3.183 \(\int \frac{x^{13/2} (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=278 \[ -\frac{b^{7/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{15/4}}+\frac{b^{7/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{15/4}}+\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{15/4}}-\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} c^{15/4}}-\frac{2 x^{7/2} (b B-A c)}{7 c^2}+\frac{2 b x^{3/2} (b B-A c)}{3 c^3}+\frac{2 B x^{11/2}}{11 c} \]

[Out]

(2*b*(b*B - A*c)*x^(3/2))/(3*c^3) - (2*(b*B - A*c)*x^(7/2))/(7*c^2) + (2*B*x^(11/2))/(11*c) + (b^(7/4)*(b*B -
A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(15/4)) - (b^(7/4)*(b*B - A*c)*ArcTan[1 + (Sqrt
[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(15/4)) - (b^(7/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)
*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(15/4)) + (b^(7/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(15/4))

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Rubi [A]  time = 0.274033, antiderivative size = 278, 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, 297, 1162, 617, 204, 1165, 628} \[ -\frac{b^{7/4} (b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{15/4}}+\frac{b^{7/4} (b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} c^{15/4}}+\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{15/4}}-\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} c^{15/4}}-\frac{2 x^{7/2} (b B-A c)}{7 c^2}+\frac{2 b x^{3/2} (b B-A c)}{3 c^3}+\frac{2 B x^{11/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(2*b*(b*B - A*c)*x^(3/2))/(3*c^3) - (2*(b*B - A*c)*x^(7/2))/(7*c^2) + (2*B*x^(11/2))/(11*c) + (b^(7/4)*(b*B -
A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(15/4)) - (b^(7/4)*(b*B - A*c)*ArcTan[1 + (Sqrt
[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(15/4)) - (b^(7/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)
*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(15/4)) + (b^(7/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(15/4))

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 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{x^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac{x^{9/2} \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac{2 B x^{11/2}}{11 c}-\frac{\left (2 \left (\frac{11 b B}{2}-\frac{11 A c}{2}\right )\right ) \int \frac{x^{9/2}}{b+c x^2} \, dx}{11 c}\\ &=-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}+\frac{(b (b B-A c)) \int \frac{x^{5/2}}{b+c x^2} \, dx}{c^2}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}-\frac{\left (b^2 (b B-A c)\right ) \int \frac{\sqrt{x}}{b+c x^2} \, dx}{c^3}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}-\frac{\left (2 b^2 (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{c^3}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}+\frac{\left (b^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^{7/2}}-\frac{\left (b^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^{7/2}}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}-\frac{\left (b^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^4}-\frac{\left (b^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^4}-\frac{\left (b^{7/4} (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^{15/4}}-\frac{\left (b^{7/4} (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^{15/4}}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}-\frac{b^{7/4} (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^{15/4}}+\frac{b^{7/4} (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^{15/4}}-\frac{\left (b^{7/4} (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^{15/4}}+\frac{\left (b^{7/4} (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^{15/4}}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{7/2}}{7 c^2}+\frac{2 B x^{11/2}}{11 c}+\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{15/4}}-\frac{b^{7/4} (b B-A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} c^{15/4}}-\frac{b^{7/4} (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^{15/4}}+\frac{b^{7/4} (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^{15/4}}\\ \end{align*}

Mathematica [A]  time = 0.230452, size = 133, normalized size = 0.48 \[ \frac{2 x^{3/2} \left (-11 b c \left (7 A+3 B x^2\right )+3 c^2 x^2 \left (11 A+7 B x^2\right )+77 b^2 B\right )}{231 c^3}+\frac{b (-b)^{3/4} (b B-A c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac{(-b)^{7/4} (b B-A c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{c^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(2*x^(3/2)*(77*b^2*B - 11*b*c*(7*A + 3*B*x^2) + 3*c^2*x^2*(11*A + 7*B*x^2)))/(231*c^3) + ((-b)^(3/4)*b*(b*B -
A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/c^(15/4) + ((-b)^(7/4)*(b*B - A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(
1/4)])/c^(15/4)

