3.3.16 \(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=343 \[ -\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}+\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.28, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1584, 457, 290, 325, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}+\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{17/4}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(9*(5*b*B - 13*A*c))/(80*b^3*c*x^(5/2)) - (9*(5*b*B - 13*A*c))/(16*b^4*Sqrt[x]) - (b*B - A*c)/(4*b*c*x^(5/2)*(
b + c*x^2)^2) - (5*b*B - 13*A*c)/(16*b^2*c*x^(5/2)*(b + c*x^2)) + (9*c^(1/4)*(5*b*B - 13*A*c)*ArcTan[1 - (Sqrt
[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(17/4)) - (9*c^(1/4)*(5*b*B - 13*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)
*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(17/4)) - (9*c^(1/4)*(5*b*B - 13*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*
Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(17/4)) + (9*c^(1/4)*(5*b*B - 13*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4
)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(17/4))

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 290

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

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 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^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^{7/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}+\frac {\left (-\frac {5 b B}{2}+\frac {13 A c}{2}\right ) \int \frac {1}{x^{7/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {(9 (5 b B-13 A c)) \int \frac {1}{x^{7/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}+\frac {(9 (5 b B-13 A c)) \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {(9 c (5 b B-13 A c)) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 b^4}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {(9 c (5 b B-13 A c)) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^4}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}+\frac {\left (9 \sqrt {c} (5 b B-13 A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^4}-\frac {\left (9 \sqrt {c} (5 b B-13 A c)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^4}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {(9 (5 b B-13 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 )}{64 b^4}-\frac {(9 (5 b B-13 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 )}{64 b^4}-\frac {\left (9 \sqrt [4]{c} (5 b B-13 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 )}{64 \sqrt {2} b^{17/4}}-\frac {\left (9 \sqrt [4]{c} (5 b B-13 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 )}{64 \sqrt {2} b^{17/4}}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}-\frac {\left (9 \sqrt [4]{c} (5 b B-13 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 )}{32 \sqrt {2} b^{17/4}}+\frac {\left (9 \sqrt [4]{c} (5 b B-13 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 )}{32 \sqrt {2} b^{17/4}}\\ &=\frac {9 (5 b B-13 A c)}{80 b^3 c x^{5/2}}-\frac {9 (5 b B-13 A c)}{16 b^4 \sqrt {x}}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}-\frac {5 b B-13 A c}{16 b^2 c x^{5/2} \left (b+c x^2\right )}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{17/4}}\\ \end {align*}

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Mathematica [C]  time = 0.52, size = 189, normalized size = 0.55 \begin {gather*} -\frac {2 c x^{3/2} (b B-2 A c) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^5}+\frac {2 c x^{3/2} (A c-b B) \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^5}+\frac {6 A c-2 b B}{b^4 \sqrt {x}}-\frac {2 A}{5 b^3 x^{5/2}}+\frac {\sqrt [4]{c} (3 A c-b B) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{(-b)^{17/4}}+\frac {\sqrt [4]{c} (b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{(-b)^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(-2*A)/(5*b^3*x^(5/2)) + (-2*b*B + 6*A*c)/(b^4*Sqrt[x]) + (c^(1/4)*(-(b*B) + 3*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(
-b)^(1/4)])/(-b)^(17/4) + (c^(1/4)*(b*B - 3*A*c)*ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(17/4) - (2*c*(b*
B - 2*A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(3*b^5) + (2*c*(-(b*B) + A*c)*x^(3/2)*Hyperge
ometric2F1[3/4, 3, 7/4, -((c*x^2)/b)])/(3*b^5)

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IntegrateAlgebraic [A]  time = 0.66, size = 224, normalized size = 0.65 \begin {gather*} \frac {9 \left (5 b B \sqrt [4]{c}-13 A c^{5/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{32 \sqrt {2} b^{17/4}}+\frac {9 \left (5 b B \sqrt [4]{c}-13 A c^{5/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{17/4}}+\frac {-32 A b^3+416 A b^2 c x^2+1053 A b c^2 x^4+585 A c^3 x^6-160 b^3 B x^2-405 b^2 B c x^4-225 b B c^2 x^6}{80 b^4 x^{5/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(-32*A*b^3 - 160*b^3*B*x^2 + 416*A*b^2*c*x^2 - 405*b^2*B*c*x^4 + 1053*A*b*c^2*x^4 - 225*b*B*c^2*x^6 + 585*A*c^
3*x^6)/(80*b^4*x^(5/2)*(b + c*x^2)^2) + (9*(5*b*B*c^(1/4) - 13*A*c^(5/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2
]*b^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*b^(17/4)) + (9*(5*b*B*c^(1/4) - 13*A*c^(5/4))*ArcTanh[(Sqrt[2]*b^(1/4
)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(17/4))

