3.2.84 \(\int \frac {x^{11/2} (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=276 \[ \frac {b^{5/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{13/4}}-\frac {b^{5/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{13/4}}+\frac {b^{5/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{13/4}}-\frac {b^{5/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{13/4}}+\frac {2 b \sqrt {x} (b B-A c)}{c^3}-\frac {2 x^{5/2} (b B-A c)}{5 c^2}+\frac {2 B x^{9/2}}{9 c} \]

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Rubi [A]  time = 0.25, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1584, 459, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {b^{5/4} (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^{13/4}}-\frac {b^{5/4} (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^{13/4}}+\frac {b^{5/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{13/4}}-\frac {b^{5/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{13/4}}-\frac {2 x^{5/2} (b B-A c)}{5 c^2}+\frac {2 b \sqrt {x} (b B-A c)}{c^3}+\frac {2 B x^{9/2}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(2*b*(b*B - A*c)*Sqrt[x])/c^3 - (2*(b*B - A*c)*x^(5/2))/(5*c^2) + (2*B*x^(9/2))/(9*c) + (b^(5/4)*(b*B - A*c)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*ArcTan[1 + (Sqrt[2]*c^
(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(13/4)) + (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[
x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(13/4)) - (b^(5/4)*(b*B - A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + S
qrt[c]*x])/(2*Sqrt[2]*c^(13/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

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

Rule 321

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

Rule 329

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

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

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Mathematica [A]  time = 0.31, size = 227, normalized size = 0.82 \begin {gather*} \frac {\frac {45 \sqrt {2} b^{5/4} (b B-A c) \left (\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )\right )}{\sqrt [4]{c}}+\frac {90 \sqrt {2} b^{5/4} (b B-A c) \left (\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )\right )}{\sqrt [4]{c}}+72 c x^{5/2} (A c-b B)+360 b \sqrt {x} (b B-A c)+40 B c^2 x^{9/2}}{180 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(360*b*(b*B - A*c)*Sqrt[x] + 72*c*(-(b*B) + A*c)*x^(5/2) + 40*B*c^2*x^(9/2) + (90*Sqrt[2]*b^(5/4)*(b*B - A*c)*
(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]))/c^(1/4) + (45
*Sqrt[2]*b^(5/4)*(b*B - A*c)*(Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - Log[Sqrt[b] + Sqrt[
2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/c^(1/4))/(180*c^3)

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IntegrateAlgebraic [A]  time = 0.30, size = 181, normalized size = 0.66 \begin {gather*} \frac {\left (b^{9/4} B-A b^{5/4} 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^{13/4}}-\frac {\left (b^{9/4} B-A b^{5/4} 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^{13/4}}+\frac {2 \sqrt {x} \left (-45 A b c+9 A c^2 x^2+45 b^2 B-9 b B c x^2+5 B c^2 x^4\right )}{45 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(2*Sqrt[x]*(45*b^2*B - 45*A*b*c - 9*b*B*c*x^2 + 9*A*c^2*x^2 + 5*B*c^2*x^4))/(45*c^3) + ((b^(9/4)*B - A*b^(5/4)
*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*c^(13/4)) - ((b^(9/4)*B - A*b^(5
/4)*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt[2]*c^(13/4))

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fricas [B]  time = 0.43, size = 714, normalized size = 2.59 \begin {gather*} \frac {180 \, c^{3} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{6} \sqrt {-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}} + {\left (B^{2} b^{4} - 2 \, A B b^{3} c + A^{2} b^{2} c^{2}\right )} x} c^{10} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {3}{4}} + {\left (B b^{2} c^{10} - A b c^{11}\right )} \sqrt {x} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {3}{4}}}{B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}\right ) + 45 \, c^{3} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {1}{4}} \log \left (c^{3} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {1}{4}} - {\left (B b^{2} - A b c\right )} \sqrt {x}\right ) - 45 \, c^{3} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {1}{4}} \log \left (-c^{3} \left (-\frac {B^{4} b^{9} - 4 \, A B^{3} b^{8} c + 6 \, A^{2} B^{2} b^{7} c^{2} - 4 \, A^{3} B b^{6} c^{3} + A^{4} b^{5} c^{4}}{c^{13}}\right )^{\frac {1}{4}} - {\left (B b^{2} - A b c\right )} \sqrt {x}\right ) + 4 \, {\left (5 \, B c^{2} x^{4} + 45 \, B b^{2} - 45 \, A b c - 9 \, {\left (B b c - A c^{2}\right )} x^{2}\right )} \sqrt {x}}{90 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/90*(180*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4)*arct
an((sqrt(c^6*sqrt(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13) + (B^2*
b^4 - 2*A*B*b^3*c + A^2*b^2*c^2)*x)*c^10*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^
4*b^5*c^4)/c^13)^(3/4) + (B*b^2*c^10 - A*b*c^11)*sqrt(x)*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^
3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(3/4))/(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b
^5*c^4)) + 45*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4)*
log(c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4) - (B*b^2 -
A*b*c)*sqrt(x)) - 45*c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13
)^(1/4)*log(-c^3*(-(B^4*b^9 - 4*A*B^3*b^8*c + 6*A^2*B^2*b^7*c^2 - 4*A^3*B*b^6*c^3 + A^4*b^5*c^4)/c^13)^(1/4) -
(B*b^2 - A*b*c)*sqrt(x)) + 4*(5*B*c^2*x^4 + 45*B*b^2 - 45*A*b*c - 9*(B*b*c - A*c^2)*x^2)*sqrt(x))/c^3

