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

Optimal. Leaf size=365 \[ -\frac{15 c^{7/4} (11 b B-19 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}-\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{23/4}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2} \]

[Out]

(15*(11*b*B - 19*A*c))/(176*b^3*c*x^(11/2)) - (15*(11*b*B - 19*A*c))/(112*b^4*x^(7/2)) + (5*c*(11*b*B - 19*A*c
))/(16*b^5*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(11/2)*(b + c*x^2)^2) - (11*b*B - 19*A*c)/(16*b^2*c*x^(11/2)*(b + c
*x^2)) - (15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) +
(15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) - (15*c^(7/
4)*(11*b*B - 19*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4)) + (15*c
^(7/4)*(11*b*B - 19*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4))

________________________________________________________________________________________

Rubi [A]  time = 0.320727, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ -\frac{15 c^{7/4} (11 b B-19 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}-\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{23/4}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(11*b*B - 19*A*c))/(176*b^3*c*x^(11/2)) - (15*(11*b*B - 19*A*c))/(112*b^4*x^(7/2)) + (5*c*(11*b*B - 19*A*c
))/(16*b^5*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(11/2)*(b + c*x^2)^2) - (11*b*B - 19*A*c)/(16*b^2*c*x^(11/2)*(b + c
*x^2)) - (15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) +
(15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) - (15*c^(7/
4)*(11*b*B - 19*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4)) + (15*c
^(7/4)*(11*b*B - 19*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4))

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{A+B x^2}{\sqrt{x} \left (b x^2+c x^4\right )^3} \, dx &=\int \frac{A+B x^2}{x^{13/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}+\frac{\left (-\frac{11 b B}{2}+\frac{19 A c}{2}\right ) \int \frac{1}{x^{13/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac{(15 (11 b B-19 A c)) \int \frac{1}{x^{13/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{(15 (11 b B-19 A c)) \int \frac{1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac{(15 c (11 b B-19 A c)) \int \frac{1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{\left (15 c^2 (11 b B-19 A c)\right ) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{32 b^5}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{\left (15 c^2 (11 b B-19 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 b^5}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{\left (15 c^2 (11 b B-19 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{11/2}}+\frac{\left (15 c^2 (11 b B-19 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{11/2}}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac{\left (15 c^{3/2} (11 b B-19 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^{11/2}}+\frac{\left (15 c^{3/2} (11 b B-19 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^{11/2}}-\frac{\left (15 c^{7/4} (11 b B-19 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^{23/4}}-\frac{\left (15 c^{7/4} (11 b B-19 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^{23/4}}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}+\frac{\left (15 c^{7/4} (11 b B-19 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^{23/4}}-\frac{\left (15 c^{7/4} (11 b B-19 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^{23/4}}\\ &=\frac{15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac{15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac{5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac{b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac{11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{23/4}}-\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}+\frac{15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{23/4}}\\ \end{align*}

Mathematica [A]  time = 0.581209, size = 467, normalized size = 1.28 \[ \frac{-\frac{19096 A b^{3/4} c^3 \sqrt{x}}{b+c x^2}-\frac{2464 A b^{7/4} c^3 \sqrt{x}}{\left (b+c x^2\right )^2}-\frac{39424 A b^{3/4} c^2}{x^{3/2}}+\frac{8448 A b^{7/4} c}{x^{7/2}}-\frac{1792 A b^{11/4}}{x^{11/2}}+2310 \sqrt{2} c^{7/4} (19 A c-11 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )+2310 \sqrt{2} c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )+21945 \sqrt{2} A c^{11/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-21945 \sqrt{2} A c^{11/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+\frac{14168 b^{7/4} B c^2 \sqrt{x}}{b+c x^2}+\frac{2464 b^{11/4} B c^2 \sqrt{x}}{\left (b+c x^2\right )^2}+\frac{19712 b^{7/4} B c}{x^{3/2}}-\frac{2816 b^{11/4} B}{x^{7/2}}-12705 \sqrt{2} b B c^{7/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+12705 \sqrt{2} b B c^{7/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{9856 b^{23/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1792*A*b^(11/4))/x^(11/2) - (2816*b^(11/4)*B)/x^(7/2) + (8448*A*b^(7/4)*c)/x^(7/2) + (19712*b^(7/4)*B*c)/x^
(3/2) - (39424*A*b^(3/4)*c^2)/x^(3/2) + (2464*b^(11/4)*B*c^2*Sqrt[x])/(b + c*x^2)^2 - (2464*A*b^(7/4)*c^3*Sqrt
[x])/(b + c*x^2)^2 + (14168*b^(7/4)*B*c^2*Sqrt[x])/(b + c*x^2) - (19096*A*b^(3/4)*c^3*Sqrt[x])/(b + c*x^2) + 2
310*Sqrt[2]*c^(7/4)*(-11*b*B + 19*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 2310*Sqrt[2]*c^(7/4)*(1
1*b*B - 19*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 12705*Sqrt[2]*b*B*c^(7/4)*Log[Sqrt[b] - Sqrt[2
]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 21945*Sqrt[2]*A*c^(11/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x
] + Sqrt[c]*x] + 12705*Sqrt[2]*b*B*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 21945*
Sqrt[2]*A*c^(11/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(9856*b^(23/4))

