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

Optimal. Leaf size=343 \[ \frac {11 c^{3/4} (7 b B-15 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{19/4}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}+\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.28, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1584, 457, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {11 c^{3/4} (7 b B-15 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{19/4}}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}+\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(7*b*B - 15*A*c))/(112*b^3*c*x^(7/2)) - (11*(7*b*B - 15*A*c))/(48*b^4*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(7/2
)*(b + c*x^2)^2) - (7*b*B - 15*A*c)/(16*b^2*c*x^(7/2)*(b + c*x^2)) + (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 - (
Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) + (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4)) - (11*c^(3/4)*(7*b*B - 15*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)
*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 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^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^{9/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}+\frac {\left (-\frac {7 b B}{2}+\frac {15 A c}{2}\right ) \int \frac {1}{x^{9/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {(11 (7 b B-15 A c)) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac {(11 (7 b B-15 A c)) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {(11 c (7 b B-15 A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {(11 c (7 b B-15 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^4}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {(11 c (7 b B-15 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{9/2}}-\frac {(11 c (7 b B-15 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{9/2}}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}-\frac {\left (11 \sqrt {c} (7 b B-15 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 )}{64 b^{9/2}}-\frac {\left (11 \sqrt {c} (7 b B-15 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 )}{64 b^{9/2}}+\frac {\left (11 c^{3/4} (7 b B-15 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 )}{64 \sqrt {2} b^{19/4}}+\frac {\left (11 c^{3/4} (7 b B-15 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 )}{64 \sqrt {2} b^{19/4}}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {\left (11 c^{3/4} (7 b B-15 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 )}{32 \sqrt {2} b^{19/4}}+\frac {\left (11 c^{3/4} (7 b B-15 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 )}{32 \sqrt {2} b^{19/4}}\\ &=\frac {11 (7 b B-15 A c)}{112 b^3 c x^{7/2}}-\frac {11 (7 b B-15 A c)}{48 b^4 x^{3/2}}-\frac {b B-A c}{4 b c x^{7/2} \left (b+c x^2\right )^2}-\frac {7 b B-15 A c}{16 b^2 c x^{7/2} \left (b+c x^2\right )}+\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {11 c^{3/4} (7 b B-15 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 433, normalized size = 1.26 \begin {gather*} \frac {\frac {672 A b^{7/4} c^2 \sqrt {x}}{\left (b+c x^2\right )^2}+\frac {3864 A b^{3/4} c^2 \sqrt {x}}{b+c x^2}+\frac {5376 A b^{3/4} c}{x^{3/2}}-\frac {768 A b^{7/4}}{x^{7/2}}+462 \sqrt {2} c^{3/4} (7 b B-15 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )+462 \sqrt {2} c^{3/4} (15 A c-7 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )-3465 \sqrt {2} A c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+3465 \sqrt {2} A c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\frac {672 b^{11/4} B c \sqrt {x}}{\left (b+c x^2\right )^2}-\frac {2520 b^{7/4} B c \sqrt {x}}{b+c x^2}-\frac {1792 b^{7/4} B}{x^{3/2}}+1617 \sqrt {2} b B c^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-1617 \sqrt {2} b B c^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2688 b^{19/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-768*A*b^(7/4))/x^(7/2) - (1792*b^(7/4)*B)/x^(3/2) + (5376*A*b^(3/4)*c)/x^(3/2) - (672*b^(11/4)*B*c*Sqrt[x])
/(b + c*x^2)^2 + (672*A*b^(7/4)*c^2*Sqrt[x])/(b + c*x^2)^2 - (2520*b^(7/4)*B*c*Sqrt[x])/(b + c*x^2) + (3864*A*
b^(3/4)*c^2*Sqrt[x])/(b + c*x^2) + 462*Sqrt[2]*c^(3/4)*(7*b*B - 15*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b
^(1/4)] + 462*Sqrt[2]*c^(3/4)*(-7*b*B + 15*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 1617*Sqrt[2]*b
*B*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 3465*Sqrt[2]*A*c^(7/4)*Log[Sqrt[b] - S
qrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 1617*Sqrt[2]*b*B*c^(3/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*S
qrt[x] + Sqrt[c]*x] + 3465*Sqrt[2]*A*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2688
*b^(19/4))

