3.196 \(\int \frac{x^{17/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=310 \[ \frac{b^{3/4} (11 b B-7 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{15/4}}+\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{15/4}}+\frac{x^{7/2} (11 b B-7 A c)}{14 b c^2}-\frac{x^{3/2} (11 b B-7 A c)}{6 c^3}-\frac{x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

-((11*b*B - 7*A*c)*x^(3/2))/(6*c^3) + ((11*b*B - 7*A*c)*x^(7/2))/(14*b*c^2) - ((b*B - A*c)*x^(11/2))/(2*b*c*(b
+ c*x^2)) - (b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (
b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (b^(3/4)*(11*b*
B - 7*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(15/4)) - (b^(3/4)*(11*b*B
- 7*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(15/4))

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Rubi [A]  time = 0.244669, antiderivative size = 310, 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, 457, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{b^{3/4} (11 b B-7 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{15/4}}+\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{15/4}}+\frac{x^{7/2} (11 b B-7 A c)}{14 b c^2}-\frac{x^{3/2} (11 b B-7 A c)}{6 c^3}-\frac{x^{11/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-((11*b*B - 7*A*c)*x^(3/2))/(6*c^3) + ((11*b*B - 7*A*c)*x^(7/2))/(14*b*c^2) - ((b*B - A*c)*x^(11/2))/(2*b*c*(b
+ c*x^2)) - (b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (
b^(3/4)*(11*b*B - 7*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(15/4)) + (b^(3/4)*(11*b*
B - 7*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(15/4)) - (b^(3/4)*(11*b*B
- 7*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*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 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 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^{17/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^{9/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac{\left (\frac{11 b B}{2}-\frac{7 A c}{2}\right ) \int \frac{x^{9/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac{(11 b B-7 A c) \int \frac{x^{5/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac{(b (11 b B-7 A c)) \int \frac{\sqrt{x}}{b+c x^2} \, dx}{4 c^3}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac{(b (11 b B-7 A c)) \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 c^3}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac{(b (11 b B-7 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^{7/2}}+\frac{(b (11 b B-7 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^{7/2}}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac{(b (11 b B-7 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 c^4}+\frac{(b (11 b B-7 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 c^4}+\frac{\left (b^{3/4} (11 b B-7 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} c^{15/4}}+\frac{\left (b^{3/4} (11 b B-7 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} c^{15/4}}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}+\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}+\frac{\left (b^{3/4} (11 b B-7 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} c^{15/4}}-\frac{\left (b^{3/4} (11 b B-7 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} c^{15/4}}\\ &=-\frac{(11 b B-7 A c) x^{3/2}}{6 c^3}+\frac{(11 b B-7 A c) x^{7/2}}{14 b c^2}-\frac{(b B-A c) x^{11/2}}{2 b c \left (b+c x^2\right )}-\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{15/4}}+\frac{b^{3/4} (11 b B-7 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{15/4}}+\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}-\frac{b^{3/4} (11 b B-7 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{15/4}}\\ \end{align*}

Mathematica [C]  time = 0.220721, size = 154, normalized size = 0.5 \[ \frac{2 x^{3/2} (A c-b B) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{c x^2}{b}\right )}{3 c^3}+\frac{2 x^{3/2} (A c-2 b B)}{3 c^3}-\frac{(-b)^{3/4} (3 b B-2 A c) \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac{(-b)^{3/4} (3 b B-2 A c) \tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac{2 B x^{7/2}}{7 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(17/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(2*(-2*b*B + A*c)*x^(3/2))/(3*c^3) + (2*B*x^(7/2))/(7*c^2) - ((-b)^(3/4)*(3*b*B - 2*A*c)*ArcTan[(c^(1/4)*Sqrt[
x])/(-b)^(1/4)])/c^(15/4) + ((-b)^(3/4)*(3*b*B - 2*A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/c^(15/4) + (2*(
-(b*B) + A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3*c^3)

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Maple [A]  time = 0.015, size = 348, normalized size = 1.1 \begin{align*}{\frac{2\,B}{7\,{c}^{2}}{x}^{{\frac{7}{2}}}}+{\frac{2\,A}{3\,{c}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{4\,Bb}{3\,{c}^{3}}{x}^{{\frac{3}{2}}}}+{\frac{Ab}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}-{\frac{B{b}^{2}}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }{x}^{{\frac{3}{2}}}}-{\frac{7\,b\sqrt{2}A}{16\,{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{7\,b\sqrt{2}A}{8\,{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{7\,b\sqrt{2}A}{8\,{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{11\,{b}^{2}\sqrt{2}B}{16\,{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}}}}}}+{\frac{11\,{b}^{2}\sqrt{2}B}{8\,{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{11\,{b}^{2}\sqrt{2}B}{8\,{c}^{4}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

