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

Optimal. Leaf size=310 \[ \frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{13/4}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{13/4}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )} \]

[Out]

(5*b*B - 9*A*c)/(10*b^2*c*x^(5/2)) - (5*b*B - 9*A*c)/(2*b^3*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(5/2)*(b + c*x^2))
+ (c^(1/4)*(5*b*B - 9*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(13/4)) - (c^(1/4)*(5*
b*B - 9*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(13/4)) - (c^(1/4)*(5*b*B - 9*A*c)*Lo
g[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(13/4)) + (c^(1/4)*(5*b*B - 9*A*c)*Log[
Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(13/4))

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Rubi [A]  time = 0.271474, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1584, 457, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{13/4}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{13/4}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*B - 9*A*c)/(10*b^2*c*x^(5/2)) - (5*b*B - 9*A*c)/(2*b^3*Sqrt[x]) - (b*B - A*c)/(2*b*c*x^(5/2)*(b + c*x^2))
+ (c^(1/4)*(5*b*B - 9*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(13/4)) - (c^(1/4)*(5*
b*B - 9*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(13/4)) - (c^(1/4)*(5*b*B - 9*A*c)*Lo
g[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(13/4)) + (c^(1/4)*(5*b*B - 9*A*c)*Log[
Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(13/4))

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 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 297

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

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])

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{x} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{A+B x^2}{x^{7/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}+\frac{\left (-\frac{5 b B}{2}+\frac{9 A c}{2}\right ) \int \frac{1}{x^{7/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}+\frac{(5 b B-9 A c) \int \frac{1}{x^{3/2} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}-\frac{(c (5 b B-9 A c)) \int \frac{\sqrt{x}}{b+c x^2} \, dx}{4 b^3}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}-\frac{(c (5 b B-9 A c)) \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 b^3}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}+\frac{\left (\sqrt{c} (5 b B-9 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 b^3}-\frac{\left (\sqrt{c} (5 b B-9 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}-\frac{(5 b B-9 A c) \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 )}{8 b^3}-\frac{(5 b B-9 A c) \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 )}{8 b^3}-\frac{\left (\sqrt [4]{c} (5 b B-9 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 )}{8 \sqrt{2} b^{13/4}}-\frac{\left (\sqrt [4]{c} (5 b B-9 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 )}{8 \sqrt{2} b^{13/4}}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}-\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}-\frac{\left (\sqrt [4]{c} (5 b B-9 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 )}{4 \sqrt{2} b^{13/4}}+\frac{\left (\sqrt [4]{c} (5 b B-9 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 )}{4 \sqrt{2} b^{13/4}}\\ &=\frac{5 b B-9 A c}{10 b^2 c x^{5/2}}-\frac{5 b B-9 A c}{2 b^3 \sqrt{x}}-\frac{b B-A c}{2 b c x^{5/2} \left (b+c x^2\right )}+\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{13/4}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{13/4}}-\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}+\frac{\sqrt [4]{c} (5 b B-9 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{13/4}}\\ \end{align*}

Mathematica [C]  time = 0.44617, size = 151, normalized size = 0.49 \[ \frac{2 c x^{3/2} (A c-b B) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{c x^2}{b}\right )}{3 b^4}+\frac{4 A c-2 b B}{b^3 \sqrt{x}}-\frac{2 A}{5 b^2 x^{5/2}}+\frac{\sqrt [4]{c} (b B-2 A c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{(-b)^{13/4}}+\frac{b \sqrt [4]{c} (b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{(-b)^{17/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(5*b^2*x^(5/2)) + (-2*b*B + 4*A*c)/(b^3*Sqrt[x]) + (c^(1/4)*(b*B - 2*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)
^(1/4)])/(-b)^(13/4) + (b*c^(1/4)*(b*B - 2*A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(17/4) + (2*c*(-(b
*B) + A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3*b^4)