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fricas [B]  time = 0.46, size = 1043, normalized size = 3.04 \begin {gather*} -\frac {180 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {{\left (15625 \, B^{6} b^{6} c^{2} - 243750 \, A B^{5} b^{5} c^{3} + 1584375 \, A^{2} B^{4} b^{4} c^{4} - 5492500 \, A^{3} B^{3} b^{3} c^{5} + 10710375 \, A^{4} B^{2} b^{2} c^{6} - 11138790 \, A^{5} B b c^{7} + 4826809 \, A^{6} c^{8}\right )} x - {\left (625 \, B^{4} b^{13} c - 6500 \, A B^{3} b^{12} c^{2} + 25350 \, A^{2} B^{2} b^{11} c^{3} - 43940 \, A^{3} B b^{10} c^{4} + 28561 \, A^{4} b^{9} c^{5}\right )} \sqrt {-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}}} b^{4} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {1}{4}} + {\left (125 \, B^{3} b^{7} c - 975 \, A B^{2} b^{6} c^{2} + 2535 \, A^{2} B b^{5} c^{3} - 2197 \, A^{3} b^{4} c^{4}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {1}{4}}}{625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}\right ) - 45 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {1}{4}} \log \left (729 \, b^{13} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {3}{4}} - 729 \, {\left (125 \, B^{3} b^{3} c - 975 \, A B^{2} b^{2} c^{2} + 2535 \, A^{2} B b c^{3} - 2197 \, A^{3} c^{4}\right )} \sqrt {x}\right ) + 45 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {1}{4}} \log \left (-729 \, b^{13} \left (-\frac {625 \, B^{4} b^{4} c - 6500 \, A B^{3} b^{3} c^{2} + 25350 \, A^{2} B^{2} b^{2} c^{3} - 43940 \, A^{3} B b c^{4} + 28561 \, A^{4} c^{5}}{b^{17}}\right )^{\frac {3}{4}} - 729 \, {\left (125 \, B^{3} b^{3} c - 975 \, A B^{2} b^{2} c^{2} + 2535 \, A^{2} B b c^{3} - 2197 \, A^{3} c^{4}\right )} \sqrt {x}\right ) + 4 \, {\left (45 \, {\left (5 \, B b c^{2} - 13 \, A c^{3}\right )} x^{6} + 81 \, {\left (5 \, B b^{2} c - 13 \, A b c^{2}\right )} x^{4} + 32 \, A b^{3} + 32 \, {\left (5 \, B b^{3} - 13 \, A b^{2} c\right )} x^{2}\right )} \sqrt {x}}{320 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/320*(180*(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^
3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(1/4)*arctan((sqrt((15625*B^6*b^6*c^2 - 243750*A*B^5*b^5*c^3 + 15
84375*A^2*B^4*b^4*c^4 - 5492500*A^3*B^3*b^3*c^5 + 10710375*A^4*B^2*b^2*c^6 - 11138790*A^5*B*b*c^7 + 4826809*A^
6*c^8)*x - (625*B^4*b^13*c - 6500*A*B^3*b^12*c^2 + 25350*A^2*B^2*b^11*c^3 - 43940*A^3*B*b^10*c^4 + 28561*A^4*b
^9*c^5)*sqrt(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)
/b^17))*b^4*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)
/b^17)^(1/4) + (125*B^3*b^7*c - 975*A*B^2*b^6*c^2 + 2535*A^2*B*b^5*c^3 - 2197*A^3*b^4*c^4)*sqrt(x)*(-(625*B^4*
b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(1/4))/(625*B^4*
b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)) - 45*(b^4*c^2*x^7 + 2
*b^5*c*x^5 + b^6*x^3)*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 2856
1*A^4*c^5)/b^17)^(1/4)*log(729*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*
B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(125*B^3*b^3*c - 975*A*B^2*b^2*c^2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^
4)*sqrt(x)) + 45*(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^2*b
^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(1/4)*log(-729*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 +
25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(125*B^3*b^3*c - 975*A*B^2*b^2*c^
2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^4)*sqrt(x)) + 4*(45*(5*B*b*c^2 - 13*A*c^3)*x^6 + 81*(5*B*b^2*c - 13*A*b*c^2)
*x^4 + 32*A*b^3 + 32*(5*B*b^3 - 13*A*b^2*c)*x^2)*sqrt(x))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)