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Maple [A]  time = 0.033, size = 336, normalized size = 1.2 \begin{align*}{\frac{2\,B}{11\,c}{x}^{{\frac{11}{2}}}}+{\frac{2\,A}{7\,c}{x}^{{\frac{7}{2}}}}-{\frac{2\,Bb}{7\,{c}^{2}}{x}^{{\frac{7}{2}}}}-{\frac{2\,Ab}{3\,{c}^{2}}{x}^{{\frac{3}{2}}}}+{\frac{2\,B{b}^{2}}{3\,{c}^{3}}{x}^{{\frac{3}{2}}}}+{\frac{{b}^{2}\sqrt{2}A}{2\,{c}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{{b}^{2}\sqrt{2}A}{2\,{c}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{{b}^{2}\sqrt{2}A}{4\,{c}^{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{{b}^{3}\sqrt{2}B}{2\,{c}^{4}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{{b}^{3}\sqrt{2}B}{2\,{c}^{4}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-{\frac{{b}^{3}\sqrt{2}B}{4\,{c}^{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}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

2/11*B*x^(11/2)/c+2/7/c*x^(7/2)*A-2/7/c^2*x^(7/2)*B*b-2/3/c^2*x^(3/2)*A*b+2/3/c^3*x^(3/2)*B*b^2+1/2*b^2/c^3/(b
/c)^(1/4)*2^(1/2)*A*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)*A*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)*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^3/c^4/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/
2*b^3/c^4/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4*b^3/c^4/(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^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.26914, size = 1914, normalized size = 6.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/462*(924*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*
arctan((sqrt((B^6*b^16 - 6*A*B^5*b^15*c + 15*A^2*B^4*b^14*c^2 - 20*A^3*B^3*b^13*c^3 + 15*A^4*B^2*b^12*c^4 - 6*
A^5*B*b^11*c^5 + A^6*b^10*c^6)*x - (B^4*b^11*c^7 - 4*A*B^3*b^10*c^8 + 6*A^2*B^2*b^9*c^9 - 4*A^3*B*b^8*c^10 + A
^4*b^7*c^11)*sqrt(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15))*c^4*
(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4) + (B^3*b^8*c^4 -
3*A*B^2*b^7*c^5 + 3*A^2*B*b^6*c^6 - A^3*b^5*c^7)*sqrt(x)*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4
*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4))/(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 +
A^4*b^7*c^4)) - 231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15
)^(1/4)*log(c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4)
- (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b^5*c^3)*sqrt(x)) + 231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c +
6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*log(-c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2
*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4) - (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b
^5*c^3)*sqrt(x)) - 4*(21*B*c^2*x^5 - 33*(B*b*c - A*c^2)*x^3 + 77*(B*b^2 - A*b*c)*x)*sqrt(x))/c^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(x**(13/2)*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

Timed out

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Giac [A]  time = 1.16441, size = 402, normalized size = 1.45 \begin{align*} -\frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{3}{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^{6}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{3}{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^{6}} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{3}{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^{6}} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac{3}{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^{6}} + \frac{2 \,{\left (21 \, B c^{10} x^{\frac{11}{2}} - 33 \, B b c^{9} x^{\frac{7}{2}} + 33 \, A c^{10} x^{\frac{7}{2}} + 77 \, B b^{2} c^{8} x^{\frac{3}{2}} - 77 \, A b c^{9} x^{\frac{3}{2}}\right )}}{231 \, c^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/
(b/c)^(1/4))/c^6 - 1/2*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^6 + 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 - 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*log(-sqrt(2)*sq
rt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/231*(21*B*c^10*x^(11/2) - 33*B*b*c^9*x^(7/2) + 33*A*c^10*x^(7/2) +
77*B*b^2*c^8*x^(3/2) - 77*A*b*c^9*x^(3/2))/c^11