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

Optimal. Leaf size=257 \[ \frac {b^{3/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{3/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{11/4}}+\frac {b^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{11/4}}-\frac {2 x^{3/2} (b B-A c)}{3 c^2}+\frac {2 B x^{7/2}}{7 c} \]

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Rubi [A]  time = 0.22, antiderivative size = 257, 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, 459, 321, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {b^{3/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{3/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{3/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{11/4}}+\frac {b^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{11/4}}-\frac {2 x^{3/2} (b B-A c)}{3 c^2}+\frac {2 B x^{7/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 459

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

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

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Mathematica [A]  time = 0.15, size = 110, normalized size = 0.43 \begin {gather*} \frac {2 c^{3/4} x^{3/2} \left (7 A c-7 b B+3 B c x^2\right )-21 (-b)^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+21 (-b)^{3/4} (b B-A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{21 c^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^(3/4)*x^(3/2)*(-7*b*B + 7*A*c + 3*B*c*x^2) - 21*(-b)^(3/4)*(b*B - A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4
)] + 21*(-b)^(3/4)*(b*B - A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(21*c^(11/4))

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IntegrateAlgebraic [A]  time = 0.21, size = 160, normalized size = 0.62 \begin {gather*} -\frac {\left (b^{7/4} B-A b^{3/4} c\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} c^{11/4}}-\frac {\left (b^{7/4} B-A b^{3/4} c\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{11/4}}+\frac {2 x^{3/2} \left (7 A c-7 b B+3 B c x^2\right )}{21 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(-7*b*B + 7*A*c + 3*B*c*x^2))/(21*c^2) - ((b^(7/4)*B - A*b^(3/4)*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(S
qrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*c^(11/4)) - ((b^(7/4)*B - A*b^(3/4)*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(
1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*c^(11/4))

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fricas [B]  time = 0.45, size = 899, normalized size = 3.50 \begin {gather*} \frac {84 \, c^{2} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (B^{6} b^{10} - 6 \, A B^{5} b^{9} c + 15 \, A^{2} B^{4} b^{8} c^{2} - 20 \, A^{3} B^{3} b^{7} c^{3} + 15 \, A^{4} B^{2} b^{6} c^{4} - 6 \, A^{5} B b^{5} c^{5} + A^{6} b^{4} c^{6}\right )} x - {\left (B^{4} b^{7} c^{5} - 4 \, A B^{3} b^{6} c^{6} + 6 \, A^{2} B^{2} b^{5} c^{7} - 4 \, A^{3} B b^{4} c^{8} + A^{4} b^{3} c^{9}\right )} \sqrt {-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}}} c^{3} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {1}{4}} + {\left (B^{3} b^{5} c^{3} - 3 \, A B^{2} b^{4} c^{4} + 3 \, A^{2} B b^{3} c^{5} - A^{3} b^{2} c^{6}\right )} \sqrt {x} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {1}{4}}}{B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}\right ) - 21 \, c^{2} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {1}{4}} \log \left (c^{8} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{5} - 3 \, A B^{2} b^{4} c + 3 \, A^{2} B b^{3} c^{2} - A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 21 \, c^{2} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {1}{4}} \log \left (-c^{8} \left (-\frac {B^{4} b^{7} - 4 \, A B^{3} b^{6} c + 6 \, A^{2} B^{2} b^{5} c^{2} - 4 \, A^{3} B b^{4} c^{3} + A^{4} b^{3} c^{4}}{c^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{5} - 3 \, A B^{2} b^{4} c + 3 \, A^{2} B b^{3} c^{2} - A^{3} b^{2} c^{3}\right )} \sqrt {x}\right ) + 4 \, {\left (3 \, B c x^{3} - 7 \, {\left (B b - A c\right )} x\right )} \sqrt {x}}{42 \, c^{2}} \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),x, algorithm="fricas")

[Out]

