3.183 \(\int \frac {x^{13/2} (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=278 \[ -\frac {b^{7/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}-\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{15/4}}+\frac {2 b x^{3/2} (b B-A c)}{3 c^3}-\frac {2 x^{7/2} (b B-A c)}{7 c^2}+\frac {2 B x^{11/2}}{11 c} \]

[Out]

2/3*b*(-A*c+B*b)*x^(3/2)/c^3-2/7*(-A*c+B*b)*x^(7/2)/c^2+2/11*B*x^(11/2)/c+1/2*b^(7/4)*(-A*c+B*b)*arctan(1-c^(1
/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(15/4)*2^(1/2)-1/2*b^(7/4)*(-A*c+B*b)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))
/c^(15/4)*2^(1/2)-1/4*b^(7/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(15/4)*2^(1/2
)+1/4*b^(7/4)*(-A*c+B*b)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(15/4)*2^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 278, 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, 297, 1162, 617, 204, 1165, 628} \[ -\frac {b^{7/4} (b B-A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{15/4}}+\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}-\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{15/4}}-\frac {2 x^{7/2} (b B-A c)}{7 c^2}+\frac {2 b x^{3/2} (b B-A c)}{3 c^3}+\frac {2 B x^{11/2}}{11 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

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^{13/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {x^{9/2} \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {2 B x^{11/2}}{11 c}-\frac {\left (2 \left (\frac {11 b B}{2}-\frac {11 A c}{2}\right )\right ) \int \frac {x^{9/2}}{b+c x^2} \, dx}{11 c}\\ &=-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}+\frac {(b (b B-A c)) \int \frac {x^{5/2}}{b+c x^2} \, dx}{c^2}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}-\frac {\left (b^2 (b B-A c)\right ) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{c^3}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}-\frac {\left (2 b^2 (b B-A c)\right ) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{c^3}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}+\frac {\left (b^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^{7/2}}-\frac {\left (b^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^{7/2}}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}-\frac {\left (b^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^4}-\frac {\left (b^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^4}-\frac {\left (b^{7/4} (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^{15/4}}-\frac {\left (b^{7/4} (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^{15/4}}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}-\frac {b^{7/4} (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^{15/4}}+\frac {b^{7/4} (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^{15/4}}-\frac {\left (b^{7/4} (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^{15/4}}+\frac {\left (b^{7/4} (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^{15/4}}\\ &=\frac {2 b (b B-A c) x^{3/2}}{3 c^3}-\frac {2 (b B-A c) x^{7/2}}{7 c^2}+\frac {2 B x^{11/2}}{11 c}+\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}-\frac {b^{7/4} (b B-A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{15/4}}-\frac {b^{7/4} (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^{15/4}}+\frac {b^{7/4} (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^{15/4}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 133, normalized size = 0.48 \[ \frac {2 x^{3/2} \left (-11 b c \left (7 A+3 B x^2\right )+3 c^2 x^2 \left (11 A+7 B x^2\right )+77 b^2 B\right )}{231 c^3}+\frac {b (-b)^{3/4} (b B-A c) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{c^{15/4}}+\frac {(-b)^{7/4} (b B-A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{c^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(2*x^(3/2)*(77*b^2*B - 11*b*c*(7*A + 3*B*x^2) + 3*c^2*x^2*(11*A + 7*B*x^2)))/(231*c^3) + ((-b)^(3/4)*b*(b*B -
A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/c^(15/4) + ((-b)^(7/4)*(b*B - A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(
1/4)])/c^(15/4)

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fricas [B]  time = 0.91, size = 920, normalized size = 3.31 \[ -\frac {924 \, c^{3} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (B^{6} b^{16} - 6 \, A B^{5} b^{15} c + 15 \, A^{2} B^{4} b^{14} c^{2} - 20 \, A^{3} B^{3} b^{13} c^{3} + 15 \, A^{4} B^{2} b^{12} c^{4} - 6 \, A^{5} B b^{11} c^{5} + A^{6} b^{10} c^{6}\right )} x - {\left (B^{4} b^{11} c^{7} - 4 \, A B^{3} b^{10} c^{8} + 6 \, A^{2} B^{2} b^{9} c^{9} - 4 \, A^{3} B b^{8} c^{10} + A^{4} b^{7} c^{11}\right )} \sqrt {-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}}} c^{4} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {1}{4}} + {\left (B^{3} b^{8} c^{4} - 3 \, A B^{2} b^{7} c^{5} + 3 \, A^{2} B b^{6} c^{6} - A^{3} b^{5} c^{7}\right )} \sqrt {x} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {1}{4}}}{B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}\right ) - 231 \, c^{3} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (c^{11} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{8} - 3 \, A B^{2} b^{7} c + 3 \, A^{2} B b^{6} c^{2} - A^{3} b^{5} c^{3}\right )} \sqrt {x}\right ) + 231 \, c^{3} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {1}{4}} \log \left (-c^{11} \left (-\frac {B^{4} b^{11} - 4 \, A B^{3} b^{10} c + 6 \, A^{2} B^{2} b^{9} c^{2} - 4 \, A^{3} B b^{8} c^{3} + A^{4} b^{7} c^{4}}{c^{15}}\right )^{\frac {3}{4}} - {\left (B^{3} b^{8} - 3 \, A B^{2} b^{7} c + 3 \, A^{2} B b^{6} c^{2} - A^{3} b^{5} c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (21 \, B c^{2} x^{5} - 33 \, {\left (B b c - A c^{2}\right )} x^{3} + 77 \, {\left (B b^{2} - A b c\right )} x\right )} \sqrt {x}}{462 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/462*(924*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*
arctan((sqrt((B^6*b^16 - 6*A*B^5*b^15*c + 15*A^2*B^4*b^14*c^2 - 20*A^3*B^3*b^13*c^3 + 15*A^4*B^2*b^12*c^4 - 6*
A^5*B*b^11*c^5 + A^6*b^10*c^6)*x - (B^4*b^11*c^7 - 4*A*B^3*b^10*c^8 + 6*A^2*B^2*b^9*c^9 - 4*A^3*B*b^8*c^10 + A
^4*b^7*c^11)*sqrt(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15))*c^4*
(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4) + (B^3*b^8*c^4 -
3*A*B^2*b^7*c^5 + 3*A^2*B*b^6*c^6 - A^3*b^5*c^7)*sqrt(x)*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4
*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4))/(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 +
A^4*b^7*c^4)) - 231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15
)^(1/4)*log(c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4)
- (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b^5*c^3)*sqrt(x)) + 231*c^3*(-(B^4*b^11 - 4*A*B^3*b^10*c +
6*A^2*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(1/4)*log(-c^11*(-(B^4*b^11 - 4*A*B^3*b^10*c + 6*A^2
*B^2*b^9*c^2 - 4*A^3*B*b^8*c^3 + A^4*b^7*c^4)/c^15)^(3/4) - (B^3*b^8 - 3*A*B^2*b^7*c + 3*A^2*B*b^6*c^2 - A^3*b
^5*c^3)*sqrt(x)) - 4*(21*B*c^2*x^5 - 33*(B*b*c - A*c^2)*x^3 + 77*(B*b^2 - A*b*c)*x)*sqrt(x))/c^3