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Maple [A]  time = 0.017, size = 339, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{5\,{b}^{2}}{x}^{-{\frac{5}{2}}}}+4\,{\frac{Ac}{{b}^{3}\sqrt{x}}}-2\,{\frac{B}{{b}^{2}\sqrt{x}}}+{\frac{A{c}^{2}}{2\,{b}^{3} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}-{\frac{Bc}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}+{\frac{9\,c\sqrt{2}A}{16\,{b}^{3}}\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{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{9\,c\sqrt{2}A}{8\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{9\,c\sqrt{2}A}{8\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{5\,\sqrt{2}B}{16\,{b}^{2}}\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{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{5\,\sqrt{2}B}{8\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{5\,\sqrt{2}B}{8\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*A/b^2/x^(5/2)+4/b^3/x^(1/2)*A*c-2/b^2/x^(1/2)*B+1/2/b^3*c^2*x^(3/2)/(c*x^2+b)*A-1/2/b^2*c*x^(3/2)/(c*x^2+
b)*B+9/16/b^3*c/(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)))+9/8/b^3*c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+9/8/b^3*c/(b/c)^(1/4
)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-5/16/b^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)))-5/8/b^2/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b
/c)^(1/4)*x^(1/2)+1)-5/8/b^2/(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((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.65527, size = 2310, normalized size = 7.45 \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)*x^(1/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/40*(20*(b^3*c*x^5 + b^4*x^3)*(-(625*B^4*b^4*c - 4500*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*
c^4 + 6561*A^4*c^5)/b^13)^(1/4)*arctan((sqrt((15625*B^6*b^6*c^2 - 168750*A*B^5*b^5*c^3 + 759375*A^2*B^4*b^4*c^
4 - 1822500*A^3*B^3*b^3*c^5 + 2460375*A^4*B^2*b^2*c^6 - 1771470*A^5*B*b*c^7 + 531441*A^6*c^8)*x - (625*B^4*b^1
1*c - 4500*A*B^3*b^10*c^2 + 12150*A^2*B^2*b^9*c^3 - 14580*A^3*B*b^8*c^4 + 6561*A^4*b^7*c^5)*sqrt(-(625*B^4*b^4
*c - 4500*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13))*b^3*(-(625*B^4*b^4*
c - 4500*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(1/4) + (125*B^3*b^6*
c - 675*A*B^2*b^5*c^2 + 1215*A^2*B*b^4*c^3 - 729*A^3*b^3*c^4)*sqrt(x)*(-(625*B^4*b^4*c - 4500*A*B^3*b^3*c^2 +
12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(1/4))/(625*B^4*b^4*c - 4500*A*B^3*b^3*c^2 + 1
2150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)) - 5*(b^3*c*x^5 + b^4*x^3)*(-(625*B^4*b^4*c - 4500*A*
B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(1/4)*log(b^10*(-(625*B^4*b^4*c
- 4500*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(3/4) - (125*B^3*b^3*c
- 675*A*B^2*b^2*c^2 + 1215*A^2*B*b*c^3 - 729*A^3*c^4)*sqrt(x)) + 5*(b^3*c*x^5 + b^4*x^3)*(-(625*B^4*b^4*c - 45
00*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(1/4)*log(-b^10*(-(625*B^4*
b^4*c - 4500*A*B^3*b^3*c^2 + 12150*A^2*B^2*b^2*c^3 - 14580*A^3*B*b*c^4 + 6561*A^4*c^5)/b^13)^(3/4) - (125*B^3*
b^3*c - 675*A*B^2*b^2*c^2 + 1215*A^2*B*b*c^3 - 729*A^3*c^4)*sqrt(x)) + 4*(5*(5*B*b*c - 9*A*c^2)*x^4 + 4*A*b^2
+ 4*(5*B*b^2 - 9*A*b*c)*x^2)*sqrt(x))/(b^3*c*x^5 + b^4*x^3)

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

[Out]

Timed out

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Giac [A]  time = 1.35634, size = 409, normalized size = 1.32 \begin{align*} -\frac{B b c x^{\frac{3}{2}} - A c^{2} x^{\frac{3}{2}}}{2 \,{\left (c x^{2} + b\right )} b^{3}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 9 \, \left (b c^{3}\right )^{\frac{3}{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 )}{8 \, b^{4} c^{2}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 9 \, \left (b c^{3}\right )^{\frac{3}{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 )}{8 \, b^{4} c^{2}} + \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 9 \, \left (b c^{3}\right )^{\frac{3}{4}} A c\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{16 \, b^{4} c^{2}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 9 \, \left (b c^{3}\right )^{\frac{3}{4}} A c\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{16 \, b^{4} c^{2}} - \frac{2 \,{\left (5 \, B b x^{2} - 10 \, A c x^{2} + A b\right )}}{5 \, b^{3} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*b*c*x^(3/2) - A*c^2*x^(3/2))/((c*x^2 + b)*b^3) - 1/8*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 9*(b*c^3)^(3/4)*A*
c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^4*c^2) - 1/8*sqrt(2)*(5*(b*c^3)^(3/4)*
B*b - 9*(b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^4*c^2) + 1/16
*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 9*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c^2
) - 1/16*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 9*(b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))
/(b^4*c^2) - 2/5*(5*B*b*x^2 - 10*A*c*x^2 + A*b)/(b^3*x^(5/2))