3.3.18 \(\int \frac {\sqrt {x} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=365 \[ \frac {13 c^{5/4} (9 b B-17 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{21/4}}+\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{21/4}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.32, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 16, 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 {13 c^{5/4} (9 b B-17 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{21/4}}+\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{21/4}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(13*(9*b*B - 17*A*c))/(144*b^3*c*x^(9/2)) - (13*(9*b*B - 17*A*c))/(80*b^4*x^(5/2)) + (13*c*(9*b*B - 17*A*c))/(
16*b^5*Sqrt[x]) - (b*B - A*c)/(4*b*c*x^(9/2)*(b + c*x^2)^2) - (9*b*B - 17*A*c)/(16*b^2*c*x^(9/2)*(b + c*x^2))
- (13*c^(5/4)*(9*b*B - 17*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(21/4)) + (13*c^(5
/4)*(9*b*B - 17*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(21/4)) + (13*c^(5/4)*(9*b*B
- 17*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(21/4)) - (13*c^(5/4)*(9*
b*B - 17*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(21/4))

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 {\sqrt {x} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^{11/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}+\frac {\left (-\frac {9 b B}{2}+\frac {17 A c}{2}\right ) \int \frac {1}{x^{11/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}-\frac {(13 (9 b B-17 A c)) \int \frac {1}{x^{11/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {(13 (9 b B-17 A c)) \int \frac {1}{x^{7/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}-\frac {(13 c (9 b B-17 A c)) \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {\left (13 c^2 (9 b B-17 A c)\right ) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{32 b^5}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {\left (13 c^2 (9 b B-17 A c)\right ) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^5}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}-\frac {\left (13 c^{3/2} (9 b B-17 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^5}+\frac {\left (13 c^{3/2} (9 b B-17 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^5}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {(13 c (9 b B-17 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^5}+\frac {(13 c (9 b B-17 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^5}+\frac {\left (13 c^{5/4} (9 b B-17 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^{21/4}}+\frac {\left (13 c^{5/4} (9 b B-17 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^{21/4}}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}+\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}+\frac {\left (13 c^{5/4} (9 b B-17 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^{21/4}}-\frac {\left (13 c^{5/4} (9 b B-17 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^{21/4}}\\ &=\frac {13 (9 b B-17 A c)}{144 b^3 c x^{9/2}}-\frac {13 (9 b B-17 A c)}{80 b^4 x^{5/2}}+\frac {13 c (9 b B-17 A c)}{16 b^5 \sqrt {x}}-\frac {b B-A c}{4 b c x^{9/2} \left (b+c x^2\right )^2}-\frac {9 b B-17 A c}{16 b^2 c x^{9/2} \left (b+c x^2\right )}-\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{21/4}}+\frac {13 c^{5/4} (9 b B-17 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{21/4}}+\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}-\frac {13 c^{5/4} (9 b B-17 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{21/4}}\\ \end {align*}

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Mathematica [C]  time = 0.55, size = 216, normalized size = 0.59 \begin {gather*} \frac {2 c^2 x^{3/2} (2 b B-3 A c) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^6}+\frac {2 c^2 x^{3/2} (b B-A c) \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};-\frac {c x^2}{b}\right )}{3 b^6}+\frac {6 c (b B-2 A c)}{b^5 \sqrt {x}}-\frac {2 (b B-3 A c)}{5 b^4 x^{5/2}}-\frac {2 A}{9 b^3 x^{9/2}}-\frac {3 c^{5/4} (b B-2 A c) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{(-b)^{21/4}}+\frac {3 c^{5/4} (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{(-b)^{21/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(9*b^3*x^(9/2)) - (2*(b*B - 3*A*c))/(5*b^4*x^(5/2)) + (6*c*(b*B - 2*A*c))/(b^5*Sqrt[x]) - (3*c^(5/4)*(b
*B - 2*A*c)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(-b)^(21/4) + (3*c^(5/4)*(b*B - 2*A*c)*ArcTanh[(c^(1/4)*Sqrt
[x])/(-b)^(1/4)])/(-b)^(21/4) + (2*c^2*(2*b*B - 3*A*c)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((c*x^2)/b)])/(
3*b^6) + (2*c^2*(b*B - A*c)*x^(3/2)*Hypergeometric2F1[3/4, 3, 7/4, -((c*x^2)/b)])/(3*b^6)

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IntegrateAlgebraic [A]  time = 0.70, size = 248, normalized size = 0.68 \begin {gather*} -\frac {13 \left (9 b B c^{5/4}-17 A c^{9/4}\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^{21/4}}-\frac {13 \left (9 b B c^{5/4}-17 A c^{9/4}\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^{21/4}}+\frac {-160 A b^4+544 A b^3 c x^2-7072 A b^2 c^2 x^4-17901 A b c^3 x^6-9945 A c^4 x^8-288 b^4 B x^2+3744 b^3 B c x^4+9477 b^2 B c^2 x^6+5265 b B c^3 x^8}{720 b^5 x^{9/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-160*A*b^4 - 288*b^4*B*x^2 + 544*A*b^3*c*x^2 + 3744*b^3*B*c*x^4 - 7072*A*b^2*c^2*x^4 + 9477*b^2*B*c^2*x^6 - 1
7901*A*b*c^3*x^6 + 5265*b*B*c^3*x^8 - 9945*A*c^4*x^8)/(720*b^5*x^(9/2)*(b + c*x^2)^2) - (13*(9*b*B*c^(5/4) - 1
7*A*c^(9/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*b^(21/4)) - (13*(9*b
*B*c^(5/4) - 17*A*c^(9/4))*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(21
/4))