2/7/c^2*B*x^(7/2)+2/3/c^2*x^(3/2)*A-4/3/c^3*x^(3/2)*B*b+1/2*b/c^2*x^(3/2)/(c*x^2+b)*A-1/2*b^2/c^3*x^(3/2)/(c*x
^2+b)*B-7/16*b/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)))-7/8*b/c^3/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-7/8*b/c^3/(b/c)^(
1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+11/16*b^2/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)))+11/8*b^2/c^4/(b/c)^(1/4)*2^(1/2)*B*arct
an(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+11/8*b^2/c^4/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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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^(17/2)*(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 = 2.64511, size = 2390, normalized size = 7.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/168*(84*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3
+ 2401*A^4*b^3*c^4)/c^15)^(1/4)*arctan((sqrt((1771561*B^6*b^10 - 6764142*A*B^5*b^9*c + 10761135*A^2*B^4*b^8*c
^2 - 9130660*A^3*B^3*b^7*c^3 + 4357815*A^4*B^2*b^6*c^4 - 1109262*A^5*B*b^5*c^5 + 117649*A^6*b^4*c^6)*x - (1464
1*B^4*b^7*c^7 - 37268*A*B^3*b^6*c^8 + 35574*A^2*B^2*b^5*c^9 - 15092*A^3*B*b^4*c^10 + 2401*A^4*b^3*c^11)*sqrt(-
(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15))*c^
4*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)
^(1/4) + (1331*B^3*b^5*c^4 - 2541*A*B^2*b^4*c^5 + 1617*A^2*B*b^3*c^6 - 343*A^3*b^2*c^7)*sqrt(x)*(-(14641*B^4*b
^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(1/4))/(14641*B
^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)) - 21*(c^4*x^2 +
b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401*A^4*b^3*c^4)/
c^15)^(1/4)*log(c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 2401
*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b^2*c^3)*sqrt(x))
+ 21*(c^4*x^2 + b*c^3)*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3*B*b^4*c^3 + 24
01*A^4*b^3*c^4)/c^15)^(1/4)*log(-c^11*(-(14641*B^4*b^7 - 37268*A*B^3*b^6*c + 35574*A^2*B^2*b^5*c^2 - 15092*A^3
*B*b^4*c^3 + 2401*A^4*b^3*c^4)/c^15)^(3/4) - (1331*B^3*b^5 - 2541*A*B^2*b^4*c + 1617*A^2*B*b^3*c^2 - 343*A^3*b
^2*c^3)*sqrt(x)) + 4*(12*B*c^2*x^5 - 4*(11*B*b*c - 7*A*c^2)*x^3 - 7*(11*B*b^2 - 7*A*b*c)*x)*sqrt(x))/(c^4*x^2
+ b*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**(17/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24149, size = 404, normalized size = 1.3 \begin{align*} -\frac{B b^{2} x^{\frac{3}{2}} - A b c x^{\frac{3}{2}}}{2 \,{\left (c x^{2} + b\right )} c^{3}} + \frac{\sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 7 \, \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 \, c^{6}} + \frac{\sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 7 \, \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 \, c^{6}} - \frac{\sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 7 \, \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 \, c^{6}} + \frac{\sqrt{2}{\left (11 \, \left (b c^{3}\right )^{\frac{3}{4}} B b - 7 \, \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 \, c^{6}} + \frac{2 \,{\left (3 \, B c^{12} x^{\frac{7}{2}} - 14 \, B b c^{11} x^{\frac{3}{2}} + 7 \, A c^{12} x^{\frac{3}{2}}\right )}}{21 \, c^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-1/2*(B*b^2*x^(3/2) - A*b*c*x^(3/2))/((c*x^2 + b)*c^3) + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b - 7*(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))/c^6 + 1/8*sqrt(2)*(11*(b*c^3)^(3/4)*B*b
- 7*(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))/c^6 - 1/16*sqrt(2)*(
11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 1/16*sqrt(2
)*(11*(b*c^3)^(3/4)*B*b - 7*(b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/21*(3
*B*c^12*x^(7/2) - 14*B*b*c^11*x^(3/2) + 7*A*c^12*x^(3/2))/c^14