3.215 \(\int \frac{x^{7/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=322 \[ -\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{7 (3 b B-11 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2} \]

[Out]

(7*(3*b*B - 11*A*c))/(48*b^3*c*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(3/2)*(b + c*x^2)^2) - (3*b*B - 11*A*c)/(16*b^2
*c*x^(3/2)*(b + c*x^2)) - (7*(3*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15
/4)*c^(1/4)) + (7*(3*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15/4)*c^(1/4)
) - (7*(3*b*B - 11*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*c^(1/
4)) + (7*(3*b*B - 11*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*c^(
1/4))

________________________________________________________________________________________

Rubi [A]  time = 0.250694, 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, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{7 (3 b B-11 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(3*b*B - 11*A*c))/(48*b^3*c*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(3/2)*(b + c*x^2)^2) - (3*b*B - 11*A*c)/(16*b^2
*c*x^(3/2)*(b + c*x^2)) - (7*(3*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15
/4)*c^(1/4)) + (7*(3*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(15/4)*c^(1/4)
) - (7*(3*b*B - 11*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*c^(1/
4)) + (7*(3*b*B - 11*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(15/4)*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 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 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^{7/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{A+B x^2}{x^{5/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}+\frac{\left (-\frac{3 b B}{2}+\frac{11 A c}{2}\right ) \int \frac{1}{x^{5/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac{(7 (3 b B-11 A c)) \int \frac{1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{(7 (3 b B-11 A c)) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{(7 (3 b B-11 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 b^3}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{(7 (3 b B-11 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{7/2}}+\frac{(7 (3 b B-11 A c)) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b^{7/2}}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}+\frac{(7 (3 b B-11 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^{7/2} \sqrt{c}}+\frac{(7 (3 b B-11 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^{7/2} \sqrt{c}}-\frac{(7 (3 b B-11 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^{15/4} \sqrt [4]{c}}-\frac{(7 (3 b B-11 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^{15/4} \sqrt [4]{c}}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac{7 (3 b B-11 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{(7 (3 b B-11 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^{15/4} \sqrt [4]{c}}-\frac{(7 (3 b B-11 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^{15/4} \sqrt [4]{c}}\\ &=\frac{7 (3 b B-11 A c)}{48 b^3 c x^{3/2}}-\frac{b B-A c}{4 b c x^{3/2} \left (b+c x^2\right )^2}-\frac{3 b B-11 A c}{16 b^2 c x^{3/2} \left (b+c x^2\right )}-\frac{7 (3 b B-11 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{15/4} \sqrt [4]{c}}-\frac{7 (3 b B-11 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}+\frac{7 (3 b B-11 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{15/4} \sqrt [4]{c}}\\ \end{align*}

Mathematica [A]  time = 0.447229, size = 400, normalized size = 1.24 \[ \frac{-\frac{96 A b^{7/4} c \sqrt{x}}{\left (b+c x^2\right )^2}-\frac{360 A b^{3/4} c \sqrt{x}}{b+c x^2}-\frac{256 A b^{3/4}}{x^{3/2}}+\frac{42 \sqrt{2} (11 A c-3 b B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac{42 \sqrt{2} (3 b B-11 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}+231 \sqrt{2} A c^{3/4} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-231 \sqrt{2} A c^{3/4} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+\frac{96 b^{11/4} B \sqrt{x}}{\left (b+c x^2\right )^2}+\frac{168 b^{7/4} B \sqrt{x}}{b+c x^2}-\frac{63 \sqrt{2} b B \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}+\frac{63 \sqrt{2} b B \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{\sqrt [4]{c}}}{384 b^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-256*A*b^(3/4))/x^(3/2) + (96*b^(11/4)*B*Sqrt[x])/(b + c*x^2)^2 - (96*A*b^(7/4)*c*Sqrt[x])/(b + c*x^2)^2 + (
168*b^(7/4)*B*Sqrt[x])/(b + c*x^2) - (360*A*b^(3/4)*c*Sqrt[x])/(b + c*x^2) + (42*Sqrt[2]*(-3*b*B + 11*A*c)*Arc
Tan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) + (42*Sqrt[2]*(3*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)
*Sqrt[x])/b^(1/4)])/c^(1/4) - (63*Sqrt[2]*b*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1
/4) + 231*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (63*Sqrt[2]*b*B*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) - 231*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] + Sqrt[2]*b
^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(384*b^(15/4))