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fricas [B]  time = 0.46, size = 1093, normalized size = 2.99

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/2880*(2340*(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*
b^2*c^7 - 176868*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21)^(1/4)*arctan((sqrt((531441*B^6*b^6*c^8 - 6022998*A*B^5*b^5
*c^9 + 28441935*A^2*B^4*b^4*c^10 - 71631540*A^3*B^3*b^3*c^11 + 101478015*A^4*B^2*b^2*c^12 - 76672278*A^5*B*b*c
^13 + 24137569*A^6*c^14)*x - (6561*B^4*b^15*c^5 - 49572*A*B^3*b^14*c^6 + 140454*A^2*B^2*b^13*c^7 - 176868*A^3*
B*b^12*c^8 + 83521*A^4*b^11*c^9)*sqrt(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 1768
68*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21))*b^5*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7
- 176868*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21)^(1/4) + (729*B^3*b^8*c^4 - 4131*A*B^2*b^7*c^5 + 7803*A^2*B*b^6*c^6
- 4913*A^3*b^5*c^7)*sqrt(x)*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176868*A^3*B
*b*c^8 + 83521*A^4*c^9)/b^21)^(1/4))/(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176868
*A^3*B*b*c^8 + 83521*A^4*c^9)) - 585*(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b
^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176868*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21)^(1/4)*log(2197*b^16*(-(6561*B^4*b^
4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176868*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21)^(3/4) - 2197*
(729*B^3*b^3*c^4 - 4131*A*B^2*b^2*c^5 + 7803*A^2*B*b*c^6 - 4913*A^3*c^7)*sqrt(x)) + 585*(b^5*c^2*x^9 + 2*b^6*c
*x^7 + b^7*x^5)*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176868*A^3*B*b*c^8 + 8352
1*A^4*c^9)/b^21)^(1/4)*log(-2197*b^16*(-(6561*B^4*b^4*c^5 - 49572*A*B^3*b^3*c^6 + 140454*A^2*B^2*b^2*c^7 - 176
868*A^3*B*b*c^8 + 83521*A^4*c^9)/b^21)^(3/4) - 2197*(729*B^3*b^3*c^4 - 4131*A*B^2*b^2*c^5 + 7803*A^2*B*b*c^6 -
4913*A^3*c^7)*sqrt(x)) + 4*(585*(9*B*b*c^3 - 17*A*c^4)*x^8 + 1053*(9*B*b^2*c^2 - 17*A*b*c^3)*x^6 - 160*A*b^4
+ 416*(9*B*b^3*c - 17*A*b^2*c^2)*x^4 - 32*(9*B*b^4 - 17*A*b^3*c)*x^2)*sqrt(x))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^
7*x^5)

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giac [A]  time = 0.26, size = 351, normalized size = 0.96 \begin {gather*} \frac {13 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 17 \, \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^{6} c} + \frac {13 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 17 \, \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^{6} c} - \frac {13 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 17 \, \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^{6} c} + \frac {13 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 17 \, \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^{6} c} + \frac {21 \, B b c^{3} x^{\frac {7}{2}} - 29 \, A c^{4} x^{\frac {7}{2}} + 25 \, B b^{2} c^{2} x^{\frac {3}{2}} - 33 \, A b c^{3} x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{5}} + \frac {2 \, {\left (135 \, B b c x^{4} - 270 \, A c^{2} x^{4} - 9 \, B b^{2} x^{2} + 27 \, A b c x^{2} - 5 \, A b^{2}\right )}}{45 \, b^{5} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

13/64*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 17*(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^6*c) + 13/64*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 17*(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^6*c) - 13/128*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 17*(b*c^3)^(3/4)*A*c)
*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^6*c) + 13/128*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 17*(b*c^3)^(
3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^6*c) + 1/16*(21*B*b*c^3*x^(7/2) - 29*A*c^4*x^(7
/2) + 25*B*b^2*c^2*x^(3/2) - 33*A*b*c^3*x^(3/2))/((c*x^2 + b)^2*b^5) + 2/45*(135*B*b*c*x^4 - 270*A*c^2*x^4 - 9
*B*b^2*x^2 + 27*A*b*c*x^2 - 5*A*b^2)/(b^5*x^(9/2))

