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

Optimal. Leaf size=289 \[ -\frac {(3 b B-7 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {(3 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1584, 457, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {(3 b B-7 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {(3 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*B - 7*A*c)/(6*b^2*c*x^(3/2)) - (b*B - A*c)/(2*b*c*x^(3/2)*(b + c*x^2)) - ((3*b*B - 7*A*c)*ArcTan[1 - (Sqr
t[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(11/4)*c^(1/4)) + ((3*b*B - 7*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sq
rt[x])/b^(1/4)])/(4*Sqrt[2]*b^(11/4)*c^(1/4)) - ((3*b*B - 7*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x]
+ Sqrt[c]*x])/(8*Sqrt[2]*b^(11/4)*c^(1/4)) + ((3*b*B - 7*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x])/(8*Sqrt[2]*b^(11/4)*c^(1/4))

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 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 )^2} \, dx &=\int \frac {A+B x^2}{x^{5/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac {\left (-\frac {3 b B}{2}+\frac {7 A c}{2}\right ) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac {(3 b B-7 A c) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac {(3 b B-7 A c) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b^2}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac {(3 b B-7 A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{5/2}}+\frac {(3 b B-7 A c) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{5/2}}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}+\frac {(3 b B-7 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 )}{8 b^{5/2} \sqrt {c}}+\frac {(3 b B-7 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 )}{8 b^{5/2} \sqrt {c}}-\frac {(3 b B-7 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 )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {(3 b B-7 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 )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}-\frac {(3 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 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 )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {(3 b B-7 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 )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}\\ &=\frac {3 b B-7 A c}{6 b^2 c x^{3/2}}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}-\frac {(3 b B-7 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {(3 b B-7 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{11/4} \sqrt [4]{c}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.58, size = 355, normalized size = 1.23 \begin {gather*} \frac {-\frac {24 A b^{3/4} c \sqrt {x}}{b+c x^2}-\frac {32 A b^{3/4}}{x^{3/2}}+\frac {6 \sqrt {2} (7 A c-3 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac {6 \sqrt {2} (3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}+21 \sqrt {2} A c^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-21 \sqrt {2} A c^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+\frac {24 b^{7/4} B \sqrt {x}}{b+c x^2}-\frac {9 \sqrt {2} b B \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {9 \sqrt {2} b B \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}}{48 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*A*b^(3/4))/x^(3/2) + (24*b^(7/4)*B*Sqrt[x])/(b + c*x^2) - (24*A*b^(3/4)*c*Sqrt[x])/(b + c*x^2) + (6*Sqrt
[2]*(-3*b*B + 7*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) + (6*Sqrt[2]*(3*b*B - 7*A*c)*ArcTa
n[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) - (9*Sqrt[2]*b*B*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/c^(1/4) + 21*Sqrt[2]*A*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + (
9*Sqrt[2]*b*B*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) - 21*Sqrt[2]*A*c^(3/4)*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(48*b^(11/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.67, size = 170, normalized size = 0.59 \begin {gather*} -\frac {(3 b B-7 A c) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {-4 A b-7 A c x^2+3 b B x^2}{6 b^2 x^{3/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*A*b + 3*b*B*x^2 - 7*A*c*x^2)/(6*b^2*x^(3/2)*(b + c*x^2)) - ((3*b*B - 7*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(
Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(4*Sqrt[2]*b^(11/4)*c^(1/4)) + ((3*b*B - 7*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2]*b^(11/4)*c^(1/4))

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 741, normalized size = 2.56 \begin {gather*} -\frac {12 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{6} \sqrt {-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}} + {\left (9 \, B^{2} b^{2} - 42 \, A B b c + 49 \, A^{2} c^{2}\right )} x} b^{8} c \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {3}{4}} + {\left (3 \, B b^{9} c - 7 \, A b^{8} c^{2}\right )} \sqrt {x} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {3}{4}}}{81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}\right ) + 3 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {1}{4}} - {\left (3 \, B b - 7 \, A c\right )} \sqrt {x}\right ) - 3 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\right )} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {81 \, B^{4} b^{4} - 756 \, A B^{3} b^{3} c + 2646 \, A^{2} B^{2} b^{2} c^{2} - 4116 \, A^{3} B b c^{3} + 2401 \, A^{4} c^{4}}{b^{11} c}\right )^{\frac {1}{4}} - {\left (3 \, B b - 7 \, A c\right )} \sqrt {x}\right ) - 4 \, {\left ({\left (3 \, B b - 7 \, A c\right )} x^{2} - 4 \, A b\right )} \sqrt {x}}{24 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\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)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 24
01*A^4*c^4)/(b^11*c))^(1/4)*arctan((sqrt(b^6*sqrt(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116
*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c)) + (9*B^2*b^2 - 42*A*B*b*c + 49*A^2*c^2)*x)*b^8*c*(-(81*B^4*b^4 - 756*A*
B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(3/4) + (3*B*b^9*c - 7*A*b^8*c^2
)*sqrt(x)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^
(3/4))/(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)) + 3*(b^2*c*x^4
+ b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c)
)^(1/4)*log(b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^1
1*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) - 3*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^
2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4)*log(-b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A
^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) - 4*((3*B*b - 7*A
*c)*x^2 - 4*A*b)*sqrt(x))/(b^2*c*x^4 + b^3*x^2)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 283, normalized size = 0.98 \begin {gather*} \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \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 \, b^{3} c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \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 \, b^{3} c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \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 \, b^{3} c} - \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \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 \, b^{3} c} + \frac {B b \sqrt {x} - A c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b^{2}} - \frac {2 \, A}{3 \, b^{2} x^{\frac {3}{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)^2,x, algorithm="giac")

