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

Optimal. Leaf size=276 \[ \frac{b^{5/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{13/4}}-\frac{b^{5/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{13/4}}+\frac{b^{5/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{13/4}}-\frac{b^{5/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} c^{13/4}}-\frac{2 x^{5/2} (b B-A c)}{5 c^2}+\frac{2 b \sqrt{x} (b B-A c)}{c^3}+\frac{2 B x^{9/2}}{9 c} \]

[Out]

(2*b*(b*B - A*c)*Sqrt[x])/c^3 - (2*(b*B - A*c)*x^(5/2))/(5*c^2) + (2*B*x^(9/2))/(9*c) + (b^(5/4)*(b*B - A*c)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) + (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + S
qrt[c]*x])/(2*Sqrt[2]*c^(13/4))

________________________________________________________________________________________

Rubi [A]  time = 0.247373, antiderivative size = 276, 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, 459, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{b^{5/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{13/4}}-\frac{b^{5/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{13/4}}+\frac{b^{5/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{13/4}}-\frac{b^{5/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} c^{13/4}}-\frac{2 x^{5/2} (b B-A c)}{5 c^2}+\frac{2 b \sqrt{x} (b B-A c)}{c^3}+\frac{2 B x^{9/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(b*B - A*c)*Sqrt[x])/c^3 - (2*(b*B - A*c)*x^(5/2))/(5*c^2) + (2*B*x^(9/2))/(9*c) + (b^(5/4)*(b*B - A*c)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) + (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + S
qrt[c]*x])/(2*Sqrt[2]*c^(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 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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

Mathematica [A]  time = 0.275619, size = 227, normalized size = 0.82 \[ \frac{\frac{45 \sqrt{2} b^{5/4} (b B-A c) \left (\log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-\log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )\right )}{\sqrt [4]{c}}+\frac{90 \sqrt{2} b^{5/4} (b B-A c) \left (\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )-\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )\right )}{\sqrt [4]{c}}+72 c x^{5/2} (A c-b B)+360 b \sqrt{x} (b B-A c)+40 B c^2 x^{9/2}}{180 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(360*b*(b*B - A*c)*Sqrt[x] + 72*c*(-(b*B) + A*c)*x^(5/2) + 40*B*c^2*x^(9/2) + (90*Sqrt[2]*b^(5/4)*(b*B - A*c)*
(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]))/c^(1/4) + (45
*Sqrt[2]*b^(5/4)*(b*B - A*c)*(Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - Log[Sqrt[b] + Sqrt[
2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/c^(1/4))/(180*c^3)

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 330, normalized size = 1.2 \begin{align*}{\frac{2\,B}{9\,c}{x}^{{\frac{9}{2}}}}+{\frac{2\,A}{5\,c}{x}^{{\frac{5}{2}}}}-{\frac{2\,Bb}{5\,{c}^{2}}{x}^{{\frac{5}{2}}}}-2\,{\frac{Ab\sqrt{x}}{{c}^{2}}}+2\,{\frac{B{b}^{2}\sqrt{x}}{{c}^{3}}}+{\frac{b\sqrt{2}A}{2\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{b\sqrt{2}A}{2\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{b\sqrt{2}A}{4\,{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 ) }-{\frac{{b}^{2}\sqrt{2}B}{2\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{{b}^{2}\sqrt{2}B}{2\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{{b}^{2}\sqrt{2}B}{4\,{c}^{3}}\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),x)

[Out]

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

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.2737, size = 1477, normalized size = 5.35 \begin{align*} \frac{180 \, c^{3} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{c^{6} \sqrt{-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}} +{\left (B^{2} b^{4} - 2 \, A B b^{3} c + A^{2} b^{2} c^{2}\right )} x} c^{10} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{3}{4}} +{\left (B b^{2} c^{10} - A b c^{11}\right )} \sqrt{x} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{3}{4}}}{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}\right ) + 45 \, c^{3} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{1}{4}} \log \left (c^{3} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{1}{4}} -{\left (B b^{2} - A b c\right )} \sqrt{x}\right ) - 45 \, c^{3} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{1}{4}} \log \left (-c^{3} \left (-\frac{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac{1}{4}} -{\left (B b^{2} - A b c\right )} \sqrt{x}\right ) + 4 \,{\left (5 \, B c^{2} x^{4} + 45 \, B b^{2} - 45 \, A b c - 9 \,{\left (B b c - A c^{2}\right )} x^{2}\right )} \sqrt{x}}{90 \, c^{3}} \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),x, algorithm="fricas")

[Out]

1/90*(180*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4)*arct
an((sqrt(c^6*sqrt(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13) + (B^2*
b^4 - 2*A*B*b^3*c + A^2*b^2*c^2)*x)*c^10*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^
4*b^5*c^4)/c^13)^(3/4) + (B*b^2*c^10 - A*b*c^11)*sqrt(x)*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^
3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(3/4))/(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b
^5*c^4)) + 45*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4)*
log(c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4) - (B*b^2 -
A*b*c)*sqrt(x)) - 45*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13
)^(1/4)*log(-c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4) -
(B*b^2 - A*b*c)*sqrt(x)) + 4*(5*B*c^2*x^4 + 45*B*b^2 - 45*A*b*c - 9*(B*b*c - A*c^2)*x^2)*sqrt(x))/c^3

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.16315, size = 402, normalized size = 1.46 \begin{align*} -\frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{1}{4}} A b 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 )}{2 \, c^{4}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{1}{4}} A b 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 )}{2 \, c^{4}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{1}{4}} A b c\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, c^{4}} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{1}{4}} A b c\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, c^{4}} + \frac{2 \,{\left (5 \, B c^{8} x^{\frac{9}{2}} - 9 \, B b c^{7} x^{\frac{5}{2}} + 9 \, A c^{8} x^{\frac{5}{2}} + 45 \, B b^{2} c^{6} \sqrt{x} - 45 \, A b c^{7} \sqrt{x}\right )}}{45 \, c^{9}} \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),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/
(b/c)^(1/4))/c^4 - 1/2*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 - 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*log(-sqrt(2)*sq
rt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 2/45*(5*B*c^8*x^(9/2) - 9*B*b*c^7*x^(5/2) + 9*A*c^8*x^(5/2) + 45*B*b^
2*c^6*sqrt(x) - 45*A*b*c^7*sqrt(x))/c^9