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

Optimal. Leaf size=332 \[ \frac {b^{5/4} (13 b B-9 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}+\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{17/4}}+\frac {b \sqrt {x} (13 b B-9 A c)}{2 c^4}-\frac {x^{5/2} (13 b B-9 A c)}{10 c^3}+\frac {x^{9/2} (13 b B-9 A c)}{18 b c^2}-\frac {x^{13/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

[Out]

-1/10*(-9*A*c+13*B*b)*x^(5/2)/c^3+1/18*(-9*A*c+13*B*b)*x^(9/2)/b/c^2-1/2*(-A*c+B*b)*x^(13/2)/b/c/(c*x^2+b)+1/8
*b^(5/4)*(-9*A*c+13*B*b)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(17/4)*2^(1/2)-1/8*b^(5/4)*(-9*A*c+13*B*b
)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(17/4)*2^(1/2)+1/16*b^(5/4)*(-9*A*c+13*B*b)*ln(b^(1/2)+x*c^(1/2)
-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(17/4)*2^(1/2)-1/16*b^(5/4)*(-9*A*c+13*B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c
^(1/4)*2^(1/2)*x^(1/2))/c^(17/4)*2^(1/2)+1/2*b*(-9*A*c+13*B*b)*x^(1/2)/c^4

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Rubi [A]  time = 0.28, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ \frac {b^{5/4} (13 b B-9 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}+\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{17/4}}+\frac {x^{9/2} (13 b B-9 A c)}{18 b c^2}-\frac {x^{5/2} (13 b B-9 A c)}{10 c^3}+\frac {b \sqrt {x} (13 b B-9 A c)}{2 c^4}-\frac {x^{13/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(13*b*B - 9*A*c)*Sqrt[x])/(2*c^4) - ((13*b*B - 9*A*c)*x^(5/2))/(10*c^3) + ((13*b*B - 9*A*c)*x^(9/2))/(18*b*
c^2) - ((b*B - A*c)*x^(13/2))/(2*b*c*(b + c*x^2)) + (b^(5/4)*(13*b*B - 9*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt
[x])/b^(1/4)])/(4*Sqrt[2]*c^(17/4)) - (b^(5/4)*(13*b*B - 9*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])
/(4*Sqrt[2]*c^(17/4)) + (b^(5/4)*(13*b*B - 9*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/
(8*Sqrt[2]*c^(17/4)) - (b^(5/4)*(13*b*B - 9*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(
8*Sqrt[2]*c^(17/4))

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^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}+\frac {\left (\frac {13 b B}{2}-\frac {9 A c}{2}\right ) \int \frac {x^{11/2}}{b+c x^2} \, dx}{2 b c}\\ &=\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}-\frac {(13 b B-9 A c) \int \frac {x^{7/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}+\frac {(b (13 b B-9 A c)) \int \frac {x^{3/2}}{b+c x^2} \, dx}{4 c^3}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}-\frac {\left (b^2 (13 b B-9 A c)\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 c^4}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}-\frac {\left (b^2 (13 b B-9 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^4}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}-\frac {\left (b^{3/2} (13 b B-9 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^4}-\frac {\left (b^{3/2} (13 b B-9 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^4}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}-\frac {\left (b^{3/2} (13 b B-9 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^{9/2}}-\frac {\left (b^{3/2} (13 b B-9 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^{9/2}}+\frac {\left (b^{5/4} (13 b B-9 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^{17/4}}+\frac {\left (b^{5/4} (13 b B-9 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^{17/4}}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}+\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}-\frac {\left (b^{5/4} (13 b B-9 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^{17/4}}+\frac {\left (b^{5/4} (13 b B-9 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^{17/4}}\\ &=\frac {b (13 b B-9 A c) \sqrt {x}}{2 c^4}-\frac {(13 b B-9 A c) x^{5/2}}{10 c^3}+\frac {(13 b B-9 A c) x^{9/2}}{18 b c^2}-\frac {(b B-A c) x^{13/2}}{2 b c \left (b+c x^2\right )}+\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{17/4}}+\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}-\frac {b^{5/4} (13 b B-9 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{17/4}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 417, normalized size = 1.26 \[ \frac {90 \sqrt {2} b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-90 \sqrt {2} b^{5/4} (13 b B-9 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )-405 \sqrt {2} A b^{5/4} c \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+405 \sqrt {2} A b^{5/4} c \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\frac {360 A b^2 c^{5/4} \sqrt {x}}{b+c x^2}-2880 A b c^{5/4} \sqrt {x}+288 A c^{9/4} x^{5/2}+585 \sqrt {2} b^{9/4} B \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-585 \sqrt {2} b^{9/4} B \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+\frac {360 b^3 B \sqrt [4]{c} \sqrt {x}}{b+c x^2}+4320 b^2 B \sqrt [4]{c} \sqrt {x}-576 b B c^{5/4} x^{5/2}+160 B c^{9/4} x^{9/2}}{720 c^{17/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4320*b^2*B*c^(1/4)*Sqrt[x] - 2880*A*b*c^(5/4)*Sqrt[x] - 576*b*B*c^(5/4)*x^(5/2) + 288*A*c^(9/4)*x^(5/2) + 160
*B*c^(9/4)*x^(9/2) + (360*b^3*B*c^(1/4)*Sqrt[x])/(b + c*x^2) - (360*A*b^2*c^(5/4)*Sqrt[x])/(b + c*x^2) + 90*Sq
rt[2]*b^(5/4)*(13*b*B - 9*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 90*Sqrt[2]*b^(5/4)*(13*b*B - 9*
A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 585*Sqrt[2]*b^(9/4)*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/
4)*Sqrt[x] + Sqrt[c]*x] - 405*Sqrt[2]*A*b^(5/4)*c*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] -
585*Sqrt[2]*b^(9/4)*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 405*Sqrt[2]*A*b^(5/4)*c*Lo
g[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(720*c^(17/4))

