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

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

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Rubi [A]  time = 0.25, antiderivative size = 322, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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^{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*}

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Mathematica [A]  time = 0.63, size = 403, normalized size = 1.25 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.73, size = 257, normalized size = 0.80 \begin {gather*} \left (\frac {45 \sqrt [4]{b} B}{32 \sqrt {2} c^{13/4}}-\frac {5 A}{32 \sqrt {2} b^{3/4} c^{9/4}}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+\frac {5 A \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{3/4} c^{9/4}}+\frac {\sqrt {x} \left (45 b^2 B-5 A b c\right )+x^{5/2} \left (81 b B c-9 A c^2\right )+32 B c^2 x^{9/2}}{16 c^3 \left (b+c x^2\right )^2}-\frac {45 \sqrt [4]{b} B \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} c^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((45*b^2*B - 5*A*b*c)*Sqrt[x] + (81*b*B*c - 9*A*c^2)*x^(5/2) + 32*B*c^2*x^(9/2))/(16*c^3*(b + c*x^2)^2) + ((45
*b^(1/4)*B)/(32*Sqrt[2]*c^(13/4)) - (5*A)/(32*Sqrt[2]*b^(3/4)*c^(9/4)))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*
b^(1/4)*c^(1/4)*Sqrt[x])] - (45*b^(1/4)*B*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(3
2*Sqrt[2]*c^(13/4)) + (5*A*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(3/
4)*c^(9/4))

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fricas [B]  time = 0.44, size = 793, normalized size = 2.46 \begin {gather*} \frac {20 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{2} c^{6} \sqrt {-\frac {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}}} + {\left (81 \, B^{2} b^{2} - 18 \, A B b c + A^{2} c^{2}\right )} x} b^{2} c^{10} \left (-\frac {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}}\right )^{\frac {3}{4}} + {\left (9 \, B b^{3} c^{10} - A b^{2} c^{11}\right )} \sqrt {x} \left (-\frac {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}}\right )^{\frac {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}}\right ) + 5 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (5 \, b c^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} - 5 \, {\left (9 \, B b - A c\right )} \sqrt {x}\right ) - 5 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {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}}\right )^{\frac {1}{4}} \log \left (-5 \, b c^{3} \left (-\frac {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}}\right )^{\frac {1}{4}} - 5 \, {\left (9 \, B b - A c\right )} \sqrt {x}\right ) + 4 \, {\left (32 \, B c^{2} x^{4} + 45 \, B b^{2} - 5 \, A b c + 9 \, {\left (9 \, B b c - A c^{2}\right )} x^{2}\right )} \sqrt {x}}{64 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} \end {gather*}

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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giac [A]  time = 0.20, size = 304, normalized size = 0.94 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.07, size = 363, normalized size = 1.13 \begin {gather*} -\frac {9 A \,x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c}+\frac {17 B b \,x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c^{2}}-\frac {5 A b \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} c^{2}}+\frac {13 B \,b^{2} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} 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 )}{64 b \,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 )}{64 b \,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 )}{128 b \,c^{2}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 c^{3}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 c^{3}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, 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 )}{128 c^{3}}+\frac {2 B \sqrt {x}}{c^{3}} \end {gather*}

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)*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)))+5/64/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)*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)))-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 [A]  time = 3.06, size = 283, normalized size = 0.88 \begin {gather*} \frac {{\left (17 \, B b c - 9 \, A c^{2}\right )} x^{\frac {5}{2}} + {\left (13 \, B b^{2} - 5 \, A b c\right )} \sqrt {x}}{16 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {2 \, B \sqrt {x}}{c^{3}} - \frac {5 \, {\left (\frac {2 \, \sqrt {2} {\left (9 \, B b - 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 - 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 - 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 - 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 )}}{128 \, c^{3}} \end {gather*}

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]

1/16*((17*B*b*c - 9*A*c^2)*x^(5/2) + (13*B*b^2 - 5*A*b*c)*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3) + 2*B*sqr
t(x)/c^3 - 5/128*(2*sqrt(2)*(9*B*b - A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqr
t(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(9*B*b - 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))) + sqrt(2)*(9*B*b - 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 - A*c)*log(-s
qrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/c^3

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mupad [B]  time = 0.23, size = 760, normalized size = 2.36 \begin {gather*} \frac {\sqrt {x}\,\left (\frac {13\,B\,b^2}{16}-\frac {5\,A\,b\,c}{16}\right )-x^{5/2}\,\left (\frac {9\,A\,c^2}{16}-\frac {17\,B\,b\,c}{16}\right )}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {2\,B\,\sqrt {x}}{c^3}-\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}-\frac {5\,\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}+\frac {\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}+\frac {5\,\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}}{\frac {5\,\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}-\frac {5\,\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}-\frac {5\,\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}+\frac {5\,\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}}\right )\,\left (A\,c-9\,B\,b\right )\,5{}\mathrm {i}}{32\,{\left (-b\right )}^{3/4}\,c^{13/4}}-\frac {5\,\mathrm {atan}\left (\frac {\frac {5\,\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}-\frac {\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}+\frac {5\,\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}+\frac {\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}}{\frac {\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}-\frac {\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}-\frac {\left (A\,c-9\,B\,b\right )\,\left (\frac {25\,\sqrt {x}\,\left (A^2\,c^2-18\,A\,B\,b\,c+81\,B^2\,b^2\right )}{64\,c^3}+\frac {\left (45\,B\,b^2-5\,A\,b\,c\right )\,\left (A\,c-9\,B\,b\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}\right )\,5{}\mathrm {i}}{64\,{\left (-b\right )}^{3/4}\,c^{13/4}}}\right )\,\left (A\,c-9\,B\,b\right )}{32\,{\left (-b\right )}^{3/4}\,c^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^(1/2)*((13*B*b^2)/16 - (5*A*b*c)/16) - x^(5/2)*((9*A*c^2)/16 - (17*B*b*c)/16))/(b^2*c^3 + c^5*x^4 + 2*b*c^4
*x^2) + (2*B*x^(1/2))/c^3 - (atan((((A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) -
(5*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)^(3/4)*c^(13/4)))*5i)/(64*(-b)^(3/4)*c^(13/4)) + ((A*c - 9*B*b
)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) + (5*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)
^(3/4)*c^(13/4)))*5i)/(64*(-b)^(3/4)*c^(13/4)))/((5*(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*
b*c))/(64*c^3) - (5*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)^(3/4)*c^(13/4)) -
(5*(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) + (5*(45*B*b^2 - 5*A*b*c)*(A*c - 9
*B*b))/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)^(3/4)*c^(13/4))))*(A*c - 9*B*b)*5i)/(32*(-b)^(3/4)*c^(13/4)) - (5*a
tan(((5*(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - ((45*B*b^2 - 5*A*b*c)*(A*c
- 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)^(3/4)*c^(13/4)) + (5*(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81
*B^2*b^2 - 18*A*B*b*c))/(64*c^3) + ((45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)
^(3/4)*c^(13/4)))/(((A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - ((45*B*b^2 - 5*
A*b*c)*(A*c - 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4)))*5i)/(64*(-b)^(3/4)*c^(13/4)) - ((A*c - 9*B*b)*((25*x^(1/2)*
(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) + ((45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4
)))*5i)/(64*(-b)^(3/4)*c^(13/4))))*(A*c - 9*B*b))/(32*(-b)^(3/4)*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**(19/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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