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giac [A]  time = 0.24, size = 298, normalized size = 1.07 \[ -\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{6}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, c^{6}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{6}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b^{2} - \left (b c^{3}\right )^{\frac {3}{4}} A b c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{6}} + \frac {2 \, {\left (21 \, B c^{10} x^{\frac {11}{2}} - 33 \, B b c^{9} x^{\frac {7}{2}} + 33 \, A c^{10} x^{\frac {7}{2}} + 77 \, B b^{2} c^{8} x^{\frac {3}{2}} - 77 \, A b c^{9} x^{\frac {3}{2}}\right )}}{231 \, c^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(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^2 - (b*c^3)^(3/4)*A*b*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/
(b/c)^(1/4))/c^6 - 1/2*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^6 + 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*log(sqrt(2)*sqrt
(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 - 1/4*sqrt(2)*((b*c^3)^(3/4)*B*b^2 - (b*c^3)^(3/4)*A*b*c)*log(-sqrt(2)*sq
rt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^6 + 2/231*(21*B*c^10*x^(11/2) - 33*B*b*c^9*x^(7/2) + 33*A*c^10*x^(7/2) +
77*B*b^2*c^8*x^(3/2) - 77*A*b*c^9*x^(3/2))/c^11

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maple [A]  time = 0.08, size = 336, normalized size = 1.21 \[ \frac {2 B \,x^{\frac {11}{2}}}{11 c}+\frac {2 A \,x^{\frac {7}{2}}}{7 c}-\frac {2 B b \,x^{\frac {7}{2}}}{7 c^{2}}-\frac {2 A b \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {2 B \,b^{2} x^{\frac {3}{2}}}{3 c^{3}}+\frac {\sqrt {2}\, A \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\sqrt {2}\, A \,b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\sqrt {2}\, A \,b^{2} \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}-\frac {\sqrt {2}\, B \,b^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}-\frac {\sqrt {2}\, B \,b^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}}-\frac {\sqrt {2}\, B \,b^{3} \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

2/11*B*x^(11/2)/c+2/7/c*x^(7/2)*A-2/7/c^2*x^(7/2)*b*B-2/3/c^2*x^(3/2)*A*b+2/3/c^3*x^(3/2)*B*b^2+1/4*b^2/c^3/(b
/c)^(1/4)*2^(1/2)*A*ln((x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))
)+1/2*b^2/c^3/(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)*A*ar
ctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-1/4*b^3/c^4/(b/c)^(1/4)*2^(1/2)*B*ln((x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^
(1/2))/(x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))-1/2*b^3/c^4/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/
4)*x^(1/2)+1)-1/2*b^3/c^4/(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 = 3.07, size = 237, normalized size = 0.85 \[ -\frac {{\left (B b^{3} - A b^{2} c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, c^{3}} + \frac {2 \, {\left (21 \, B c^{2} x^{\frac {11}{2}} - 33 \, {\left (B b c - A c^{2}\right )} x^{\frac {7}{2}} + 77 \, {\left (B b^{2} - A b c\right )} x^{\frac {3}{2}}\right )}}{231 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

-1/4*(B*b^3 - A*b^2*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)))/c^3 + 2/231*(21*B*c^2*x^(11/2) - 33*(B*b*c - A*c^2)*x^(7/2) + 77*(B*b^2 - A*b*c)*x^(3/2
))/c^3

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mupad [B]  time = 0.23, size = 115, normalized size = 0.41 \[ x^{7/2}\,\left (\frac {2\,A}{7\,c}-\frac {2\,B\,b}{7\,c^2}\right )+\frac {2\,B\,x^{11/2}}{11\,c}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )}{c^{15/4}}-\frac {b\,x^{3/2}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )}{3\,c}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{c^{15/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(13/2)*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^(7/2)*((2*A)/(7*c) - (2*B*b)/(7*c^2)) + (2*B*x^(11/2))/(11*c) + ((-b)^(7/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4
))*(A*c - B*b))/c^(15/4) + ((-b)^(7/4)*atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*(A*c - B*b)*1i)/c^(15/4) - (b*x^(
3/2)*((2*A)/c - (2*B*b)/c^2))/(3*c)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

Timed out

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