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IntegrateAlgebraic [A]  time = 0.73, size = 224, normalized size = 0.65 \begin {gather*} \frac {11 \left (7 b B c^{3/4}-15 A c^{7/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {11 \left (7 b B c^{3/4}-15 A c^{7/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{19/4}}+\frac {-96 A b^3+480 A b^2 c x^2+1815 A b c^2 x^4+1155 A c^3 x^6-224 b^3 B x^2-847 b^2 B c x^4-539 b B c^2 x^6}{336 b^4 x^{7/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-96*A*b^3 - 224*b^3*B*x^2 + 480*A*b^2*c*x^2 - 847*b^2*B*c*x^4 + 1815*A*b*c^2*x^4 - 539*b*B*c^2*x^6 + 1155*A*c
^3*x^6)/(336*b^4*x^(7/2)*(b + c*x^2)^2) + (11*(7*b*B*c^(3/4) - 15*A*c^(7/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqr
t[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*b^(19/4)) - (11*(7*b*B*c^(3/4) - 15*A*c^(7/4))*ArcTanh[(Sqrt[2]*b^
(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(19/4))

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fricas [B]  time = 0.46, size = 864, normalized size = 2.52 \begin {gather*} \frac {924 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{10} \sqrt {-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}} + {\left (49 \, B^{2} b^{2} c^{2} - 210 \, A B b c^{3} + 225 \, A^{2} c^{4}\right )} x} b^{14} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {3}{4}} + {\left (7 \, B b^{15} c - 15 \, A b^{14} c^{2}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {3}{4}}}{2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}\right ) + 231 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (11 \, b^{5} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {1}{4}} - 11 \, {\left (7 \, B b c - 15 \, A c^{2}\right )} \sqrt {x}\right ) - 231 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (-11 \, b^{5} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 20580 \, A B^{3} b^{3} c^{4} + 66150 \, A^{2} B^{2} b^{2} c^{5} - 94500 \, A^{3} B b c^{6} + 50625 \, A^{4} c^{7}}{b^{19}}\right )^{\frac {1}{4}} - 11 \, {\left (7 \, B b c - 15 \, A c^{2}\right )} \sqrt {x}\right ) - 4 \, {\left (77 \, {\left (7 \, B b c^{2} - 15 \, A c^{3}\right )} x^{6} + 121 \, {\left (7 \, B b^{2} c - 15 \, A b c^{2}\right )} x^{4} + 96 \, A b^{3} + 32 \, {\left (7 \, B b^{3} - 15 \, A b^{2} c\right )} x^{2}\right )} \sqrt {x}}{1344 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1344*(924*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^
2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1/4)*arctan((sqrt(b^10*sqrt(-(2401*B^4*b^4*c^3 - 20580*A*B^3
*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19) + (49*B^2*b^2*c^2 - 210*A*B*b*c^3
+ 225*A^2*c^4)*x)*b^14*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 +
50625*A^4*c^7)/b^19)^(3/4) + (7*B*b^15*c - 15*A*b^14*c^2)*sqrt(x)*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 +
66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(3/4))/(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c
^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)) + 231*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(
-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1
/4)*log(11*b^5*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A
^4*c^7)/b^19)^(1/4) - 11*(7*B*b*c - 15*A*c^2)*sqrt(x)) - 231*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-(2401*B^4
*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b^19)^(1/4)*log(-1
1*b^5*(-(2401*B^4*b^4*c^3 - 20580*A*B^3*b^3*c^4 + 66150*A^2*B^2*b^2*c^5 - 94500*A^3*B*b*c^6 + 50625*A^4*c^7)/b
^19)^(1/4) - 11*(7*B*b*c - 15*A*c^2)*sqrt(x)) - 4*(77*(7*B*b*c^2 - 15*A*c^3)*x^6 + 121*(7*B*b^2*c - 15*A*b*c^2
)*x^4 + 96*A*b^3 + 32*(7*B*b^3 - 15*A*b^2*c)*x^2)*sqrt(x))/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)

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giac [A]  time = 0.22, size = 315, normalized size = 0.92 \begin {gather*} -\frac {11 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 15 \, \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^{5}} - \frac {11 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 15 \, \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^{5}} - \frac {11 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 15 \, \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^{5}} + \frac {11 \, \sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 15 \, \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^{5}} - \frac {15 \, B b c^{2} x^{\frac {5}{2}} - 23 \, A c^{3} x^{\frac {5}{2}} + 19 \, B b^{2} c \sqrt {x} - 27 \, A b c^{2} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{4}} - \frac {2 \, {\left (7 \, B b x^{2} - 21 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/64*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(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^5 - 11/64*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(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^5 - 11/128*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*log(sq
rt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 + 11/128*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 15*(b*c^3)^(1/4)*A*c)*l
og(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 1/16*(15*B*b*c^2*x^(5/2) - 23*A*c^3*x^(5/2) + 19*B*b^2*
c*sqrt(x) - 27*A*b*c^2*sqrt(x))/((c*x^2 + b)^2*b^4) - 2/21*(7*B*b*x^2 - 21*A*c*x^2 + 3*A*b)/(b^4*x^(7/2))

