3.199 \(\int \frac{x^{11/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=289 \[ \frac{(5 b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}+\frac{(5 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}+\frac{\sqrt{x} (5 b B-A c)}{2 b c^2}-\frac{x^{5/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.233747, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {1584, 457, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{(5 b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}+\frac{(5 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}+\frac{\sqrt{x} (5 b B-A c)}{2 b c^2}-\frac{x^{5/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^{3/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac{\left (\frac{5 b B}{2}-\frac{A c}{2}\right ) \int \frac{x^{3/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac{(5 b B-A c) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{4 c^2}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac{(5 b B-A c) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 c^2}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac{(5 b B-A c) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 \sqrt{b} c^2}-\frac{(5 b B-A c) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 \sqrt{b} c^2}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}-\frac{(5 b B-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 \sqrt{b} c^{5/2}}-\frac{(5 b B-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 \sqrt{b} c^{5/2}}+\frac{(5 b B-A c) \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^{3/4} c^{9/4}}+\frac{(5 b B-A c) \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^{3/4} c^{9/4}}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac{(5 b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \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^{3/4} c^{9/4}}+\frac{(5 b B-A c) \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^{3/4} c^{9/4}}\\ &=\frac{(5 b B-A c) \sqrt{x}}{2 b c^2}-\frac{(b B-A c) x^{5/2}}{2 b c \left (b+c x^2\right )}+\frac{(5 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{3/4} c^{9/4}}+\frac{(5 b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}-\frac{(5 b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} b^{3/4} c^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.390944, size = 354, normalized size = 1.22 \[ \frac{\frac{2 \sqrt{2} (5 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{b^{3/4}}-\frac{2 \sqrt{2} (5 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{b^{3/4}}-\frac{\sqrt{2} A c \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}+\frac{\sqrt{2} A c \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}-\frac{8 A c^{5/4} \sqrt{x}}{b+c x^2}+\frac{8 b B \sqrt [4]{c} \sqrt{x}}{b+c x^2}+5 \sqrt{2} \sqrt [4]{b} B \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-5 \sqrt{2} \sqrt [4]{b} B \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+32 B \sqrt [4]{c} \sqrt{x}}{16 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(32*B*c^(1/4)*Sqrt[x] + (8*b*B*c^(1/4)*Sqrt[x])/(b + c*x^2) - (8*A*c^(5/4)*Sqrt[x])/(b + c*x^2) + (2*Sqrt[2]*(
5*b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) - (2*Sqrt[2]*(5*b*B - A*c)*ArcTan[1 + (Sqr
t[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) + 5*Sqrt[2]*b^(1/4)*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x] - (Sqrt[2]*A*c*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4) - 5*Sqrt[2]*b^(1
/4)*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (Sqrt[2]*A*c*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*
c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4))/(16*c^(9/4))

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Maple [A]  time = 0.014, size = 323, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{{c}^{2}}}-{\frac{A}{2\,c \left ( c{x}^{2}+b \right ) }\sqrt{x}}+{\frac{Bb}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }\sqrt{x}}+{\frac{\sqrt{2}A}{8\,bc}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{\sqrt{2}A}{8\,bc}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{\sqrt{2}A}{16\,bc}\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{5\,\sqrt{2}B}{8\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{5\,\sqrt{2}B}{8\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{5\,\sqrt{2}B}{16\,{c}^{2}}\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(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

2*B/c^2*x^(1/2)-1/2/c*x^(1/2)/(c*x^2+b)*A+1/2/c^2*x^(1/2)/(c*x^2+b)*B*b+1/8/c*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2
^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/8/c*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/16/c*(b/c)
^(1/4)/b*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)))
-5/8/c^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-5/8/c^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1
/2)/(b/c)^(1/4)*x^(1/2)-1)-5/16/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(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.52275, size = 1594, normalized size = 5.52 \begin{align*} \frac{4 \,{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{b^{2} c^{4} \sqrt{-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}} +{\left (25 \, B^{2} b^{2} - 10 \, A B b c + A^{2} c^{2}\right )} x} b^{2} c^{7} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{3}{4}} +{\left (5 \, B b^{3} c^{7} - A b^{2} c^{8}\right )} \sqrt{x} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{3}{4}}}{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}\right ) +{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{1}{4}} \log \left (b c^{2} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{1}{4}} -{\left (5 \, B b - A c\right )} \sqrt{x}\right ) -{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{1}{4}} \log \left (-b c^{2} \left (-\frac{625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac{1}{4}} -{\left (5 \, B b - A c\right )} \sqrt{x}\right ) + 4 \,{\left (4 \, B c x^{2} + 5 \, B b - A c\right )} \sqrt{x}}{8 \,{\left (c^{3} x^{2} + b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/8*(4*(c^3*x^2 + b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b
^3*c^9))^(1/4)*arctan((sqrt(b^2*c^4*sqrt(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^
3 + A^4*c^4)/(b^3*c^9)) + (25*B^2*b^2 - 10*A*B*b*c + A^2*c^2)*x)*b^2*c^7*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 15
0*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(3/4) + (5*B*b^3*c^7 - A*b^2*c^8)*sqrt(x)*(-(625*B^4*
b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(3/4))/(625*B^4*b^4 - 500*A
*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)) + (c^3*x^2 + b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b
^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log(b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b
^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) - (5*B*b - A*c)*sqrt(x)) - (c^3*x^2 +
b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log
(-b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) -
(5*B*b - A*c)*sqrt(x)) + 4*(4*B*c*x^2 + 5*B*b - A*c)*sqrt(x))/(c^3*x^2 + b*c^2)

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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**(11/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.22107, size = 382, normalized size = 1.32 \begin{align*} \frac{2 \, B \sqrt{x}}{c^{2}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{8 \, b c^{3}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{8 \, b c^{3}} - \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{16 \, b c^{3}} + \frac{\sqrt{2}{\left (5 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{16 \, b c^{3}} + \frac{B b \sqrt{x} - A c \sqrt{x}}{2 \,{\left (c x^{2} + b\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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