3.2.100 \(\int \frac {x^{9/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=261 \[ \frac {(A c+3 b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(A c+3 b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(A c+3 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}+\frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}-\frac {x^{3/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1584, 457, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {(A c+3 b B) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(A c+3 b B) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(A c+3 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}+\frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}-\frac {x^{3/2} (b B-A c)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-((b*B - A*c)*x^(3/2))/(2*b*c*(b + c*x^2)) - ((3*b*B + A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*
Sqrt[2]*b^(5/4)*c^(7/4)) + ((3*b*B + A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(5/4)*c^
(7/4)) + ((3*b*B + A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(5/4)*c^(7/4)
) - ((3*b*B + A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(5/4)*c^(7/4))

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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

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Mathematica [C]  time = 0.18, size = 95, normalized size = 0.36 \begin {gather*} \frac {2 x^{3/2} (A c-b B) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^2 c}+\frac {B \left (\tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+\tanh ^{-1}\left (\frac {b \sqrt [4]{c} \sqrt {x}}{(-b)^{5/4}}\right )\right )}{\sqrt [4]{-b} c^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(B*(ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)] + ArcTanh[(b*c^(1/4)*Sqrt[x])/(-b)^(5/4)]))/((-b)^(1/4)*c^(7/4)) + (2
*(-(b*B) + A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3*b^2*c)

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IntegrateAlgebraic [A]  time = 0.68, size = 160, normalized size = 0.61 \begin {gather*} -\frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(A c+3 b B) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}+\frac {x^{3/2} (A c-b B)}{2 b c \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

((-(b*B) + A*c)*x^(3/2))/(2*b*c*(b + c*x^2)) - ((3*b*B + A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x])])/(4*Sqrt[2]*b^(5/4)*c^(7/4)) - ((3*b*B + A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[
b] + Sqrt[c]*x)])/(4*Sqrt[2]*b^(5/4)*c^(7/4))

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fricas [B]  time = 0.45, size = 912, normalized size = 3.49 \begin {gather*} -\frac {4 \, {\left (B b - A c\right )} x^{\frac {3}{2}} + 4 \, {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (729 \, B^{6} b^{6} + 1458 \, A B^{5} b^{5} c + 1215 \, A^{2} B^{4} b^{4} c^{2} + 540 \, A^{3} B^{3} b^{3} c^{3} + 135 \, A^{4} B^{2} b^{2} c^{4} + 18 \, A^{5} B b c^{5} + A^{6} c^{6}\right )} x - {\left (81 \, B^{4} b^{7} c^{3} + 108 \, A B^{3} b^{6} c^{4} + 54 \, A^{2} B^{2} b^{5} c^{5} + 12 \, A^{3} B b^{4} c^{6} + A^{4} b^{3} c^{7}\right )} \sqrt {-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}}} b c^{2} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {1}{4}} - {\left (27 \, B^{3} b^{4} c^{2} + 27 \, A B^{2} b^{3} c^{3} + 9 \, A^{2} B b^{2} c^{4} + A^{3} b c^{5}\right )} \sqrt {x} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {1}{4}}}{81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}\right ) - {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (b^{4} c^{5} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} b^{3} + 27 \, A B^{2} b^{2} c + 9 \, A^{2} B b c^{2} + A^{3} c^{3}\right )} \sqrt {x}\right ) + {\left (b c^{2} x^{2} + b^{2} c\right )} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {1}{4}} \log \left (-b^{4} c^{5} \left (-\frac {81 \, B^{4} b^{4} + 108 \, A B^{3} b^{3} c + 54 \, A^{2} B^{2} b^{2} c^{2} + 12 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{5} c^{7}}\right )^{\frac {3}{4}} + {\left (27 \, B^{3} b^{3} + 27 \, A B^{2} b^{2} c + 9 \, A^{2} B b c^{2} + A^{3} c^{3}\right )} \sqrt {x}\right )}{8 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/8*(4*(B*b - A*c)*x^(3/2) + 4*(b*c^2*x^2 + b^2*c)*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*
A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4)*arctan((sqrt((729*B^6*b^6 + 1458*A*B^5*b^5*c + 1215*A^2*B^4*b^4*c^2 +
540*A^3*B^3*b^3*c^3 + 135*A^4*B^2*b^2*c^4 + 18*A^5*B*b*c^5 + A^6*c^6)*x - (81*B^4*b^7*c^3 + 108*A*B^3*b^6*c^4
+ 54*A^2*B^2*b^5*c^5 + 12*A^3*B*b^4*c^6 + A^4*b^3*c^7)*sqrt(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^
2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7)))*b*c^2*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*
B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4) - (27*B^3*b^4*c^2 + 27*A*B^2*b^3*c^3 + 9*A^2*B*b^2*c^4 + A^3*b*c^5)*sqrt(x
)*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4))/(81*B^4*b
^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)) - (b*c^2*x^2 + b^2*c)*(-(81*B^4*b^4 + 1
08*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4)*log(b^4*c^5*(-(81*B^4*b^4 + 1
08*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A*B^2*b^2*
c + 9*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) + (b*c^2*x^2 + b^2*c)*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c
^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4)*log(-b^4*c^5*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*
c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A*B^2*b^2*c + 9*A^2*B*b*c^2 + A^3*c^3)*sqr
t(x)))/(b*c^2*x^2 + b^2*c)

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giac [A]  time = 0.21, size = 273, normalized size = 1.05 \begin {gather*} -\frac {B b x^{\frac {3}{2}} - A c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} b c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{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^{2} c^{4}} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{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^{2} c^{4}} - \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{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^{2} c^{4}} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{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^{2} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-1/2*(B*b*x^(3/2) - A*c*x^(3/2))/((c*x^2 + b)*b*c) + 1/8*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*arc
tan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) + 1/8*sqrt(2)*(3*(b*c^3)^(3/4)*B*b +
(b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) - 1/16*sqrt(2)
*(3*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^4) + 1/16*s
qrt(2)*(3*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^4)

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maple [A]  time = 0.06, size = 305, normalized size = 1.17 \begin {gather*} \frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {\sqrt {2}\, A \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 \left (\frac {b}{c}\right )^{\frac {1}{4}} b c}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {3 \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 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {\left (A c -b B \right ) x^{\frac {3}{2}}}{2 \left (c \,x^{2}+b \right ) b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

1/2*(A*c-B*b)/b/c*x^(3/2)/(c*x^2+b)+1/8/b/c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/16/b
/c/(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)))+1/8/b/c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+3/8/c^2/(b/c)^(1/4)*2^(1/2)*B*arcta
n(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+3/16/c^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/c^2/(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 = 3.11, size = 217, normalized size = 0.83 \begin {gather*} -\frac {{\left (B b - A c\right )} x^{\frac {3}{2}}}{2 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} + \frac {{\left (3 \, B b + A c\right )} {\left (\frac {2 \, \sqrt {2} \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 {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \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 {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.23, size = 91, normalized size = 0.35 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,{\left (-b\right )}^{5/4}\,c^{7/4}}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,{\left (-b\right )}^{5/4}\,c^{7/4}}+\frac {x^{3/2}\,\left (A\,c-B\,b\right )}{2\,b\,c\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

(atanh((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 3*B*b))/(4*(-b)^(5/4)*c^(7/4)) - (atan((c^(1/4)*x^(1/2))/(-b)^(1/4
))*(A*c + 3*B*b))/(4*(-b)^(5/4)*c^(7/4)) + (x^(3/2)*(A*c - B*b))/(2*b*c*(b + c*x^2))

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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**(9/2)*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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