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

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

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Rubi [A]  time = 0.25, antiderivative size = 310, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 )} \end {gather*}

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 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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.55, size = 385, normalized size = 1.24 \begin {gather*} \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}-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 )-\frac {40 b^2 B \sqrt [4]{c} \sqrt {x}}{b+c x^2}-320 b B \sqrt [4]{c} \sqrt {x}+32 B c^{5/4} x^{5/2}}{80 c^{13/4}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.68, size = 208, normalized size = 0.67 \begin {gather*} -\frac {\left (9 b^{5/4} B-5 A \sqrt [4]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {\left (9 b^{5/4} B-5 A \sqrt [4]{b} c\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} c^{13/4}}+\frac {25 A b c \sqrt {x}+20 A c^2 x^{5/2}-45 b^2 B \sqrt {x}-36 b B c x^{5/2}+4 B c^2 x^{9/2}}{10 c^3 \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-45*b^2*B*Sqrt[x] + 25*A*b*c*Sqrt[x] - 36*b*B*c*x^(5/2) + 20*A*c^2*x^(5/2) + 4*B*c^2*x^(9/2))/(10*c^3*(b + c*
x^2)) - ((9*b^(5/4)*B - 5*A*b^(1/4)*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(4*Sqr
t[2]*c^(13/4)) + ((9*b^(5/4)*B - 5*A*b^(1/4)*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x
)])/(4*Sqrt[2]*c^(13/4))

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fricas [B]  time = 0.45, size = 748, normalized size = 2.41 \begin {gather*} -\frac {20 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{6} \sqrt {-\frac {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}}} + {\left (81 \, B^{2} b^{2} - 90 \, A B b c + 25 \, A^{2} c^{2}\right )} x} c^{10} \left (-\frac {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}}\right )^{\frac {3}{4}} + {\left (9 \, B b c^{10} - 5 \, A c^{11}\right )} \sqrt {x} \left (-\frac {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}}\right )^{\frac {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}}\right ) + 5 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (c^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} - {\left (9 \, B b - 5 \, A c\right )} \sqrt {x}\right ) - 5 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (-c^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} - {\left (9 \, B b - 5 \, A c\right )} \sqrt {x}\right ) - 4 \, {\left (4 \, B c^{2} x^{4} - 45 \, B b^{2} + 25 \, A b c - 4 \, {\left (9 \, B b c - 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {x}}{40 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \end {gather*}

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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giac [A]  time = 0.19, size = 298, normalized size = 0.96 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.07, size = 339, normalized size = 1.09 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 c^{2}}+\frac {A b \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{2}}-\frac {B \,b^{2} \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{3}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{2}}-\frac {5 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{2}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B b \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{3}}+\frac {2 A \sqrt {x}}{c^{2}}-\frac {4 B b \sqrt {x}}{c^{3}} \end {gather*}

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)-5/16/c^2*(b/c)^(1/4)*2^(1/2)*A*ln((x+(
b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))-5/8/c^2*(b/c)^(1/4)*2^(1/
2)*A*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)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*2^(1/2)*x^(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)

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maxima [A]  time = 3.03, size = 271, normalized size = 0.87 \begin {gather*} -\frac {{\left (B b^{2} - A b c\right )} \sqrt {x}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (\frac {2 \, \sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (9 \, B b - 5 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b}{16 \, c^{3}} + \frac {2 \, {\left (B c x^{\frac {5}{2}} - 5 \, {\left (2 \, B b - A c\right )} \sqrt {x}\right )}}{5 \, c^{3}} \end {gather*}

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]

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

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mupad [B]  time = 0.26, size = 823, normalized size = 2.65 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{c^2}-\frac {4\,B\,b}{c^3}\right )+\frac {2\,B\,x^{5/2}}{5\,c^2}-\frac {\sqrt {x}\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}-\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{13/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}+\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{13/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}-\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{13/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}+\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{13/4}}}\right )\,\left (5\,A\,c-9\,B\,b\right )\,1{}\mathrm {i}}{4\,c^{13/4}}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}-\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )\,1{}\mathrm {i}}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{13/4}}+\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}+\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )\,1{}\mathrm {i}}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{13/4}}}{\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}-\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )\,1{}\mathrm {i}}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{13/4}}-\frac {{\left (-b\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,b^2\,c^2-90\,A\,B\,b^3\,c+81\,B^2\,b^4\right )}{c^3}+\frac {{\left (-b\right )}^{1/4}\,\left (5\,A\,c-9\,B\,b\right )\,\left (72\,B\,b^3-40\,A\,b^2\,c\right )\,1{}\mathrm {i}}{8\,c^{13/4}}\right )\,\left (5\,A\,c-9\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{13/4}}}\right )\,\left (5\,A\,c-9\,B\,b\right )}{4\,c^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/2)*((2*A)/c^2 - (4*B*b)/c^3) + (2*B*x^(5/2))/(5*c^2) - (x^(1/2)*((B*b^2)/2 - (A*b*c)/2))/(b*c^3 + c^4*x^2
) + ((-b)^(1/4)*atan((((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)*(5
*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b)*1i)/(8*c^(13/4)) + ((-b)^(1/4)*((x^(1/2)*
(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^
(13/4)))*(5*A*c - 9*B*b)*1i)/(8*c^(13/4)))/(((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c)
)/c^3 - ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b))/(8*c^(13/4)) - ((-
b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 -
40*A*b^2*c))/(8*c^(13/4)))*(5*A*c - 9*B*b))/(8*c^(13/4))))*(5*A*c - 9*B*b)*1i)/(4*c^(13/4)) + ((-b)^(1/4)*ata
n((((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B
*b^3 - 40*A*b^2*c)*1i)/(8*c^(13/4)))*(5*A*c - 9*B*b))/(8*c^(13/4)) + ((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^
2*b^2*c^2 - 90*A*B*b^3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c)*1i)/(8*c^(13/4)))*(5*A*c
- 9*B*b))/(8*c^(13/4)))/(((-b)^(1/4)*((x^(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 - ((-b)^(1/4)
*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c)*1i)/(8*c^(13/4)))*(5*A*c - 9*B*b)*1i)/(8*c^(13/4)) - ((-b)^(1/4)*((x^
(1/2)*(81*B^2*b^4 + 25*A^2*b^2*c^2 - 90*A*B*b^3*c))/c^3 + ((-b)^(1/4)*(5*A*c - 9*B*b)*(72*B*b^3 - 40*A*b^2*c)*
1i)/(8*c^(13/4)))*(5*A*c - 9*B*b)*1i)/(8*c^(13/4))))*(5*A*c - 9*B*b))/(4*c^(13/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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