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

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

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Rubi [A]  time = 0.27, antiderivative size = 310, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 )} \end {gather*}

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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.64, size = 151, normalized size = 0.49 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.68, size = 200, normalized size = 0.65 \begin {gather*} \frac {\left (5 b B \sqrt [4]{c}-9 A c^{5/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {\left (5 b B \sqrt [4]{c}-9 A c^{5/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{13/4}}+\frac {-4 A b^2+36 A b c x^2+45 A c^2 x^4-20 b^2 B x^2-25 b B c x^4}{10 b^3 x^{5/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*A*b^2 - 20*b^2*B*x^2 + 36*A*b*c*x^2 - 25*b*B*c*x^4 + 45*A*c^2*x^4)/(10*b^3*x^(5/2)*(b + c*x^2)) + ((5*b*B*
c^(1/4) - 9*A*c^(5/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(4*Sqrt[2]*b^(13/4)) +
((5*b*B*c^(1/4) - 9*A*c^(5/4))*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*b
^(13/4))

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fricas [B]  time = 0.46, size = 974, normalized size = 3.14 \begin {gather*} -\frac {20 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (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}\right )} x - {\left (625 \, B^{4} b^{11} 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}\right )} \sqrt {-\frac {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} \left (-\frac {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}}\right )^{\frac {1}{4}} + {\left (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}\right )} \sqrt {x} \left (-\frac {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}}\right )^{\frac {1}{4}}}{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}}\right ) - 5 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (b^{10} \left (-\frac {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}}\right )^{\frac {3}{4}} - {\left (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}\right )} \sqrt {x}\right ) + 5 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (-b^{10} \left (-\frac {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}}\right )^{\frac {3}{4}} - {\left (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}\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, {\left (5 \, B b c - 9 \, A c^{2}\right )} x^{4} + 4 \, A b^{2} + 4 \, {\left (5 \, B b^{2} - 9 \, A b c\right )} x^{2}\right )} \sqrt {x}}{40 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}} \end {gather*}

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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giac [A]  time = 0.21, size = 303, normalized size = 0.98 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.06, size = 339, normalized size = 1.09 \begin {gather*} \frac {A \,c^{2} x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b^{3}}-\frac {B c \,x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b^{2}}+\frac {9 \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}+\frac {9 \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}+\frac {9 \sqrt {2}\, A c \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {5 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}-\frac {5 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}-\frac {5 \sqrt {2}\, B \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{2}}+\frac {4 A c}{b^{3} \sqrt {x}}-\frac {2 B}{b^{2} \sqrt {x}}-\frac {2 A}{5 b^{2} x^{\frac {5}{2}}} \end {gather*}

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]

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)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))+9/8/b^3*c/(b/c)^(1/4)*2^(1/2)*A*a
rctan(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)-5/1
6/b^2/(b/c)^(1/4)*2^(1/2)*B*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(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*ar
ctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/5/b^2*A/x^(5/2)+4/b^3/x^(1/2)*A*c-2/b^2/x^(1/2)*B

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maxima [A]  time = 3.03, size = 250, normalized size = 0.81 \begin {gather*} -\frac {5 \, {\left (5 \, B b c - 9 \, A c^{2}\right )} x^{4} + 4 \, A b^{2} + 4 \, {\left (5 \, B b^{2} - 9 \, A b c\right )} x^{2}}{10 \, {\left (b^{3} c x^{\frac {9}{2}} + b^{4} x^{\frac {5}{2}}\right )}} - \frac {{\left (5 \, B b c - 9 \, A c^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b^{3}} \end {gather*}

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]

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

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mupad [B]  time = 0.25, size = 121, normalized size = 0.39 \begin {gather*} \frac {\frac {2\,x^2\,\left (9\,A\,c-5\,B\,b\right )}{5\,b^2}-\frac {2\,A}{5\,b}+\frac {c\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{2\,b^3}}{b\,x^{5/2}+c\,x^{9/2}}+\frac {{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^{13/4}}-\frac {{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{4\,b^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x^2*(9*A*c - 5*B*b))/(5*b^2) - (2*A)/(5*b) + (c*x^4*(9*A*c - 5*B*b))/(2*b^3))/(b*x^(5/2) + c*x^(9/2)) + ((
-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4))*(9*A*c - 5*B*b))/(4*b^(13/4)) - ((-c)^(1/4)*atanh(((-c)^(1/4)*x^(
1/2))/b^(1/4))*(9*A*c - 5*B*b))/(4*b^(13/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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