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Maple [A]  time = 0.017, size = 357, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{3\,{b}^{3}}{x}^{-{\frac{3}{2}}}}-{\frac{15\,A{c}^{2}}{16\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}+{\frac{7\,Bc}{16\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{19\,Ac}{16\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}+{\frac{11\,B}{16\,b \left ( c{x}^{2}+b \right ) ^{2}}\sqrt{x}}-{\frac{77\,\sqrt{2}Ac}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{77\,\sqrt{2}Ac}{64\,{b}^{4}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{77\,\sqrt{2}Ac}{128\,{b}^{4}}\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{21\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{21\,\sqrt{2}B}{64\,{b}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{21\,\sqrt{2}B}{128\,{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

-2/3*A/b^3/x^(3/2)-15/16/b^3/(c*x^2+b)^2*x^(5/2)*A*c^2+7/16/b^2/(c*x^2+b)^2*x^(5/2)*B*c-19/16/b^2/(c*x^2+b)^2*
A*x^(1/2)*c+11/16/b/(c*x^2+b)^2*B*x^(1/2)-77/64/b^4*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1
)*c-77/64/b^4*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)*c-77/128/b^4*(b/c)^(1/4)*2^(1/2)*A*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)))*c+21/64/b^3*(b/c)^(
1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+21/64/b^3*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4
)*x^(1/2)-1)+21/128/b^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)))

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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^(7/2)*(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 = 2.75335, size = 1901, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/192*(84*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 159
72*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*arctan((sqrt(b^8*sqrt(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A
^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c)) + (9*B^2*b^2 - 66*A*B*b*c + 121*A^2*c^2)*x)*b^11
*c*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(3/4
) + (3*B*b^12*c - 11*A*b^11*c^2)*sqrt(x)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B
*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(3/4))/(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b
*c^3 + 14641*A^4*c^4)) + 21*(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*
B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(7*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c +
6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 21*(
b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2)*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972*A^3*B*b*c
^3 + 14641*A^4*c^4)/(b^15*c))^(1/4)*log(-7*b^4*(-(81*B^4*b^4 - 1188*A*B^3*b^3*c + 6534*A^2*B^2*b^2*c^2 - 15972
*A^3*B*b*c^3 + 14641*A^4*c^4)/(b^15*c))^(1/4) - 7*(3*B*b - 11*A*c)*sqrt(x)) - 4*(7*(3*B*b*c - 11*A*c^2)*x^4 -
32*A*b^2 + 11*(3*B*b^2 - 11*A*b*c)*x^2)*sqrt(x))/(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*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**(7/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.31035, size = 410, normalized size = 1.27 \begin{align*} \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \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^{4} c} + \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \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^{4} c} + \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \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^{4} c} - \frac{7 \, \sqrt{2}{\left (3 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 11 \, \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^{4} c} - \frac{2 \, A}{3 \, b^{3} x^{\frac{3}{2}}} + \frac{7 \, B b c x^{\frac{5}{2}} - 15 \, A c^{2} x^{\frac{5}{2}} + 11 \, B b^{2} \sqrt{x} - 19 \, A b c \sqrt{x}}{16 \,{\left (c x^{2} + b\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

7/64*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(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^4*c) + 7/64*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(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^4*c) + 7/128*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*lo
g(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) - 7/128*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)
*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^4*c) - 2/3*A/(b^3*x^(3/2)) + 1/16*(7*B*b*c*x^(5/2)
- 15*A*c^2*x^(5/2) + 11*B*b^2*sqrt(x) - 19*A*b*c*sqrt(x))/((c*x^2 + b)^2*b^3)