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giac [A]  time = 0.20, size = 298, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{4}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{4}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{4}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {1}{4}} A b c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{4}} + \frac {2 \, {\left (5 \, B c^{8} x^{\frac {9}{2}} - 9 \, B b c^{7} x^{\frac {5}{2}} + 9 \, A c^{8} x^{\frac {5}{2}} + 45 \, B b^{2} c^{6} \sqrt {x} - 45 \, A b c^{7} \sqrt {x}\right )}}{45 \, c^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/
(b/c)^(1/4))/c^4 - 1/2*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 - 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b^2 - (b*c^3)^(1/4)*A*b*c)*log(-sqrt(2)*sq
rt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 2/45*(5*B*c^8*x^(9/2) - 9*B*b*c^7*x^(5/2) + 9*A*c^8*x^(5/2) + 45*B*b^
2*c^6*sqrt(x) - 45*A*b*c^7*sqrt(x))/c^9

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maple [A]  time = 0.05, size = 330, normalized size = 1.20 \begin {gather*} \frac {2 B \,x^{\frac {9}{2}}}{9 c}+\frac {2 A \,x^{\frac {5}{2}}}{5 c}-\frac {2 B b \,x^{\frac {5}{2}}}{5 c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2}}+\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A b \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 c^{2}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c^{3}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c^{3}}-\frac {\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,b^{2} \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 c^{3}}-\frac {2 A b \sqrt {x}}{c^{2}}+\frac {2 B \,b^{2} \sqrt {x}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 3.01, size = 259, normalized size = 0.94 \begin {gather*} -\frac {{\left (\frac {2 \, \sqrt {2} {\left (B b - 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 (B b - 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 (B b - 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 (B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )} b^{2}}{4 \, c^{3}} + \frac {2 \, {\left (5 \, B c^{2} x^{\frac {9}{2}} - 9 \, {\left (B b c - A c^{2}\right )} x^{\frac {5}{2}} + 45 \, {\left (B b^{2} - A b c\right )} \sqrt {x}\right )}}{45 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.25, size = 788, normalized size = 2.86 \begin {gather*} x^{5/2}\,\left (\frac {2\,A}{5\,c}-\frac {2\,B\,b}{5\,c^2}\right )+\frac {2\,B\,x^{9/2}}{9\,c}-\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}-\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )\,1{}\mathrm {i}}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )}{2\,c^{13/4}}+\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}+\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )\,1{}\mathrm {i}}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )}{2\,c^{13/4}}}{\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}-\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )\,1{}\mathrm {i}}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{13/4}}-\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}+\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )\,1{}\mathrm {i}}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{13/4}}}\right )\,\left (A\,c-B\,b\right )}{c^{13/4}}-\frac {b\,\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{c}-\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}-\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{13/4}}+\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}+\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{2\,c^{13/4}}}{\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}-\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )}{2\,c^{13/4}}-\frac {{\left (-b\right )}^{5/4}\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,b^4\,c^2-2\,A\,B\,b^5\,c+B^2\,b^6\right )}{c^3}+\frac {{\left (-b\right )}^{5/4}\,\left (A\,c-B\,b\right )\,\left (32\,B\,b^4-32\,A\,b^3\,c\right )}{2\,c^{13/4}}\right )\,\left (A\,c-B\,b\right )}{2\,c^{13/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^(5/2)*((2*A)/(5*c) - (2*B*b)/(5*c^2)) + (2*B*x^(9/2))/(9*c) - ((-b)^(5/4)*atan((((-b)^(5/4)*((16*x^(1/2)*(B^
2*b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 - ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4 - 32*A*b^3*c))/(2*c^(13/4)))*(A*c
- B*b)*1i)/(2*c^(13/4)) + ((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 + ((-b)^(5/4)*(A
*c - B*b)*(32*B*b^4 - 32*A*b^3*c))/(2*c^(13/4)))*(A*c - B*b)*1i)/(2*c^(13/4)))/(((-b)^(5/4)*((16*x^(1/2)*(B^2*
b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 - ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4 - 32*A*b^3*c))/(2*c^(13/4)))*(A*c -
B*b))/(2*c^(13/4)) - ((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 + ((-b)^(5/4)*(A*c -
B*b)*(32*B*b^4 - 32*A*b^3*c))/(2*c^(13/4)))*(A*c - B*b))/(2*c^(13/4))))*(A*c - B*b)*1i)/c^(13/4) - ((-b)^(5/4)
*atan((((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 - ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4
- 32*A*b^3*c)*1i)/(2*c^(13/4)))*(A*c - B*b))/(2*c^(13/4)) + ((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2 -
2*A*B*b^5*c))/c^3 + ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4 - 32*A*b^3*c)*1i)/(2*c^(13/4)))*(A*c - B*b))/(2*c^(13/4
)))/(((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2 - 2*A*B*b^5*c))/c^3 - ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4 -
32*A*b^3*c)*1i)/(2*c^(13/4)))*(A*c - B*b)*1i)/(2*c^(13/4)) - ((-b)^(5/4)*((16*x^(1/2)*(B^2*b^6 + A^2*b^4*c^2
- 2*A*B*b^5*c))/c^3 + ((-b)^(5/4)*(A*c - B*b)*(32*B*b^4 - 32*A*b^3*c)*1i)/(2*c^(13/4)))*(A*c - B*b)*1i)/(2*c^(
13/4))))*(A*c - B*b))/c^(13/4) - (b*x^(1/2)*((2*A)/c - (2*B*b)/c^2))/c

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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**(11/2)*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

Timed out

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