1/42*(84*c^2*(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)/c^11)^(1/4)*arcta
n((sqrt((B^6*b^10 - 6*A*B^5*b^9*c + 15*A^2*B^4*b^8*c^2 - 20*A^3*B^3*b^7*c^3 + 15*A^4*B^2*b^6*c^4 - 6*A^5*B*b^5
*c^5 + A^6*b^4*c^6)*x - (B^4*b^7*c^5 - 4*A*B^3*b^6*c^6 + 6*A^2*B^2*b^5*c^7 - 4*A^3*B*b^4*c^8 + A^4*b^3*c^9)*sq
rt(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)/c^11))*c^3*(-(B^4*b^7 - 4*A*
B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)/c^11)^(1/4) + (B^3*b^5*c^3 - 3*A*B^2*b^4*c^4 +
3*A^2*B*b^3*c^5 - A^3*b^2*c^6)*sqrt(x)*(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*
b^3*c^4)/c^11)^(1/4))/(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)) - 21*c^2*
(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)/c^11)^(1/4)*log(c^8*(-(B^4*b^7
- 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^4*b^3*c^4)/c^11)^(3/4) - (B^3*b^5 - 3*A*B^2*b^4*c +
3*A^2*B*b^3*c^2 - A^3*b^2*c^3)*sqrt(x)) + 21*c^2*(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4
*c^3 + A^4*b^3*c^4)/c^11)^(1/4)*log(-c^8*(-(B^4*b^7 - 4*A*B^3*b^6*c + 6*A^2*B^2*b^5*c^2 - 4*A^3*B*b^4*c^3 + A^
4*b^3*c^4)/c^11)^(3/4) - (B^3*b^5 - 3*A*B^2*b^4*c + 3*A^2*B*b^3*c^2 - A^3*b^2*c^3)*sqrt(x)) + 4*(3*B*c*x^3 - 7
*(B*b - A*c)*x)*sqrt(x))/c^2

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giac [A]  time = 0.21, size = 264, normalized size = 1.03 \begin {gather*} \frac {\sqrt {2} {\left (\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 )}{2 \, c^{5}} + \frac {\sqrt {2} {\left (\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 )}{2 \, c^{5}} - \frac {\sqrt {2} {\left (\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 )}{4 \, c^{5}} + \frac {\sqrt {2} {\left (\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 )}{4 \, c^{5}} + \frac {2 \, {\left (3 \, B c^{6} x^{\frac {7}{2}} - 7 \, B b c^{5} x^{\frac {3}{2}} + 7 \, A c^{6} x^{\frac {3}{2}}\right )}}{21 \, c^{7}} \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),x, algorithm="giac")

[Out]

1/2*sqrt(2)*((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))/c^5 + 1/2*sqrt(2)*((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))/c^5 - 1/4*sqrt(2)*((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))/c^5 + 1/4*sqrt(2)*((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))/c^5 + 2/21*(3*B*c^6*x^(7/2) - 7*B*b*c^5*x^(3/2) + 7*A*c^6*x^(3/2))/c^7

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maple [A]  time = 0.05, size = 308, normalized size = 1.20 \begin {gather*} \frac {2 B \,x^{\frac {7}{2}}}{7 c}+\frac {2 A \,x^{\frac {3}{2}}}{3 c}-\frac {2 B b \,x^{\frac {3}{2}}}{3 c^{2}}-\frac {\sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {\sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}-\frac {\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 )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{2}}+\frac {\sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\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 )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}} \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),x)

[Out]

2/7*B*x^(7/2)/c+2/3/c*x^(3/2)*A-2/3/c^2*x^(3/2)*b*B-1/2*b/c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)
*x^(1/2)-1)-1/4*b/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)))-1/2*b/c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+1/2*b^2/c^3/(b
/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+1/4*b^2/c^3/(b/c)^(1/4)*2^(1/2)*B*ln((x-(b/c)^(1/4)*
2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))+1/2*b^2/c^3/(b/c)^(1/4)*2^(1/2)*B*ar
ctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)

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maxima [A]  time = 3.13, size = 214, normalized size = 0.83 \begin {gather*} \frac {{\left (B b^{2} - A b 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 )}}{4 \, c^{2}} + \frac {2 \, {\left (3 \, B c x^{\frac {7}{2}} - 7 \, {\left (B b - A c\right )} x^{\frac {3}{2}}\right )}}{21 \, c^{2}} \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),x, algorithm="maxima")

[Out]

1/4*(B*b^2 - A*b*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)*s
qrt(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)))/c^2 + 2/21*(3*B*c*x^(7/2) - 7*(B*b - A*c)*x^(3/2))/c^2

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mupad [B]  time = 0.24, size = 92, normalized size = 0.36 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,c}-\frac {2\,B\,b}{3\,c^2}\right )+\frac {2\,B\,x^{7/2}}{7\,c}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )}{c^{11/4}}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{11/4}} \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),x)

[Out]

x^(3/2)*((2*A)/(3*c) - (2*B*b)/(3*c^2)) + (2*B*x^(7/2))/(7*c) + ((-b)^(3/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4))
*(A*c - B*b))/c^(11/4) + ((-b)^(3/4)*atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*(A*c - B*b)*1i)/c^(11/4)

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

[Out]

Timed out

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