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fricas [B]  time = 0.78, size = 804, normalized size = 2.42 \[ \frac {180 \, {\left (c^{5} x^{2} + b c^{4}\right )} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{8} \sqrt {-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}} + {\left (169 \, B^{2} b^{4} - 234 \, A B b^{3} c + 81 \, A^{2} b^{2} c^{2}\right )} x} c^{13} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {3}{4}} + {\left (13 \, B b^{2} c^{13} - 9 \, A b c^{14}\right )} \sqrt {x} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {3}{4}}}{28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}\right ) + 45 \, {\left (c^{5} x^{2} + b c^{4}\right )} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {1}{4}} \log \left (c^{4} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {1}{4}} - {\left (13 \, B b^{2} - 9 \, A b c\right )} \sqrt {x}\right ) - 45 \, {\left (c^{5} x^{2} + b c^{4}\right )} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {1}{4}} \log \left (-c^{4} \left (-\frac {28561 \, B^{4} b^{9} - 79092 \, A B^{3} b^{8} c + 82134 \, A^{2} B^{2} b^{7} c^{2} - 37908 \, A^{3} B b^{6} c^{3} + 6561 \, A^{4} b^{5} c^{4}}{c^{17}}\right )^{\frac {1}{4}} - {\left (13 \, B b^{2} - 9 \, A b c\right )} \sqrt {x}\right ) + 4 \, {\left (20 \, B c^{3} x^{6} - 4 \, {\left (13 \, B b c^{2} - 9 \, A c^{3}\right )} x^{4} + 585 \, B b^{3} - 405 \, A b^{2} c + 36 \, {\left (13 \, B b^{2} c - 9 \, A b c^{2}\right )} x^{2}\right )} \sqrt {x}}{360 \, {\left (c^{5} x^{2} + b c^{4}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

1/360*(180*(c^5*x^2 + b*c^4)*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2 - 37908*A^3*B*b^6*c^
3 + 6561*A^4*b^5*c^4)/c^17)^(1/4)*arctan((sqrt(c^8*sqrt(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^
7*c^2 - 37908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17) + (169*B^2*b^4 - 234*A*B*b^3*c + 81*A^2*b^2*c^2)*x)*c^13
*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2 - 37908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17)^
(3/4) + (13*B*b^2*c^13 - 9*A*b*c^14)*sqrt(x)*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2 - 37
908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17)^(3/4))/(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2
- 37908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)) + 45*(c^5*x^2 + b*c^4)*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134
*A^2*B^2*b^7*c^2 - 37908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17)^(1/4)*log(c^4*(-(28561*B^4*b^9 - 79092*A*B^3*
b^8*c + 82134*A^2*B^2*b^7*c^2 - 37908*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17)^(1/4) - (13*B*b^2 - 9*A*b*c)*sqr
t(x)) - 45*(c^5*x^2 + b*c^4)*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2 - 37908*A^3*B*b^6*c^
3 + 6561*A^4*b^5*c^4)/c^17)^(1/4)*log(-c^4*(-(28561*B^4*b^9 - 79092*A*B^3*b^8*c + 82134*A^2*B^2*b^7*c^2 - 3790
8*A^3*B*b^6*c^3 + 6561*A^4*b^5*c^4)/c^17)^(1/4) - (13*B*b^2 - 9*A*b*c)*sqrt(x)) + 4*(20*B*c^3*x^6 - 4*(13*B*b*
c^2 - 9*A*c^3)*x^4 + 585*B*b^3 - 405*A*b^2*c + 36*(13*B*b^2*c - 9*A*b*c^2)*x^2)*sqrt(x))/(c^5*x^2 + b*c^4)

