3.209 \(\int \frac{x^{19/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=322 \[ \frac{5 (9 b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}+\frac{5 (9 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}-\frac{x^{5/2} (9 b B-A c)}{16 b c^2 \left (b+c x^2\right )}+\frac{5 \sqrt{x} (9 b B-A c)}{16 b c^3}-\frac{x^{9/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

[Out]

(5*(9*b*B - A*c)*Sqrt[x])/(16*b*c^3) - ((b*B - A*c)*x^(9/2))/(4*b*c*(b + c*x^2)^2) - ((9*b*B - A*c)*x^(5/2))/(
16*b*c^2*(b + c*x^2)) + (5*(9*b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^
(13/4)) - (5*(9*b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^(13/4)) + (5*(
9*b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(13/4)) - (5*(9
*b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(13/4))

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Rubi [A]  time = 0.253757, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {1584, 457, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{5 (9 b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}+\frac{5 (9 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}-\frac{x^{5/2} (9 b B-A c)}{16 b c^2 \left (b+c x^2\right )}+\frac{5 \sqrt{x} (9 b B-A c)}{16 b c^3}-\frac{x^{9/2} (b B-A c)}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(19/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(5*(9*b*B - A*c)*Sqrt[x])/(16*b*c^3) - ((b*B - A*c)*x^(9/2))/(4*b*c*(b + c*x^2)^2) - ((9*b*B - A*c)*x^(5/2))/(
16*b*c^2*(b + c*x^2)) + (5*(9*b*B - A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^
(13/4)) - (5*(9*b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(3/4)*c^(13/4)) + (5*(
9*b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*c^(13/4)) - (5*(9
*b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(3/4)*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 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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^{7/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}+\frac{\left (\frac{9 b B}{2}-\frac{A c}{2}\right ) \int \frac{x^{7/2}}{\left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}+\frac{(5 (9 b B-A c)) \int \frac{x^{3/2}}{b+c x^2} \, dx}{32 b c^2}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}-\frac{(5 (9 b B-A c)) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{32 c^3}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}-\frac{(5 (9 b B-A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 c^3}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}-\frac{(5 (9 b B-A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{b} c^3}-\frac{(5 (9 b B-A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{b} c^3}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}-\frac{(5 (9 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 )}{64 \sqrt{b} c^{7/2}}-\frac{(5 (9 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 )}{64 \sqrt{b} c^{7/2}}+\frac{(5 (9 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 )}{64 \sqrt{2} b^{3/4} c^{13/4}}+\frac{(5 (9 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 )}{64 \sqrt{2} b^{3/4} c^{13/4}}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}+\frac{5 (9 b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}-\frac{(5 (9 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 )}{32 \sqrt{2} b^{3/4} c^{13/4}}+\frac{(5 (9 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 )}{32 \sqrt{2} b^{3/4} c^{13/4}}\\ &=\frac{5 (9 b B-A c) \sqrt{x}}{16 b c^3}-\frac{(b B-A c) x^{9/2}}{4 b c \left (b+c x^2\right )^2}-\frac{(9 b B-A c) x^{5/2}}{16 b c^2 \left (b+c x^2\right )}+\frac{5 (9 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{3/4} c^{13/4}}+\frac{5 (9 b B-A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}-\frac{5 (9 b B-A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{3/4} c^{13/4}}\\ \end{align*}

Mathematica [A]  time = 0.457858, size = 403, normalized size = 1.25 \[ \frac{\frac{10 \sqrt{2} (9 b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{b^{3/4}}-\frac{10 \sqrt{2} (9 b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{b^{3/4}}-\frac{5 \sqrt{2} A c \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}+\frac{5 \sqrt{2} A c \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{b^{3/4}}+\frac{32 A b c^{5/4} \sqrt{x}}{\left (b+c x^2\right )^2}-\frac{72 A c^{5/4} \sqrt{x}}{b+c x^2}-\frac{32 b^2 B \sqrt [4]{c} \sqrt{x}}{\left (b+c x^2\right )^2}+\frac{136 b B \sqrt [4]{c} \sqrt{x}}{b+c x^2}+45 \sqrt{2} \sqrt [4]{b} B \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-45 \sqrt{2} \sqrt [4]{b} B \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+256 B \sqrt [4]{c} \sqrt{x}}{128 c^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(19/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(256*B*c^(1/4)*Sqrt[x] - (32*b^2*B*c^(1/4)*Sqrt[x])/(b + c*x^2)^2 + (32*A*b*c^(5/4)*Sqrt[x])/(b + c*x^2)^2 + (
136*b*B*c^(1/4)*Sqrt[x])/(b + c*x^2) - (72*A*c^(5/4)*Sqrt[x])/(b + c*x^2) + (10*Sqrt[2]*(9*b*B - A*c)*ArcTan[1
- (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/b^(3/4) - (10*Sqrt[2]*(9*b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x]
)/b^(1/4)])/b^(3/4) + 45*Sqrt[2]*b^(1/4)*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - (5*Sqr
t[2]*A*c*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/b^(3/4) - 45*Sqrt[2]*b^(1/4)*B*Log[Sqrt[b
] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (5*Sqrt[2]*A*c*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x
] + Sqrt[c]*x])/b^(3/4))/(128*c^(13/4))

