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

Optimal. Leaf size=332 \[ \frac{c^{5/4} (9 b B-13 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{17/4}}+\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{17/4}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )} \]

[Out]

(9*b*B - 13*A*c)/(18*b^2*c*x^(9/2)) - (9*b*B - 13*A*c)/(10*b^3*x^(5/2)) + (c*(9*b*B - 13*A*c))/(2*b^4*Sqrt[x])
- (b*B - A*c)/(2*b*c*x^(9/2)*(b + c*x^2)) - (c^(5/4)*(9*b*B - 13*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^
(1/4)])/(4*Sqrt[2]*b^(17/4)) + (c^(5/4)*(9*b*B - 13*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqr
t[2]*b^(17/4)) + (c^(5/4)*(9*b*B - 13*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt
[2]*b^(17/4)) - (c^(5/4)*(9*b*B - 13*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[
2]*b^(17/4))

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Rubi [A]  time = 0.28566, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 15, 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} \[ \frac{c^{5/4} (9 b B-13 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{17/4}}+\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{17/4}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b*B - 13*A*c)/(18*b^2*c*x^(9/2)) - (9*b*B - 13*A*c)/(10*b^3*x^(5/2)) + (c*(9*b*B - 13*A*c))/(2*b^4*Sqrt[x])
- (b*B - A*c)/(2*b*c*x^(9/2)*(b + c*x^2)) - (c^(5/4)*(9*b*B - 13*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^
(1/4)])/(4*Sqrt[2]*b^(17/4)) + (c^(5/4)*(9*b*B - 13*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqr
t[2]*b^(17/4)) + (c^(5/4)*(9*b*B - 13*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt
[2]*b^(17/4)) - (c^(5/4)*(9*b*B - 13*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[
2]*b^(17/4))

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{A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx &=\int \frac{A+B x^2}{x^{11/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{\left (-\frac{9 b B}{2}+\frac{13 A c}{2}\right ) \int \frac{1}{x^{11/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{(9 b B-13 A c) \int \frac{1}{x^{7/2} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}-\frac{(c (9 b B-13 A c)) \int \frac{1}{x^{3/2} \left (b+c x^2\right )} \, dx}{4 b^3}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{\left (c^2 (9 b B-13 A c)\right ) \int \frac{\sqrt{x}}{b+c x^2} \, dx}{4 b^4}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{\left (c^2 (9 b B-13 A c)\right ) \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 b^4}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}-\frac{\left (c^{3/2} (9 b B-13 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^4}+\frac{\left (c^{3/2} (9 b B-13 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^4}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{(c (9 b B-13 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^4}+\frac{(c (9 b B-13 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^4}+\frac{\left (c^{5/4} (9 b B-13 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^{17/4}}+\frac{\left (c^{5/4} (9 b B-13 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^{17/4}}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}+\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}+\frac{\left (c^{5/4} (9 b B-13 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^{17/4}}-\frac{\left (c^{5/4} (9 b B-13 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^{17/4}}\\ &=\frac{9 b B-13 A c}{18 b^2 c x^{9/2}}-\frac{9 b B-13 A c}{10 b^3 x^{5/2}}+\frac{c (9 b B-13 A c)}{2 b^4 \sqrt{x}}-\frac{b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}-\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{17/4}}+\frac{c^{5/4} (9 b B-13 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{17/4}}+\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}-\frac{c^{5/4} (9 b B-13 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{17/4}}\\ \end{align*}

Mathematica [C]  time = 0.4141, size = 176, normalized size = 0.53 \[ \frac{2 c^2 x^{3/2} (b B-A c) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{c x^2}{b}\right )}{3 b^5}-\frac{2 (b B-2 A c)}{5 b^3 x^{5/2}}+\frac{2 c (2 b B-3 A c)}{b^4 \sqrt{x}}-\frac{2 A}{9 b^2 x^{9/2}}+\frac{c^{5/4} (2 b B-3 A c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{(-b)^{17/4}}+\frac{c^{5/4} (3 A c-2 b B) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{(-b)^{17/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(9*b^2*x^(9/2)) - (2*(b*B - 2*A*c))/(5*b^3*x^(5/2)) + (2*c*(2*b*B - 3*A*c))/(b^4*Sqrt[x]) + (c^(5/4)*(2
*b*B - 3*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(17/4) + (c^(5/4)*(-2*b*B + 3*A*c)*ArcTanh[(c^(1/4)*S
qrt[x])/(-b)^(1/4)])/(-b)^(17/4) + (2*c^2*(b*B - A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3
*b^5)