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giac [A]  time = 0.22, size = 335, normalized size = 1.01 \[ -\frac {\sqrt {2} {\left (13 \, \left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - 9 \, \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{5}} - \frac {\sqrt {2} {\left (13 \, \left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - 9 \, \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{5}} - \frac {\sqrt {2} {\left (13 \, \left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - 9 \, \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{5}} + \frac {\sqrt {2} {\left (13 \, \left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - 9 \, \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{5}} + \frac {B b^{3} \sqrt {x} - A b^{2} c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{4}} + \frac {2 \, {\left (5 \, B c^{16} x^{\frac {9}{2}} - 18 \, B b c^{15} x^{\frac {5}{2}} + 9 \, A c^{16} x^{\frac {5}{2}} + 135 \, B b^{2} c^{14} \sqrt {x} - 90 \, A b c^{15} \sqrt {x}\right )}}{45 \, c^{18}} \]

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)^2,x, algorithm="giac")

[Out]

-1/8*sqrt(2)*(13*(b*c^3)^(1/4)*B*b^2 - 9*(b*c^3)^(1/4)*A*b*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt
(x))/(b/c)^(1/4))/c^5 - 1/8*sqrt(2)*(13*(b*c^3)^(1/4)*B*b^2 - 9*(b*c^3)^(1/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt
(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^5 - 1/16*sqrt(2)*(13*(b*c^3)^(1/4)*B*b^2 - 9*(b*c^3)^(1/4)*A*b*c)*
log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5 + 1/16*sqrt(2)*(13*(b*c^3)^(1/4)*B*b^2 - 9*(b*c^3)^(1/4)*
A*b*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5 + 1/2*(B*b^3*sqrt(x) - A*b^2*c*sqrt(x))/((c*x^2 +
b)*c^4) + 2/45*(5*B*c^16*x^(9/2) - 18*B*b*c^15*x^(5/2) + 9*A*c^16*x^(5/2) + 135*B*b^2*c^14*sqrt(x) - 90*A*b*c
^15*sqrt(x))/c^18

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maple [A]  time = 0.06, size = 372, normalized size = 1.12 \[ \frac {2 B \,x^{\frac {9}{2}}}{9 c^{2}}+\frac {2 A \,x^{\frac {5}{2}}}{5 c^{2}}-\frac {4 B b \,x^{\frac {5}{2}}}{5 c^{3}}-\frac {A \,b^{2} \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{3}}+\frac {B \,b^{3} \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{4}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{3}}-\frac {13 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{4}}-\frac {13 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{4}}-\frac {13 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{4}}-\frac {4 A b \sqrt {x}}{c^{3}}+\frac {6 B \,b^{2} \sqrt {x}}{c^{4}} \]

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)^2,x)

[Out]

2/9/c^2*B*x^(9/2)+2/5/c^2*A*x^(5/2)-4/5/c^3*B*x^(5/2)*b-4/c^3*A*b*x^(1/2)+6/c^4*B*b^2*x^(1/2)-1/2*b^2/c^3*x^(1
/2)/(c*x^2+b)*A+1/2*b^3/c^4*x^(1/2)/(c*x^2+b)*B+9/8*b/c^3*(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)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+9/16*b/c^3*(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)))-13/8*b^2/c^4*(b/c
)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-13/8*b^2/c^4*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c
)^(1/4)*x^(1/2)-1)-13/16*b^2/c^4*(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)))

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maxima [A]  time = 3.07, size = 298, normalized size = 0.90 \[ \frac {{\left (B b^{3} - A b^{2} c\right )} \sqrt {x}}{2 \, {\left (c^{5} x^{2} + b c^{4}\right )}} - \frac {{\left (\frac {2 \, \sqrt {2} {\left (13 \, B b - 9 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (13 \, B b - 9 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (13 \, B b - 9 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (13 \, B b - 9 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b^{2}}{16 \, c^{4}} + \frac {2 \, {\left (5 \, B c^{2} x^{\frac {9}{2}} - 9 \, {\left (2 \, B b c - A c^{2}\right )} x^{\frac {5}{2}} + 45 \, {\left (3 \, B b^{2} - 2 \, A b c\right )} \sqrt {x}\right )}}{45 \, c^{4}} \]

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)^2,x, algorithm="maxima")

[Out]

1/2*(B*b^3 - A*b^2*c)*sqrt(x)/(c^5*x^2 + b*c^4) - 1/16*(2*sqrt(2)*(13*B*b - 9*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)*(13*B
*b - 9*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)*(13*B*b - 9*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)*(13*B*b - 9*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^2/c^4 + 2/45*(5*B*c^2*x^(9/2) - 9*(2*B*b*c - A*c^2)*x^(5/2) + 45*(3*B*b^2 - 2*A*b*c)*sqrt(x))/c
^4