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Maple [A]  time = 0.022, size = 420, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{11\,{b}^{3}}{x}^{-{\frac{11}{2}}}}+{\frac{6\,Ac}{7\,{b}^{4}}{x}^{-{\frac{7}{2}}}}-{\frac{2\,B}{7\,{b}^{3}}{x}^{-{\frac{7}{2}}}}-4\,{\frac{A{c}^{2}}{{b}^{5}{x}^{3/2}}}+2\,{\frac{Bc}{{b}^{4}{x}^{3/2}}}-{\frac{31\,{c}^{4}A}{16\,{b}^{5} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{23\,{c}^{3}B}{16\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{35\,A{c}^{3}}{16\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{27\,{c}^{2}B}{16\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}-{\frac{285\,{c}^{3}\sqrt{2}A}{64\,{b}^{6}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{285\,{c}^{3}\sqrt{2}A}{64\,{b}^{6}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{285\,{c}^{3}\sqrt{2}A}{128\,{b}^{6}}\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}B}{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}B}{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}B}{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+b*x^2)^3/x^(1/2),x)

[Out]

-2/11*A/b^3/x^(11/2)+6/7/b^4/x^(7/2)*A*c-2/7/b^3/x^(7/2)*B-4*c^2/b^5/x^(3/2)*A+2*c/b^4/x^(3/2)*B-31/16/b^5*c^4
/(c*x^2+b)^2*x^(5/2)*A+23/16/b^4*c^3/(c*x^2+b)^2*x^(5/2)*B-35/16/b^4*c^3/(c*x^2+b)^2*A*x^(1/2)+27/16/b^3*c^2/(
c*x^2+b)^2*B*x^(1/2)-285/64/b^6*c^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-285/64/b^6*c^3
*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-285/128/b^6*c^3*(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)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+165/64/b^5*c^2*(b/c)^(1/4)*2^(1/2)*B*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)*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)))

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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((B*x^2+A)/(c*x^4+b*x^2)^3/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.5948, size = 2233, normalized size = 6.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^3/x^(1/2),x, algorithm="fricas")

[Out]

-1/4928*(4620*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*
B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*arctan((sqrt(b^12*sqrt(-(14641*B^4*b^4*c^7 -
101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23) + (121*B^2*b^2*c^
4 - 418*A*B*b*c^5 + 361*A^2*c^6)*x)*b^17*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9
- 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(3/4) + (11*B*b^18*c^2 - 19*A*b^17*c^3)*sqrt(x)*(-(14641*B^4*b^
4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(3/4))/(1
4641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)) + 1
155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^
9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*log(15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8
+ 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B*b*c^2 - 19*A*c^3)*sqr
t(x)) - 1155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B
^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*log(-15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^
3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B*b*c^2 - 19*
A*c^3)*sqrt(x)) - 4*(385*(11*B*b*c^3 - 19*A*c^4)*x^8 + 605*(11*B*b^2*c^2 - 19*A*b*c^3)*x^6 - 224*A*b^4 + 160*(
11*B*b^3*c - 19*A*b^2*c^2)*x^4 - 32*(11*B*b^4 - 19*A*b^3*c)*x^2)*sqrt(x))/(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^
6)

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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((B*x**2+A)/(c*x**4+b*x**2)**3/x**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.39349, size = 474, normalized size = 1.3 \begin{align*} \frac{15 \, \sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac{1}{4}} A c^{2}\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^{6}} + \frac{15 \, \sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac{1}{4}} A c^{2}\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^{6}} + \frac{15 \, \sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac{1}{4}} A c^{2}\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{6}} - \frac{15 \, \sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac{1}{4}} A c^{2}\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{6}} + \frac{23 \, B b c^{3} x^{\frac{5}{2}} - 31 \, A c^{4} x^{\frac{5}{2}} + 27 \, B b^{2} c^{2} \sqrt{x} - 35 \, A b c^{3} \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} b^{5}} + \frac{2 \,{\left (77 \, B b c x^{4} - 154 \, A c^{2} x^{4} - 11 \, B b^{2} x^{2} + 33 \, A b c x^{2} - 7 \, A b^{2}\right )}}{77 \, b^{5} x^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^3/x^(1/2),x, algorithm="giac")

[Out]

15/64*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sq
rt(x))/(b/c)^(1/4))/b^6 + 15/64*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(-1/2*sqrt(2)*
(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^6 + 15/128*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)
*A*c^2)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 - 15/128*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*
c^3)^(1/4)*A*c^2)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 + 1/16*(23*B*b*c^3*x^(5/2) - 31*A*c^4*
x^(5/2) + 27*B*b^2*c^2*sqrt(x) - 35*A*b*c^3*sqrt(x))/((c*x^2 + b)^2*b^5) + 2/77*(77*B*b*c*x^4 - 154*A*c^2*x^4
- 11*B*b^2*x^2 + 33*A*b*c*x^2 - 7*A*b^2)/(b^5*x^(11/2))