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Maple [A]  time = 0.018, size = 372, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{9\,{b}^{2}}{x}^{-{\frac{9}{2}}}}+{\frac{4\,Ac}{5\,{b}^{3}}{x}^{-{\frac{5}{2}}}}-{\frac{2\,B}{5\,{b}^{2}}{x}^{-{\frac{5}{2}}}}-6\,{\frac{A{c}^{2}}{{b}^{4}\sqrt{x}}}+4\,{\frac{Bc}{{b}^{3}\sqrt{x}}}-{\frac{A{c}^{3}}{2\,{b}^{4} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}+{\frac{{c}^{2}B}{2\,{b}^{3} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}-{\frac{13\,{c}^{2}\sqrt{2}A}{16\,{b}^{4}}\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{13\,{c}^{2}\sqrt{2}A}{8\,{b}^{4}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{13\,{c}^{2}\sqrt{2}A}{8\,{b}^{4}}\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}B}{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}B}{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}B}{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}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*A/b^2/x^(9/2)+4/5/b^3/x^(5/2)*A*c-2/5/b^2/x^(5/2)*B-6*c^2/b^4/x^(1/2)*A+4*c/b^3/x^(1/2)*B-1/2/b^4*c^3*x^(
3/2)/(c*x^2+b)*A+1/2/b^3*c^2*x^(3/2)/(c*x^2+b)*B-13/16/b^4*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)))-13/8/b^4*c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2
^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-13/8/b^4*c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+9/16/b^
3*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)))+9/8/b^3*c/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+9/8/b^3*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((B*x^2+A)/x^(3/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.63345, size = 2475, 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^(3/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/360*(180*(b^4*c*x^7 + b^5*x^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3
*B*b*c^8 + 28561*A^4*c^9)/b^17)^(1/4)*arctan((sqrt((531441*B^6*b^6*c^8 - 4605822*A*B^5*b^5*c^9 + 16632135*A^2*
B^4*b^4*c^10 - 32032260*A^3*B^3*b^3*c^11 + 34701615*A^4*B^2*b^2*c^12 - 20049822*A^5*B*b*c^13 + 4826809*A^6*c^1
4)*x - (6561*B^4*b^13*c^5 - 37908*A*B^3*b^12*c^6 + 82134*A^2*B^2*b^11*c^7 - 79092*A^3*B*b^10*c^8 + 28561*A^4*b
^9*c^9)*sqrt(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*
c^9)/b^17))*b^4*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*
A^4*c^9)/b^17)^(1/4) + (729*B^3*b^7*c^4 - 3159*A*B^2*b^6*c^5 + 4563*A^2*B*b^5*c^6 - 2197*A^3*b^4*c^7)*sqrt(x)*
(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(
1/4))/(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)) -
45*(b^4*c*x^7 + b^5*x^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8
+ 28561*A^4*c^9)/b^17)^(1/4)*log(b^13*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 790
92*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(3/4) - (729*B^3*b^3*c^4 - 3159*A*B^2*b^2*c^5 + 4563*A^2*B*b*c^6 - 2197*
A^3*c^7)*sqrt(x)) + 45*(b^4*c*x^7 + b^5*x^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7
- 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(1/4)*log(-b^13*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*
A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(3/4) - (729*B^3*b^3*c^4 - 3159*A*B^2*b^2*c^5 + 456
3*A^2*B*b*c^6 - 2197*A^3*c^7)*sqrt(x)) + 4*(45*(9*B*b*c^2 - 13*A*c^3)*x^6 + 36*(9*B*b^2*c - 13*A*b*c^2)*x^4 -
20*A*b^3 - 4*(9*B*b^3 - 13*A*b^2*c)*x^2)*sqrt(x))/(b^4*c*x^7 + b^5*x^5)

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

[Out]

Timed out

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Giac [A]  time = 1.37606, size = 443, normalized size = 1.33 \begin{align*} \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 13 \, \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^{5} c} + \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 13 \, \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^{5} c} - \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 13 \, \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^{5} c} + \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 13 \, \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^{5} c} + \frac{B b c^{2} x^{\frac{3}{2}} - A c^{3} x^{\frac{3}{2}}}{2 \,{\left (c x^{2} + b\right )} b^{4}} + \frac{2 \,{\left (90 \, B b c x^{4} - 135 \, A c^{2} x^{4} - 9 \, B b^{2} x^{2} + 18 \, A b c x^{2} - 5 \, A b^{2}\right )}}{45 \, b^{4} x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(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^5*c) + 1/8*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(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^5*c) - 1/16*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c)*log(s
qrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c) + 1/16*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c
)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c) + 1/2*(B*b*c^2*x^(3/2) - A*c^3*x^(3/2))/((c*x^2 +
b)*b^4) + 2/45*(90*B*b*c*x^4 - 135*A*c^2*x^4 - 9*B*b^2*x^2 + 18*A*b*c*x^2 - 5*A*b^2)/(b^4*x^(9/2))