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maple [A]  time = 0.07, size = 414, normalized size = 1.13 \begin {gather*} -\frac {29 A \,c^{4} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{5}}+\frac {21 B \,c^{3} x^{\frac {7}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {33 A \,c^{3} x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}+\frac {25 B \,c^{2} x^{\frac {3}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {221 \sqrt {2}\, A \,c^{2} \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^{5}}-\frac {221 \sqrt {2}\, A \,c^{2} \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^{5}}-\frac {221 \sqrt {2}\, A \,c^{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 )}{128 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{5}}+\frac {117 \sqrt {2}\, B 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}\, B 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}\, B 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 {12 A \,c^{2}}{b^{5} \sqrt {x}}+\frac {6 B c}{b^{4} \sqrt {x}}+\frac {6 A c}{5 b^{4} x^{\frac {5}{2}}}-\frac {2 B}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 A}{9 b^{3} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2)^3,x)

[Out]

-29/16/b^5*c^4/(c*x^2+b)^2*A*x^(7/2)+21/16/b^4*c^3/(c*x^2+b)^2*B*x^(7/2)-33/16/b^4*c^3/(c*x^2+b)^2*x^(3/2)*A+2
5/16/b^3*c^2/(c*x^2+b)^2*x^(3/2)*B-221/128/b^5*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)))-221/64/b^5*c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/
c)^(1/4)*x^(1/2)+1)-221/64/b^5*c^2/(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+117/128/b^4*c/(
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)
))+117/64/b^4*c/(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+117/64/b^4*c/(b/c)^(1/4)*2^(1/2)*B
*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/9*A/b^3/x^(9/2)+6/5/b^4/x^(5/2)*A*c-2/5/b^3/x^(5/2)*B-12*c^2/b^5/x^(1
/2)*A+6*c/b^4/x^(1/2)*B

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maxima [A]  time = 3.08, size = 311, normalized size = 0.85 \begin {gather*} \frac {585 \, {\left (9 \, B b c^{3} - 17 \, A c^{4}\right )} x^{8} + 1053 \, {\left (9 \, B b^{2} c^{2} - 17 \, A b c^{3}\right )} x^{6} - 160 \, A b^{4} + 416 \, {\left (9 \, B b^{3} c - 17 \, A b^{2} c^{2}\right )} x^{4} - 32 \, {\left (9 \, B b^{4} - 17 \, A b^{3} c\right )} x^{2}}{720 \, {\left (b^{5} c^{2} x^{\frac {17}{2}} + 2 \, b^{6} c x^{\frac {13}{2}} + b^{7} x^{\frac {9}{2}}\right )}} + \frac {13 \, {\left (9 \, B b c^{2} - 17 \, A c^{3}\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^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/720*(585*(9*B*b*c^3 - 17*A*c^4)*x^8 + 1053*(9*B*b^2*c^2 - 17*A*b*c^3)*x^6 - 160*A*b^4 + 416*(9*B*b^3*c - 17*
A*b^2*c^2)*x^4 - 32*(9*B*b^4 - 17*A*b^3*c)*x^2)/(b^5*c^2*x^(17/2) + 2*b^6*c*x^(13/2) + b^7*x^(9/2)) + 13/128*(
9*B*b*c^2 - 17*A*c^3)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)
*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c
)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x
) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b
))/(b^(1/4)*c^(3/4)))/b^5

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mupad [B]  time = 0.29, size = 173, normalized size = 0.47 \begin {gather*} \frac {13\,{\left (-c\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (17\,A\,c-9\,B\,b\right )}{32\,b^{21/4}}-\frac {\frac {2\,A}{9\,b}-\frac {2\,x^2\,\left (17\,A\,c-9\,B\,b\right )}{45\,b^2}+\frac {117\,c^2\,x^6\,\left (17\,A\,c-9\,B\,b\right )}{80\,b^4}+\frac {13\,c^3\,x^8\,\left (17\,A\,c-9\,B\,b\right )}{16\,b^5}+\frac {26\,c\,x^4\,\left (17\,A\,c-9\,B\,b\right )}{45\,b^3}}{b^2\,x^{9/2}+c^2\,x^{17/2}+2\,b\,c\,x^{13/2}}-\frac {13\,{\left (-c\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (17\,A\,c-9\,B\,b\right )}{32\,b^{21/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(13*(-c)^(5/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4))*(17*A*c - 9*B*b))/(32*b^(21/4)) - ((2*A)/(9*b) - (2*x^2*(17*
A*c - 9*B*b))/(45*b^2) + (117*c^2*x^6*(17*A*c - 9*B*b))/(80*b^4) + (13*c^3*x^8*(17*A*c - 9*B*b))/(16*b^5) + (2
6*c*x^4*(17*A*c - 9*B*b))/(45*b^3))/(b^2*x^(9/2) + c^2*x^(17/2) + 2*b*c*x^(13/2)) - (13*(-c)^(5/4)*atanh(((-c)
^(1/4)*x^(1/2))/b^(1/4))*(17*A*c - 9*B*b))/(32*b^(21/4))

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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((B*x**2+A)*x**(1/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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