3.197 \(\int \frac{x^{15/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=310 \[ \frac{x^{5/2} (9 b B-5 A c)}{10 b c^2}-\frac{\sqrt{x} (9 b B-5 A c)}{2 c^3}-\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}-\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{13/4}}-\frac{x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

-((9*b*B - 5*A*c)*Sqrt[x])/(2*c^3) + ((9*b*B - 5*A*c)*x^(5/2))/(10*b*c^2) - ((b*B - A*c)*x^(9/2))/(2*b*c*(b +
c*x^2)) - (b^(1/4)*(9*b*B - 5*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(13/4)) + (b^(1
/4)*(9*b*B - 5*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(13/4)) - (b^(1/4)*(9*b*B - 5*
A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(13/4)) + (b^(1/4)*(9*b*B - 5*A*
c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(13/4))

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Rubi [A]  time = 0.253366, 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, 211, 1165, 628, 1162, 617, 204} \[ \frac{x^{5/2} (9 b B-5 A c)}{10 b c^2}-\frac{\sqrt{x} (9 b B-5 A c)}{2 c^3}-\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}-\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{13/4}}-\frac{x^{9/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-((9*b*B - 5*A*c)*Sqrt[x])/(2*c^3) + ((9*b*B - 5*A*c)*x^(5/2))/(10*b*c^2) - ((b*B - A*c)*x^(9/2))/(2*b*c*(b +
c*x^2)) - (b^(1/4)*(9*b*B - 5*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(13/4)) + (b^(1
/4)*(9*b*B - 5*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(13/4)) - (b^(1/4)*(9*b*B - 5*
A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(13/4)) + (b^(1/4)*(9*b*B - 5*A*
c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*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 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^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^{7/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}+\frac{\left (\frac{9 b B}{2}-\frac{5 A c}{2}\right ) \int \frac{x^{7/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}-\frac{(9 b B-5 A c) \int \frac{x^{3/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}+\frac{(b (9 b B-5 A c)) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{4 c^3}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}+\frac{(b (9 b B-5 A c)) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 c^3}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}+\frac{\left (\sqrt{b} (9 b B-5 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^3}+\frac{\left (\sqrt{b} (9 b B-5 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^3}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}+\frac{\left (\sqrt{b} (9 b B-5 A c)\right ) \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^{7/2}}+\frac{\left (\sqrt{b} (9 b B-5 A c)\right ) \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^{7/2}}-\frac{\left (\sqrt [4]{b} (9 b B-5 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^{13/4}}-\frac{\left (\sqrt [4]{b} (9 b B-5 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^{13/4}}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}-\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}+\frac{\left (\sqrt [4]{b} (9 b B-5 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^{13/4}}-\frac{\left (\sqrt [4]{b} (9 b B-5 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^{13/4}}\\ &=-\frac{(9 b B-5 A c) \sqrt{x}}{2 c^3}+\frac{(9 b B-5 A c) x^{5/2}}{10 b c^2}-\frac{(b B-A c) x^{9/2}}{2 b c \left (b+c x^2\right )}-\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{13/4}}-\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}+\frac{\sqrt [4]{b} (9 b B-5 A c) \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{13/4}}\\ \end{align*}

Mathematica [A]  time = 0.398497, size = 385, normalized size = 1.24 \[ \frac{-10 \sqrt{2} \sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )+10 \sqrt{2} \sqrt [4]{b} (9 b B-5 A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )+\frac{40 A b c^{5/4} \sqrt{x}}{b+c x^2}+25 \sqrt{2} A \sqrt [4]{b} c \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-25 \sqrt{2} A \sqrt [4]{b} c \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+160 A c^{5/4} \sqrt{x}-\frac{40 b^2 B \sqrt [4]{c} \sqrt{x}}{b+c x^2}-45 \sqrt{2} b^{5/4} B \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+45 \sqrt{2} b^{5/4} B \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-320 b B \sqrt [4]{c} \sqrt{x}+32 B c^{5/4} x^{5/2}}{80 c^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(-320*b*B*c^(1/4)*Sqrt[x] + 160*A*c^(5/4)*Sqrt[x] + 32*B*c^(5/4)*x^(5/2) - (40*b^2*B*c^(1/4)*Sqrt[x])/(b + c*x
^2) + (40*A*b*c^(5/4)*Sqrt[x])/(b + c*x^2) - 10*Sqrt[2]*b^(1/4)*(9*b*B - 5*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sq
rt[x])/b^(1/4)] + 10*Sqrt[2]*b^(1/4)*(9*b*B - 5*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 45*Sqrt[2
]*b^(5/4)*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 25*Sqrt[2]*A*b^(1/4)*c*Log[Sqrt[b] -
Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 45*Sqrt[2]*b^(5/4)*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt
[x] + Sqrt[c]*x] - 25*Sqrt[2]*A*b^(1/4)*c*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(80*c^(1
3/4))