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maple [A]  time = 0.07, size = 390, normalized size = 1.14 \begin {gather*} \frac {23 A \,c^{3} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {15 B \,c^{2} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {27 A \,c^{2} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {19 B c \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{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 )}{128 b^{5}}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{4}}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{4}}-\frac {77 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \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^{4}}+\frac {2 A c}{b^{4} x^{\frac {3}{2}}}-\frac {2 B}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 A}{7 b^{3} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

23/16/b^4*c^3/(c*x^2+b)^2*x^(5/2)*A-15/16/b^3*c^2/(c*x^2+b)^2*x^(5/2)*B+27/16/b^3*c^2/(c*x^2+b)^2*A*x^(1/2)-19
/16/b^2*c/(c*x^2+b)^2*B*x^(1/2)+165/64/b^5*c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+165
/128/b^5*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)))+165/64/b^5*c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-77/64/b^4*c*(b/c)^
(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-77/128/b^4*c*(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)))-77/64/b^4*c*(b/c)^(1/4)*2^(1/2)*B*arcta
n(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-2/7*A/b^3/x^(7/2)+2/b^4/x^(3/2)*A*c-2/3/b^3/x^(3/2)*B

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maxima [A]  time = 3.15, size = 321, normalized size = 0.94 \begin {gather*} -\frac {77 \, {\left (7 \, B b c^{2} - 15 \, A c^{3}\right )} x^{6} + 121 \, {\left (7 \, B b^{2} c - 15 \, A b c^{2}\right )} x^{4} + 96 \, A b^{3} + 32 \, {\left (7 \, B b^{3} - 15 \, A b^{2} c\right )} x^{2}}{336 \, {\left (b^{4} c^{2} x^{\frac {15}{2}} + 2 \, b^{5} c x^{\frac {11}{2}} + b^{6} x^{\frac {7}{2}}\right )}} - \frac {11 \, {\left (\frac {2 \, \sqrt {2} {\left (7 \, B b c - 15 \, A c^{2}\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 (7 \, B b c - 15 \, A c^{2}\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 (7 \, B b c - 15 \, A c^{2}\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 (7 \, B b c - 15 \, A c^{2}\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 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/336*(77*(7*B*b*c^2 - 15*A*c^3)*x^6 + 121*(7*B*b^2*c - 15*A*b*c^2)*x^4 + 96*A*b^3 + 32*(7*B*b^3 - 15*A*b^2*c
)*x^2)/(b^4*c^2*x^(15/2) + 2*b^5*c*x^(11/2) + b^6*x^(7/2)) - 11/128*(2*sqrt(2)*(7*B*b*c - 15*A*c^2)*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)*(7*B*b*c - 15*A*c^2)*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)*(7*B*b*c - 15*A*c^2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x)
+ sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(7*B*b*c - 15*A*c^2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x)
+ sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^4