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giac [A]  time = 0.22, size = 326, normalized size = 0.95 \begin {gather*} -\frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \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 )}{64 \, b^{5} c^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \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 )}{64 \, b^{5} c^{2}} + \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \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 )}{128 \, b^{5} c^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \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 )}{128 \, b^{5} c^{2}} - \frac {13 \, B b c^{2} x^{\frac {7}{2}} - 21 \, A c^{3} x^{\frac {7}{2}} + 17 \, B b^{2} c x^{\frac {3}{2}} - 25 \, A b c^{2} x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{4}} - \frac {2 \, {\left (5 \, B b x^{2} - 15 \, A c x^{2} + A b\right )}}{5 \, b^{4} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-9/64*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(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^5*c^2) - 9/64*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(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^5*c^2) + 9/128*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*
c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c^2) - 9/128*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(b*c^3
)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c^2) - 1/16*(13*B*b*c^2*x^(7/2) - 21*A*c^3
*x^(7/2) + 17*B*b^2*c*x^(3/2) - 25*A*b*c^2*x^(3/2))/((c*x^2 + b)^2*b^4) - 2/5*(5*B*b*x^2 - 15*A*c*x^2 + A*b)/(
b^4*x^(5/2))

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maple [A]  time = 0.07, size = 381, normalized size = 1.11 \begin {gather*} \frac {21 A \,c^{3} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {13 B \,c^{2} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {25 A \,c^{2} x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {17 B c \,x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {117 \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{4}}+\frac {117 \sqrt {2}\, A c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{4}}+\frac {117 \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 )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{4}}-\frac {45 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {45 \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {45 \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 )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}+\frac {6 A c}{b^{4} \sqrt {x}}-\frac {2 B}{b^{3} \sqrt {x}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

21/16/b^4*c^3/(c*x^2+b)^2*x^(7/2)*A-13/16/b^3*c^2/(c*x^2+b)^2*x^(7/2)*B+25/16/b^3*c^2/(c*x^2+b)^2*A*x^(3/2)-17
/16/b^2*c/(c*x^2+b)^2*B*x^(3/2)+117/128/b^4*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)))+117/64/b^4*c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4
)*x^(1/2)+1)+117/64/b^4*c/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-45/128/b^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)))-45/64/b^3
/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-45/64/b^3/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b
/c)^(1/4)*x^(1/2)-1)-2/5*A/b^3/x^(5/2)+6/b^4/x^(1/2)*A*c-2/b^3/x^(1/2)*B

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maxima [A]  time = 3.04, size = 285, normalized size = 0.83 \begin {gather*} -\frac {45 \, {\left (5 \, B b c^{2} - 13 \, A c^{3}\right )} x^{6} + 81 \, {\left (5 \, B b^{2} c - 13 \, A b c^{2}\right )} x^{4} + 32 \, A b^{3} + 32 \, {\left (5 \, B b^{3} - 13 \, A b^{2} c\right )} x^{2}}{80 \, {\left (b^{4} c^{2} x^{\frac {13}{2}} + 2 \, b^{5} c x^{\frac {9}{2}} + b^{6} x^{\frac {5}{2}}\right )}} - \frac {9 \, {\left (5 \, B b c - 13 \, A c^{2}\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 )}}{128 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

-1/80*(45*(5*B*b*c^2 - 13*A*c^3)*x^6 + 81*(5*B*b^2*c - 13*A*b*c^2)*x^4 + 32*A*b^3 + 32*(5*B*b^3 - 13*A*b^2*c)*
x^2)/(b^4*c^2*x^(13/2) + 2*b^5*c*x^(9/2) + b^6*x^(5/2)) - 9/128*(5*B*b*c - 13*A*c^2)*(2*sqrt(2)*arctan(1/2*sqr
t(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)) + sq
rt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/b^4

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mupad [B]  time = 0.19, size = 152, normalized size = 0.44 \begin {gather*} \frac {\frac {2\,x^2\,\left (13\,A\,c-5\,B\,b\right )}{5\,b^2}-\frac {2\,A}{5\,b}+\frac {9\,c^2\,x^6\,\left (13\,A\,c-5\,B\,b\right )}{16\,b^4}+\frac {81\,c\,x^4\,\left (13\,A\,c-5\,B\,b\right )}{80\,b^3}}{b^2\,x^{5/2}+c^2\,x^{13/2}+2\,b\,c\,x^{9/2}}+\frac {9\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-5\,B\,b\right )}{32\,b^{17/4}}-\frac {9\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-5\,B\,b\right )}{32\,b^{17/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

((2*x^2*(13*A*c - 5*B*b))/(5*b^2) - (2*A)/(5*b) + (9*c^2*x^6*(13*A*c - 5*B*b))/(16*b^4) + (81*c*x^4*(13*A*c -
5*B*b))/(80*b^3))/(b^2*x^(5/2) + c^2*x^(13/2) + 2*b*c*x^(9/2)) + (9*(-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/
4))*(13*A*c - 5*B*b))/(32*b^(17/4)) - (9*(-c)^(1/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4))*(13*A*c - 5*B*b))/(32*
b^(17/4))

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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**(5/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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