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Maple [A]  time = 0.013, size = 339, normalized size = 1.1 \begin{align*}{\frac{2\,B}{5\,{c}^{2}}{x}^{{\frac{5}{2}}}}+2\,{\frac{A\sqrt{x}}{{c}^{2}}}-4\,{\frac{Bb\sqrt{x}}{{c}^{3}}}+{\frac{Ab}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }\sqrt{x}}-{\frac{B{b}^{2}}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }\sqrt{x}}-{\frac{5\,\sqrt{2}A}{8\,{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}{8\,{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}{16\,{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{9\,b\sqrt{2}B}{8\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{9\,b\sqrt{2}B}{8\,{c}^{3}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{9\,b\sqrt{2}B}{16\,{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

2/5/c^2*B*x^(5/2)+2/c^2*A*x^(1/2)-4/c^3*B*b*x^(1/2)+1/2*b/c^2*x^(1/2)/(c*x^2+b)*A-1/2*b^2/c^3*x^(1/2)/(c*x^2+b
)*B-5/8/c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-5/8/c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2
^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-5/16/c^2*(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)))+9/8*b/c^3*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+
1)+9/8*b/c^3*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+9/16*b/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)))

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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^(15/2)*(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 = 2.46764, size = 1775, normalized size = 5.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/40*(20*(c^4*x^2 + b*c^3)*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 +
625*A^4*b*c^4)/c^13)^(1/4)*arctan((sqrt(c^6*sqrt(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 -
4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13) + (81*B^2*b^2 - 90*A*B*b*c + 25*A^2*c^2)*x)*c^10*(-(6561*B^4*b^5 -
14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(3/4) + (9*B*b*c^10 - 5*
A*c^11)*sqrt(x)*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c
^4)/c^13)^(3/4))/(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^
4)) + 5*(c^4*x^2 + b*c^3)*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 6
25*A^4*b*c^4)/c^13)^(1/4)*log(c^3*(-(6561*B^4*b^5 - 14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2
*c^3 + 625*A^4*b*c^4)/c^13)^(1/4) - (9*B*b - 5*A*c)*sqrt(x)) - 5*(c^4*x^2 + b*c^3)*(-(6561*B^4*b^5 - 14580*A*B
^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4)*log(-c^3*(-(6561*B^4*b^5 -
14580*A*B^3*b^4*c + 12150*A^2*B^2*b^3*c^2 - 4500*A^3*B*b^2*c^3 + 625*A^4*b*c^4)/c^13)^(1/4) - (9*B*b - 5*A*c)*
sqrt(x)) - 4*(4*B*c^2*x^4 - 45*B*b^2 + 25*A*b*c - 4*(9*B*b*c - 5*A*c^2)*x^2)*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**(15/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.32878, size = 402, normalized size = 1.3 \begin{align*} \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 5 \, \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 \, c^{4}} + \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 5 \, \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 \, c^{4}} + \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 5 \, \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 \, c^{4}} - \frac{\sqrt{2}{\left (9 \, \left (b c^{3}\right )^{\frac{1}{4}} B b - 5 \, \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 \, c^{4}} - \frac{B b^{2} \sqrt{x} - A b c \sqrt{x}}{2 \,{\left (c x^{2} + b\right )} c^{3}} + \frac{2 \,{\left (B c^{8} x^{\frac{5}{2}} - 10 \, B b c^{7} \sqrt{x} + 5 \, A c^{8} \sqrt{x}\right )}}{5 \, c^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/8*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(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))/c^4 + 1/8*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(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))/c^4 + 1/16*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 1/16*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - 5*(b*c^3)^(1/4)*A*c)*log(-sqrt(2)*s
qrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 1/2*(B*b^2*sqrt(x) - A*b*c*sqrt(x))/((c*x^2 + b)*c^3) + 2/5*(B*c^8*x
^(5/2) - 10*B*b*c^7*sqrt(x) + 5*A*c^8*sqrt(x))/c^10