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mupad [B]  time = 0.45, size = 626, normalized size = 1.83 \begin {gather*} \frac {\frac {2\,x^2\,\left (15\,A\,c-7\,B\,b\right )}{21\,b^2}-\frac {2\,A}{7\,b}+\frac {11\,c^2\,x^6\,\left (15\,A\,c-7\,B\,b\right )}{48\,b^4}+\frac {121\,c\,x^4\,\left (15\,A\,c-7\,B\,b\right )}{336\,b^3}}{b^2\,x^{7/2}+c^2\,x^{15/2}+2\,b\,c\,x^{11/2}}+\frac {11\,{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {\frac {11\,{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (446054400\,A^2\,b^{12}\,c^7-416317440\,A\,B\,b^{13}\,c^6+97140736\,B^2\,b^{14}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (173015040\,A\,b^{17}\,c^5-80740352\,B\,b^{18}\,c^4\right )\,11{}\mathrm {i}}{64\,b^{19/4}}\right )}{64\,b^{19/4}}+\frac {11\,{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (446054400\,A^2\,b^{12}\,c^7-416317440\,A\,B\,b^{13}\,c^6+97140736\,B^2\,b^{14}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (173015040\,A\,b^{17}\,c^5-80740352\,B\,b^{18}\,c^4\right )\,11{}\mathrm {i}}{64\,b^{19/4}}\right )}{64\,b^{19/4}}}{\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (446054400\,A^2\,b^{12}\,c^7-416317440\,A\,B\,b^{13}\,c^6+97140736\,B^2\,b^{14}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (173015040\,A\,b^{17}\,c^5-80740352\,B\,b^{18}\,c^4\right )\,11{}\mathrm {i}}{64\,b^{19/4}}\right )\,11{}\mathrm {i}}{64\,b^{19/4}}-\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (446054400\,A^2\,b^{12}\,c^7-416317440\,A\,B\,b^{13}\,c^6+97140736\,B^2\,b^{14}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (15\,A\,c-7\,B\,b\right )\,\left (173015040\,A\,b^{17}\,c^5-80740352\,B\,b^{18}\,c^4\right )\,11{}\mathrm {i}}{64\,b^{19/4}}\right )\,11{}\mathrm {i}}{64\,b^{19/4}}}\right )\,\left (15\,A\,c-7\,B\,b\right )}{32\,b^{19/4}}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {A^3\,c^8\,\sqrt {x}\,3375{}\mathrm {i}-B^3\,b^3\,c^5\,\sqrt {x}\,343{}\mathrm {i}-A^2\,B\,b\,c^7\,\sqrt {x}\,4725{}\mathrm {i}+A\,B^2\,b^2\,c^6\,\sqrt {x}\,2205{}\mathrm {i}}{b^{1/4}\,{\left (-c\right )}^{19/4}\,\left (c\,\left (c\,\left (3375\,A^3\,c-4725\,A^2\,B\,b\right )+2205\,A\,B^2\,b^2\right )-343\,B^3\,b^3\right )}\right )\,\left (15\,A\,c-7\,B\,b\right )\,11{}\mathrm {i}}{32\,b^{19/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x^2*(15*A*c - 7*B*b))/(21*b^2) - (2*A)/(7*b) + (11*c^2*x^6*(15*A*c - 7*B*b))/(48*b^4) + (121*c*x^4*(15*A*c
- 7*B*b))/(336*b^3))/(b^2*x^(7/2) + c^2*x^(15/2) + 2*b*c*x^(11/2)) + (11*(-c)^(3/4)*atan(((11*(-c)^(3/4)*(15*
A*c - 7*B*b)*(x^(1/2)*(446054400*A^2*b^12*c^7 + 97140736*B^2*b^14*c^5 - 416317440*A*B*b^13*c^6) - ((-c)^(3/4)*
(15*A*c - 7*B*b)*(173015040*A*b^17*c^5 - 80740352*B*b^18*c^4)*11i)/(64*b^(19/4))))/(64*b^(19/4)) + (11*(-c)^(3
/4)*(15*A*c - 7*B*b)*(x^(1/2)*(446054400*A^2*b^12*c^7 + 97140736*B^2*b^14*c^5 - 416317440*A*B*b^13*c^6) + ((-c
)^(3/4)*(15*A*c - 7*B*b)*(173015040*A*b^17*c^5 - 80740352*B*b^18*c^4)*11i)/(64*b^(19/4))))/(64*b^(19/4)))/(((-
c)^(3/4)*(15*A*c - 7*B*b)*(x^(1/2)*(446054400*A^2*b^12*c^7 + 97140736*B^2*b^14*c^5 - 416317440*A*B*b^13*c^6) -
((-c)^(3/4)*(15*A*c - 7*B*b)*(173015040*A*b^17*c^5 - 80740352*B*b^18*c^4)*11i)/(64*b^(19/4)))*11i)/(64*b^(19/
4)) - ((-c)^(3/4)*(15*A*c - 7*B*b)*(x^(1/2)*(446054400*A^2*b^12*c^7 + 97140736*B^2*b^14*c^5 - 416317440*A*B*b^
13*c^6) + ((-c)^(3/4)*(15*A*c - 7*B*b)*(173015040*A*b^17*c^5 - 80740352*B*b^18*c^4)*11i)/(64*b^(19/4)))*11i)/(
64*b^(19/4))))*(15*A*c - 7*B*b))/(32*b^(19/4)) - ((-c)^(3/4)*atan((A^3*c^8*x^(1/2)*3375i - B^3*b^3*c^5*x^(1/2)
*343i - A^2*B*b*c^7*x^(1/2)*4725i + A*B^2*b^2*c^6*x^(1/2)*2205i)/(b^(1/4)*(-c)^(19/4)*(c*(c*(3375*A^3*c - 4725
*A^2*B*b) + 2205*A*B^2*b^2) - 343*B^3*b^3)))*(15*A*c - 7*B*b)*11i)/(32*b^(19/4))

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

[Out]

Timed out

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