3.212 \(\int \frac{x^{13/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=298 \[ \frac{(5 A c+3 b B) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(5 A c+3 b B) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(5 A c+3 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(5 A c+3 b B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{x^{3/2} (5 A c+3 b B)}{16 b^2 c \left (b+c x^2\right )}-\frac{x^{3/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

[Out]

-((b*B - A*c)*x^(3/2))/(4*b*c*(b + c*x^2)^2) + ((3*b*B + 5*A*c)*x^(3/2))/(16*b^2*c*(b + c*x^2)) - ((3*b*B + 5*
A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(9/4)*c^(7/4)) + ((3*b*B + 5*A*c)*ArcTan[1 +
(Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(9/4)*c^(7/4)) + ((3*b*B + 5*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(9/4)*c^(7/4)) - ((3*b*B + 5*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/
4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(9/4)*c^(7/4))

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Rubi [A]  time = 0.22876, antiderivative size = 298, 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, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{(5 A c+3 b B) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(5 A c+3 b B) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(5 A c+3 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(5 A c+3 b B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{x^{3/2} (5 A c+3 b B)}{16 b^2 c \left (b+c x^2\right )}-\frac{x^{3/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

-((b*B - A*c)*x^(3/2))/(4*b*c*(b + c*x^2)^2) + ((3*b*B + 5*A*c)*x^(3/2))/(16*b^2*c*(b + c*x^2)) - ((3*b*B + 5*
A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(9/4)*c^(7/4)) + ((3*b*B + 5*A*c)*ArcTan[1 +
(Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(9/4)*c^(7/4)) + ((3*b*B + 5*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(9/4)*c^(7/4)) - ((3*b*B + 5*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/
4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(9/4)*c^(7/4))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -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{x^{13/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{\sqrt{x} \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{\left (\frac{3 b B}{2}+\frac{5 A c}{2}\right ) \int \frac{\sqrt{x}}{\left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}+\frac{(3 b B+5 A c) \int \frac{\sqrt{x}}{b+c x^2} \, dx}{32 b^2 c}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}+\frac{(3 b B+5 A c) \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 b^2 c}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}-\frac{(3 b B+5 A c) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^2 c^{3/2}}+\frac{(3 b B+5 A c) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^2 c^{3/2}}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}+\frac{(3 b B+5 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 )}{64 b^2 c^2}+\frac{(3 b B+5 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 )}{64 b^2 c^2}+\frac{(3 b B+5 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 )}{64 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(3 b B+5 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 )}{64 \sqrt{2} b^{9/4} c^{7/4}}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}+\frac{(3 b B+5 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(3 b B+5 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(3 b B+5 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 )}{32 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(3 b B+5 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 )}{32 \sqrt{2} b^{9/4} c^{7/4}}\\ &=-\frac{(b B-A c) x^{3/2}}{4 b c \left (b+c x^2\right )^2}+\frac{(3 b B+5 A c) x^{3/2}}{16 b^2 c \left (b+c x^2\right )}-\frac{(3 b B+5 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(3 b B+5 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{9/4} c^{7/4}}+\frac{(3 b B+5 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}-\frac{(3 b B+5 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{9/4} c^{7/4}}\\ \end{align*}

Mathematica [C]  time = 0.054874, size = 62, normalized size = 0.21 \[ \frac{2 x^{3/2} \left ((A c-b B) \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};-\frac{c x^2}{b}\right )+b B \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{c x^2}{b}\right )\right )}{3 b^3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(2*x^(3/2)*(b*B*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)] + (-(b*B) + A*c)*Hypergeometric2F1[3/4, 3, 7/4, -
((c*x^2)/b)]))/(3*b^3*c)

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Maple [A]  time = 0.016, size = 335, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{2}+b \right ) ^{2}} \left ( 1/32\,{\frac{ \left ( 5\,Ac+3\,Bb \right ){x}^{7/2}}{{b}^{2}}}+1/32\,{\frac{ \left ( 9\,Ac-Bb \right ){x}^{3/2}}{bc}} \right ) }+{\frac{5\,\sqrt{2}A}{64\,{b}^{2}c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{5\,\sqrt{2}A}{64\,{b}^{2}c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{5\,\sqrt{2}A}{128\,{b}^{2}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{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{3\,\sqrt{2}B}{64\,b{c}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{3\,\sqrt{2}B}{64\,b{c}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{3\,\sqrt{2}B}{128\,b{c}^{2}}\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)^3,x)

[Out]