[Out]

1/8*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/8*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(b*c^3)^(1/4)*A*c)*log(sqrt
(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) - 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(b*c^3)^(1/4)*A*c)*lo
g(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/2*(B*b*sqrt(x) - A*c*sqrt(x))/((c*x^2 + b)*b^2) -
2/3*A/(b^2*x^(3/2))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 317, normalized size = 1.10 \begin {gather*} -\frac {A c \sqrt {x}}{2 \left (c \,x^{2}+b \right ) b^{2}}+\frac {B \sqrt {x}}{2 \left (c \,x^{2}+b \right ) b}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{3}}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{3}}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A 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 )}{16 b^{3}}+\frac {3 \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 )}{8 b^{2}}+\frac {3 \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 )}{8 b^{2}}+\frac {3 \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 )}{16 b^{2}}-\frac {2 A}{3 b^{2} x^{\frac {3}{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)^2,x)

[Out]

-1/2/b^2*x^(1/2)/(c*x^2+b)*A*c+1/2/b*x^(1/2)/(c*x^2+b)*B-7/16/b^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)))*c-7/8/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2
^(1/2)/(b/c)^(1/4)*x^(1/2)+1)*c-7/8/b^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)*c+3/16/b^2
*(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)))+3/8/b^2*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+3/8/b^2*(b/c)^(1/4)*2^(1/2)*B*arctan(
2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/3*A/b^2/x^(3/2)

________________________________________________________________________________________

maxima [A]  time = 3.11, size = 251, normalized size = 0.87 \begin {gather*} \frac {{\left (3 \, B b - 7 \, A c\right )} x^{2} - 4 \, A b}{6 \, {\left (b^{2} c x^{\frac {7}{2}} + b^{3} x^{\frac {3}{2}}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, B b - 7 \, 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 (3 \, B b - 7 \, 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 (3 \, B b - 7 \, 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 (3 \, B b - 7 \, 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}}}}{16 \, b^{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)^2,x, algorithm="maxima")

[Out]

1/6*((3*B*b - 7*A*c)*x^2 - 4*A*b)/(b^2*c*x^(7/2) + b^3*x^(3/2)) + 1/16*(2*sqrt(2)*(3*B*b - 7*A*c)*arctan(1/2*s
qrt(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)*(3*B*b - 7*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)*(3*B*b - 7*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)*(3*B*b - 7*A*c)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqr
t(b))/(b^(3/4)*c^(1/4)))/b^2

________________________________________________________________________________________

mupad [B]  time = 0.37, size = 859, normalized size = 2.97 \begin {gather*} -\frac {\frac {2\,A}{3\,b}+\frac {x^2\,\left (7\,A\,c-3\,B\,b\right )}{6\,b^2}}{b\,x^{3/2}+c\,x^{7/2}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}}{\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}}\right )\,\left (7\,A\,c-3\,B\,b\right )\,1{}\mathrm {i}}{4\,{\left (-b\right )}^{11/4}\,c^{1/4}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}}{\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}-\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (\sqrt {x}\,\left (1568\,A^2\,b^6\,c^5-1344\,A\,B\,b^7\,c^4+288\,B^2\,b^8\,c^3\right )+\frac {\left (7\,A\,c-3\,B\,b\right )\,\left (1792\,A\,b^9\,c^4-768\,B\,b^{10}\,c^3\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-b\right )}^{11/4}\,c^{1/4}}}\right )\,\left (7\,A\,c-3\,B\,b\right )}{4\,{\left (-b\right )}^{11/4}\,c^{1/4}} \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)^2,x)

[Out]

- ((2*A)/(3*b) + (x^2*(7*A*c - 3*B*b))/(6*b^2))/(b*x^(3/2) + c*x^(7/2)) - (atan((((7*A*c - 3*B*b)*(x^(1/2)*(15
68*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*
(-b)^(11/4)*c^(1/4)))*1i)/(8*(-b)^(11/4)*c^(1/4)) + ((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*
c^3 - 1344*A*B*b^7*c^4) + ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*(-b)^(11/4)*c^(1/4)))*1i)/(8*
(-b)^(11/4)*c^(1/4)))/(((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) - ((7
*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4)) - ((7*A*c -
3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) + ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 -
768*B*b^10*c^3))/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4))))*(7*A*c - 3*B*b)*1i)/(4*(-b)^(11/4)*c^(1/4
)) - (atan((((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) - ((7*A*c - 3*B*
b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3)*1i)/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4)) + ((7*A*c - 3*B*b)*
(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) + ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^
10*c^3)*1i)/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4)))/(((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 +
288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3)*1i)/(8*(-b)^(11/4)*c^
(1/4)))*1i)/(8*(-b)^(11/4)*c^(1/4)) - ((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B
*b^7*c^4) + ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3)*1i)/(8*(-b)^(11/4)*c^(1/4)))*1i)/(8*(-b)^(11/4)
*c^(1/4))))*(7*A*c - 3*B*b))/(4*(-b)^(11/4)*c^(1/4))

________________________________________________________________________________________

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________