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Maple [A]  time = 0.016, size = 363, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{{c}^{3}}}-{\frac{9\,A}{16\,c \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{17\,Bb}{16\,{c}^{2} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{5\,Ab}{16\,{c}^{2} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{13\,B{b}^{2}}{16\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{5\,\sqrt{2}A}{64\,b{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}A}{128\,b{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{5\,\sqrt{2}A}{64\,b{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{45\,\sqrt{2}B}{64\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{45\,\sqrt{2}B}{128\,{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 ) }-{\frac{45\,\sqrt{2}B}{64\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

2*B/c^3*x^(1/2)-9/16/c/(c*x^2+b)^2*x^(5/2)*A+17/16/c^2/(c*x^2+b)^2*x^(5/2)*B*b-5/16/c^2/(c*x^2+b)^2*A*x^(1/2)*
b+13/16/c^3/(c*x^2+b)^2*B*x^(1/2)*b^2+5/64/c^2*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+5
/128/c^2*(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)))+5/64/c^2*(b/c)^(1/4)/b*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-45/64/c^3*(b/c)^(1/4)*2^
(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-45/128/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)))-45/64/c^3*(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^(19/2)*(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 = 2.60046, size = 1787, normalized size = 5.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/64*(20*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B
*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*arctan((sqrt(b^2*c^6*sqrt(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*
b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13)) + (81*B^2*b^2 - 18*A*B*b*c + A^2*c^2)*x)*b^2*c^10*(-(6561*B^4*
b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(3/4) + (9*B*b^3*c^10 - A
*b^2*c^11)*sqrt(x)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c
^13))^(3/4))/(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)) + 5*(c^5*x^4
+ 2*b*c^4*x^2 + b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/
(b^3*c^13))^(1/4)*log(5*b*c^3*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*
c^4)/(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(x)) - 5*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-(6561*B^4*b^4 - 2916
*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*log(-5*b*c^3*(-(6561*B^4*b^4
- 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(
x)) + 4*(32*B*c^2*x^4 + 45*B*b^2 - 5*A*b*c + 9*(9*B*b*c - A*c^2)*x^2)*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*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**(19/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25843, size = 410, normalized size = 1.27 \begin{align*} \frac{2 \, B \sqrt{x}}{c^{3}} - \frac{5 \, \sqrt{2}{\left (9 \, \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 )}{64 \, b c^{4}} - \frac{5 \, \sqrt{2}{\left (9 \, \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 )}{64 \, b c^{4}} - \frac{5 \, \sqrt{2}{\left (9 \, \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 )}{128 \, b c^{4}} + \frac{5 \, \sqrt{2}{\left (9 \, \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 )}{128 \, b c^{4}} + \frac{17 \, B b c x^{\frac{5}{2}} - 9 \, A c^{2} x^{\frac{5}{2}} + 13 \, B b^{2} \sqrt{x} - 5 \, A b c \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

2*B*sqrt(x)/c^3 - 5/64*sqrt(2)*(9*(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^4) - 5/64*sqrt(2)*(9*(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^4) - 5/128*sqrt(2)*(9*(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^4) + 5/128*sqrt(2)*(9*(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^4) + 1/16*(17*B*b*c*x^(5/2) - 9*A*c^2*x^(5/
2) + 13*B*b^2*sqrt(x) - 5*A*b*c*sqrt(x))/((c*x^2 + b)^2*c^3)