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

Optimal. Leaf size=289 \[ \frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{(3 b B-7 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{(3 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.225164, antiderivative size = 289, 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, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{(3 b B-7 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{(3 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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 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^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{A+B x^2}{x^{5/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac{\left (-\frac{3 b B}{2}+\frac{7 A c}{2}\right ) \int \frac{1}{x^{5/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac{(3 b B-7 A c) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac{(3 b B-7 A c) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 b^2}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac{(3 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 b^{5/2}}+\frac{(3 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 b^{5/2}}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac{(3 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 b^{5/2} \sqrt{c}}+\frac{(3 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 b^{5/2} \sqrt{c}}-\frac{(3 b B-7 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 )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{(3 b B-7 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 )}{8 \sqrt{2} b^{11/4} \sqrt [4]{c}}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}-\frac{(3 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} b^{11/4} \sqrt [4]{c}}+\frac{(3 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} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 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 )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{(3 b B-7 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 )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}\\ &=\frac{3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac{b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}-\frac{(3 b B-7 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}+\frac{(3 b B-7 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} b^{11/4} \sqrt [4]{c}}-\frac{(3 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} b^{11/4} \sqrt [4]{c}}+\frac{(3 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} b^{11/4} \sqrt [4]{c}}\\ \end{align*}

Mathematica [A]  time = 0.423149, size = 355, normalized size = 1.23 \[ \frac{-\frac{24 A b^{3/4} c \sqrt{x}}{b+c x^2}-\frac{32 A b^{3/4}}{x^{3/2}}+\frac{6 \sqrt{2} (7 A c-3 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac{6 \sqrt{2} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}+21 \sqrt{2} A c^{3/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-21 \sqrt{2} A c^{3/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+\frac{24 b^{7/4} B \sqrt{x}}{b+c x^2}-\frac{9 \sqrt{2} b B \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}+\frac{9 \sqrt{2} b B \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}}{48 b^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*A*b^(3/4))/x^(3/2) + (24*b^(7/4)*B*Sqrt[x])/(b + c*x^2) - (24*A*b^(3/4)*c*Sqrt[x])/(b + c*x^2) + (6*Sqrt
[2]*(-3*b*B + 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) + (6*Sqrt[2]*(3*b*B - 7*A*c)*ArcTa
n[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) - (9*Sqrt[2]*b*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/c^(1/4) + 21*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (
9*Sqrt[2]*b*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) - 21*Sqrt[2]*A*c^(3/4)*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(48*b^(11/4))

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Maple [A]  time = 0.015, size = 317, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{3\,{b}^{2}}{x}^{-{\frac{3}{2}}}}-{\frac{Ac}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }\sqrt{x}}+{\frac{B}{2\,b \left ( c{x}^{2}+b \right ) }\sqrt{x}}-{\frac{7\,\sqrt{2}Ac}{8\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{7\,\sqrt{2}Ac}{16\,{b}^{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{7\,\sqrt{2}Ac}{8\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{3\,\sqrt{2}B}{8\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}B}{16\,{b}^{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{3\,\sqrt{2}B}{8\,{b}^{2}}\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^(3/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

-2/3*A/b^2/x^(3/2)-1/2/b^2*x^(1/2)/(c*x^2+b)*A*c+1/2/b*x^(1/2)/(c*x^2+b)*B-7/8/b^3*(b/c)^(1/4)*2^(1/2)*A*arcta
n(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)*c-7/16/b^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)))*c-7/8/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(
1/2)+1)*c+3/8/b^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+3/16/b^2*(b/c)^(1/4)*2^(1/2)*B*l
n((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)))+3/8/b^2*(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^(3/2)*(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 = 2.4927, size = 1700, normalized size = 5.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(12*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 24
01*A^4*c^4)/(b^11*c))^(1/4)*arctan((sqrt(b^6*sqrt(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116
*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c)) + (9*B^2*b^2 - 42*A*B*b*c + 49*A^2*c^2)*x)*b^8*c*(-(81*B^4*b^4 - 756*A*
B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(3/4) + (3*B*b^9*c - 7*A*b^8*c^2
)*sqrt(x)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^
(3/4))/(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)) + 3*(b^2*c*x^4
 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c)
)^(1/4)*log(b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^1
1*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) - 3*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^
2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4)*log(-b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A
^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) - 4*((3*B*b - 7*A
*c)*x^2 - 4*A*b)*sqrt(x))/(b^2*c*x^4 + b^3*x^2)

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

[Out]

Timed out

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Giac [A]  time = 1.26993, size = 382, normalized size = 1.32 \begin{align*} \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 7 \, \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 )}{8 \, b^{3} c} + \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 7 \, \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 )}{8 \, b^{3} c} + \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 7 \, \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 )}{16 \, b^{3} c} - \frac{\sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 7 \, \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 )}{16 \, b^{3} c} + \frac{B b \sqrt{x} - A c \sqrt{x}}{2 \,{\left (c x^{2} + b\right )} b^{2}} - \frac{2 \, A}{3 \, b^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/8*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(b*c^3)^(1/4)*A*c)*log(sqrt
(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) - 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(b*c^3)^(1/4)*A*c)*lo
g(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/2*(B*b*sqrt(x) - A*c*sqrt(x))/((c*x^2 + b)*b^2) -
2/3*A/(b^2*x^(3/2))