9/2}}+\frac{\left (11 c^{3/4} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} b^{19/4}}+\frac{\left (11 c^{3/4} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} b^{19/4}}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}-\frac{\left (11 c^{3/4} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}+\frac{\left (11 c^{3/4} (7 b B-15 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}\\ &=\frac{11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac{11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac{b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac{7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{19/4}}+\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}-\frac{11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{19/4}}\\ \end{align*}

Mathematica [A]  time = 0.51115, size = 433, normalized size = 1.26 \[ \frac{\frac{672 A b^{7/4} c^2 \sqrt{x}}{\left (b+c x^2\right )^2}+\frac{3864 A b^{3/4} c^2 \sqrt{x}}{b+c x^2}+\frac{5376 A b^{3/4} c}{x^{3/2}}-\frac{768 A b^{7/4}}{x^{7/2}}+462 \sqrt{2} c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )+462 \sqrt{2} c^{3/4} (15 A c-7 b B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )-3465 \sqrt{2} A c^{7/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+3465 \sqrt{2} A c^{7/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-\frac{672 b^{11/4} B c \sqrt{x}}{\left (b+c x^2\right )^2}-\frac{2520 b^{7/4} B c \sqrt{x}}{b+c x^2}-\frac{1792 b^{7/4} B}{x^{3/2}}+1617 \sqrt{2} b B c^{3/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-1617 \sqrt{2} b B c^{3/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2688 b^{19/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-768*A*b^(7/4))/x^(7/2) - (1792*b^(7/4)*B)/x^(3/2) + (5376*A*b^(3/4)*c)/x^(3/2) - (672*b^(11/4)*B*c*Sqrt[x])
/(b + c*x^2)^2 + (672*A*b^(7/4)*c^2*Sqrt[x])/(b + c*x^2)^2 - (2520*b^(7/4)*B*c*Sqrt[x])/(b + c*x^2) + (3864*A*
b^(3/4)*c^2*Sqrt[x])/(b + c*x^2) + 462*Sqrt[2]*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b
^(1/4)] + 462*Sqrt[2]*c^(3/4)*(-7*b*B + 15*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 1617*Sqrt[2]*b
*B*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 3465*Sqrt[2]*A*c^(7/4)*Log[Sqrt[b] - S
qrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 1617*Sqrt[2]*b*B*c^(3/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*S
qrt[x] + Sqrt[c]*x] + 3465*Sqrt[2]*A*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2688
*b^(19/4))

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Maple [A]  time = 0.02, size = 390, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{7\,{b}^{3}}{x}^{-{\frac{7}{2}}}}+2\,{\frac{Ac}{{b}^{4}{x}^{3/2}}}-{\frac{2\,B}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}+{\frac{23\,A{c}^{3}}{16\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{15\,{c}^{2}B}{16\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{27\,A{c}^{2}}{16\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}-{\frac{19\,Bc}{16\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{165\,{c}^{2}\sqrt{2}A}{64\,{b}^{5}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{165\,{c}^{2}\sqrt{2}A}{128\,{b}^{5}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }+{\frac{165\,{c}^{2}\sqrt{2}A}{64\,{b}^{5}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{77\,c\sqrt{2}B}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{77\,c\sqrt{2}B}{128\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }-{\frac{77\,c\sqrt{2}B}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

-2/7*A/b^3/x^(7/2)+2/b^4/x^(3/2)*A*c-2/3/b^3/x^(3/2)*B+23/16/b^4*c^3/(c*x^2+b)^2*x^(5/2)*A-15/16/b^3*c^2/(c*x^
2+b)^2*x^(5/2)*B+27/16/b^3*c^2/(c*x^2+b)^2*A*x^(1/2)-19/16/b^2*c/(c*x^2+b)^2*B*x^(1/2)+165/64/b^5*c^2*(b/c)^(1
/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+165/128/b^5*c^2*(b/c)^(1/4)*2^(1/2)*A*ln((x+(b/c)^(1/4)*x^
(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+165/64/b^5*c^2*(b/c)^(1/4)*2^(1/2)*A*a
rctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-77/64/b^4*c*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-7
7/128/b^4*c*(b/c)^(1/4)*2^(1/2)*B*ln((x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2
)+(b/c)^(1/2)))-77/64/b^4*c*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72965, size = 2063, normalized size = 6.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/1344*(924*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^
2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1/4)*arctan((sqrt(b^10*sqrt(-(2401*B^4*b^4*c^3 - 20580*A*B^3
*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19) + (49*B^2*b^2*c^2 - 210*A*B*b*c^3
+ 225*A^2*c^4)*x)*b^14*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 +
50625*A^4*c^7)/b^19)^(3/4) + (7*B*b^15*c - 15*A*b^14*c^2)*sqrt(x)*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 +
66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(3/4))/(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c
^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)) + 231*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(
-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1
/4)*log(11*b^5*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A
^4*c^7)/b^19)^(1/4) - 11*(7*B*b*c - 15*A*c^2)*sqrt(x)) - 231*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-(2401*B^4
*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1/4)*log(-1
1*b^5*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b
^19)^(1/4) - 11*(7*B*b*c - 15*A*c^2)*sqrt(x)) - 4*(77*(7*B*b*c^2 - 15*A*c^3)*x^6 + 121*(7*B*b^2*c - 15*A*b*c^2
)*x^4 + 96*A*b^3 + 32*(7*B*b^3 - 15*A*b^2*c)*x^2)*sqrt(x))/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)

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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**(3/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.3088, size = 425, normalized size = 1.24 \begin{align*} -\frac{11 \, \sqrt{2}{\left (7 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 15 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{5}} - \frac{11 \, \sqrt{2}{\left (7 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 15 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{5}} - \frac{11 \, \sqrt{2}{\left (7 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 15 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{5}} + \frac{11 \, \sqrt{2}{\left (7 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 15 \, \left (b c^{3}\right )^{\frac{1}{4}} A c\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{5}} - \frac{15 \, B b c^{2} x^{\frac{5}{2}} - 23 \, A c^{3} x^{\frac{5}{2}} + 19 \, B b^{2} c \sqrt{x} - 27 \, A b c^{2} \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} b^{4}} - \frac{2 \,{\left (7 \, B b x^{2} - 21 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{4} x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-11/64*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x
))/(b/c)^(1/4))/b^5 - 11/64*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*
(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^5 - 11/128*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*log(sq
rt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 + 11/128*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*l
og(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 1/16*(15*B*b*c^2*x^(5/2) - 23*A*c^3*x^(5/2) + 19*B*b^2*
c*sqrt(x) - 27*A*b*c^2*sqrt(x))/((c*x^2 + b)^2*b^4) - 2/21*(7*B*b*x^2 - 21*A*c*x^2 + 3*A*b)/(b^4*x^(7/2))