2*(1/32*(5*A*c+3*B*b)/b^2*x^(7/2)+1/32*(9*A*c-B*b)/b/c*x^(3/2))/(c*x^2+b)^2+5/64/b^2/c/(b/c)^(1/4)*2^(1/2)*A*a
rctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+5/64/b^2/c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+5/
128/b^2/c/(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)))+3/64/b/c^2/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+3/64/b/c^2/(b/c)^(1/4)*2^
(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+3/128/b/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^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.76955, size = 2276, normalized size = 7.64 \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)^3,x, algorithm="fricas")

[Out]

-1/64*(4*(b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(81*B^4*b^4 + 540*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 + 1500*A
^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(1/4)*arctan((sqrt((729*B^6*b^6 + 7290*A*B^5*b^5*c + 30375*A^2*B^4*b^4*c^
2 + 67500*A^3*B^3*b^3*c^3 + 84375*A^4*B^2*b^2*c^4 + 56250*A^5*B*b*c^5 + 15625*A^6*c^6)*x - (81*B^4*b^9*c^3 + 5
40*A*B^3*b^8*c^4 + 1350*A^2*B^2*b^7*c^5 + 1500*A^3*B*b^6*c^6 + 625*A^4*b^5*c^7)*sqrt(-(81*B^4*b^4 + 540*A*B^3*
b^3*c + 1350*A^2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7)))*b^2*c^2*(-(81*B^4*b^4 + 540*A*B^3*b
^3*c + 1350*A^2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(1/4) - (27*B^3*b^5*c^2 + 135*A*B^2*b
^4*c^3 + 225*A^2*B*b^3*c^4 + 125*A^3*b^2*c^5)*sqrt(x)*(-(81*B^4*b^4 + 540*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 +
1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(1/4))/(81*B^4*b^4 + 540*A*B^3*b^3*c + 1350*A^2*B^2*b^2*c^2 + 1500
*A^3*B*b*c^3 + 625*A^4*c^4)) - (b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(81*B^4*b^4 + 540*A*B^3*b^3*c + 1350*A^
2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(1/4)*log(b^7*c^5*(-(81*B^4*b^4 + 540*A*B^3*b^3*c +
1350*A^2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(3/4) + (27*B^3*b^3 + 135*A*B^2*b^2*c + 225
*A^2*B*b*c^2 + 125*A^3*c^3)*sqrt(x)) + (b^2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)*(-(81*B^4*b^4 + 540*A*B^3*b^3*c +
1350*A^2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(1/4)*log(-b^7*c^5*(-(81*B^4*b^4 + 540*A*B^
3*b^3*c + 1350*A^2*B^2*b^2*c^2 + 1500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^7))^(3/4) + (27*B^3*b^3 + 135*A*B^2*b^
2*c + 225*A^2*B*b*c^2 + 125*A^3*c^3)*sqrt(x)) - 4*((3*B*b*c + 5*A*c^2)*x^3 - (B*b^2 - 9*A*b*c)*x)*sqrt(x))/(b^
2*c^3*x^4 + 2*b^3*c^2*x^2 + b^4*c)

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.39314, size = 402, normalized size = 1.35 \begin{align*} \frac{3 \, B b c x^{\frac{7}{2}} + 5 \, A c^{2} x^{\frac{7}{2}} - B b^{2} x^{\frac{3}{2}} + 9 \, A b c x^{\frac{3}{2}}}{16 \,{\left (c x^{2} + b\right )}^{2} b^{2} c} + \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{3}{4}} B b + 5 \, \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 )}{64 \, b^{3} c^{4}} + \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{3}{4}} B b + 5 \, \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 )}{64 \, b^{3} c^{4}} - \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{3}{4}} B b + 5 \, \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 )}{128 \, b^{3} c^{4}} + \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{3}{4}} B b + 5 \, \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 )}{128 \, b^{3} c^{4}} \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)^3,x, algorithm="giac")

[Out]

1/16*(3*B*b*c*x^(7/2) + 5*A*c^2*x^(7/2) - B*b^2*x^(3/2) + 9*A*b*c*x^(3/2))/((c*x^2 + b)^2*b^2*c) + 1/64*sqrt(2
)*(3*(b*c^3)^(3/4)*B*b + 5*(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^3*c^4) + 1/64*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + 5*(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^3*c^4) - 1/128*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + 5*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*
sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^4) + 1/128*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + 5*(b*c^3)^(3/4)*A*c)*log
(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^4)