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mupad [B]  time = 0.22, size = 857, normalized size = 2.58 \[ x^{5/2}\,\left (\frac {2\,A}{5\,c^2}-\frac {4\,B\,b}{5\,c^3}\right )-\sqrt {x}\,\left (\frac {2\,b\,\left (\frac {2\,A}{c^2}-\frac {4\,B\,b}{c^3}\right )}{c}+\frac {2\,B\,b^2}{c^4}\right )+\frac {2\,B\,x^{9/2}}{9\,c^2}+\frac {\sqrt {x}\,\left (\frac {B\,b^3}{2}-\frac {A\,b^2\,c}{2}\right )}{c^5\,x^2+b\,c^4}+\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}+\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{17/4}}+\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}-\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{17/4}}}{\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}+\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )}{8\,c^{17/4}}-\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}-\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )}{8\,c^{17/4}}}\right )\,\left (9\,A\,c-13\,B\,b\right )\,1{}\mathrm {i}}{4\,c^{17/4}}-\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}-\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )\,1{}\mathrm {i}}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )}{8\,c^{17/4}}+\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}+\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )\,1{}\mathrm {i}}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )}{8\,c^{17/4}}}{\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}-\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )\,1{}\mathrm {i}}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{17/4}}-\frac {{\left (-b\right )}^{5/4}\,\left (\frac {\sqrt {x}\,\left (81\,A^2\,b^4\,c^2-234\,A\,B\,b^5\,c+169\,B^2\,b^6\right )}{c^5}+\frac {{\left (-b\right )}^{5/4}\,\left (9\,A\,c-13\,B\,b\right )\,\left (13\,B\,b^4-9\,A\,b^3\,c\right )\,1{}\mathrm {i}}{c^{21/4}}\right )\,\left (9\,A\,c-13\,B\,b\right )\,1{}\mathrm {i}}{8\,c^{17/4}}}\right )\,\left (9\,A\,c-13\,B\,b\right )}{4\,c^{17/4}} \]

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)^2,x)

[Out]

x^(5/2)*((2*A)/(5*c^2) - (4*B*b)/(5*c^3)) - x^(1/2)*((2*b*((2*A)/c^2 - (4*B*b)/c^3))/c + (2*B*b^2)/c^4) + (2*B
*x^(9/2))/(9*c^2) + (x^(1/2)*((B*b^3)/2 - (A*b^2*c)/2))/(b*c^4 + c^5*x^2) + ((-b)^(5/4)*atan((((-b)^(5/4)*((x^
(1/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*B*b^5*c))/c^5 + ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c
))/c^(21/4))*(9*A*c - 13*B*b)*1i)/(8*c^(17/4)) + ((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*B
*b^5*c))/c^5 - ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c))/c^(21/4))*(9*A*c - 13*B*b)*1i)/(8*c^(17/4)
))/(((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*B*b^5*c))/c^5 + ((-b)^(5/4)*(9*A*c - 13*B*b)*(
13*B*b^4 - 9*A*b^3*c))/c^(21/4))*(9*A*c - 13*B*b))/(8*c^(17/4)) - ((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 81*A^2*
b^4*c^2 - 234*A*B*b^5*c))/c^5 - ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c))/c^(21/4))*(9*A*c - 13*B*b
))/(8*c^(17/4))))*(9*A*c - 13*B*b)*1i)/(4*c^(17/4)) - ((-b)^(5/4)*atan((((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 8
1*A^2*b^4*c^2 - 234*A*B*b^5*c))/c^5 - ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c)*1i)/c^(21/4))*(9*A*c
- 13*B*b))/(8*c^(17/4)) + ((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*B*b^5*c))/c^5 + ((-b)^(
5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c)*1i)/c^(21/4))*(9*A*c - 13*B*b))/(8*c^(17/4)))/(((-b)^(5/4)*((x^(1
/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*B*b^5*c))/c^5 - ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c)*
1i)/c^(21/4))*(9*A*c - 13*B*b)*1i)/(8*c^(17/4)) - ((-b)^(5/4)*((x^(1/2)*(169*B^2*b^6 + 81*A^2*b^4*c^2 - 234*A*
B*b^5*c))/c^5 + ((-b)^(5/4)*(9*A*c - 13*B*b)*(13*B*b^4 - 9*A*b^3*c)*1i)/c^(21/4))*(9*A*c - 13*B*b)*1i)/(8*c^(1
7/4))))*(9*A*c - 13*B*b))/(4*c^(17/4))

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

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)**2,x)

[Out]

Timed out

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