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

Optimal. Leaf size=310 \[ \frac {c^{3/4} (7 b B-11 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{15/4}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}+\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )} \]

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Rubi [A]  time = 0.26, antiderivative size = 310, 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, 457, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {c^{3/4} (7 b B-11 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{15/4}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}+\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*b*B - 11*A*c)/(14*b^2*c*x^(7/2)) - (7*b*B - 11*A*c)/(6*b^3*x^(3/2)) - (b*B - A*c)/(2*b*c*x^(7/2)*(b + c*x^2
)) + (c^(3/4)*(7*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(15/4)) - (c^(3/4)*
(7*b*B - 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(15/4)) + (c^(3/4)*(7*b*B - 11*A*
c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(15/4)) - (c^(3/4)*(7*b*B - 11*A*c
)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*b^(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 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 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 {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^2} \, dx &=\int \frac {A+B x^2}{x^{9/2} \left (b+c x^2\right )^2} \, dx\\ &=-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {\left (-\frac {7 b B}{2}+\frac {11 A c}{2}\right ) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{2 b c}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {(7 b B-11 A c) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 b^3}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b^3}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {(c (7 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{7/2}}-\frac {(c (7 b B-11 A c)) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^{7/2}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}-\frac {\left (\sqrt {c} (7 b B-11 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 )}{8 b^{7/2}}-\frac {\left (\sqrt {c} (7 b B-11 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 )}{8 b^{7/2}}+\frac {\left (c^{3/4} (7 b B-11 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 )}{8 \sqrt {2} b^{15/4}}+\frac {\left (c^{3/4} (7 b B-11 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 )}{8 \sqrt {2} b^{15/4}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {\left (c^{3/4} (7 b B-11 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 )}{4 \sqrt {2} b^{15/4}}+\frac {\left (c^{3/4} (7 b B-11 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 )}{4 \sqrt {2} b^{15/4}}\\ &=\frac {7 b B-11 A c}{14 b^2 c x^{7/2}}-\frac {7 b B-11 A c}{6 b^3 x^{3/2}}-\frac {b B-A c}{2 b c x^{7/2} \left (b+c x^2\right )}+\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{15/4}}+\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}-\frac {c^{3/4} (7 b B-11 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{15/4}}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 385, normalized size = 1.24 \begin {gather*} \frac {\frac {168 A b^{3/4} c^2 \sqrt {x}}{b+c x^2}+\frac {448 A b^{3/4} c}{x^{3/2}}-\frac {96 A b^{7/4}}{x^{7/2}}+42 \sqrt {2} c^{3/4} (7 b B-11 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )+42 \sqrt {2} c^{3/4} (11 A c-7 b B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )-231 \sqrt {2} A c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+231 \sqrt {2} A c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-\frac {168 b^{7/4} B c \sqrt {x}}{b+c x^2}-\frac {224 b^{7/4} B}{x^{3/2}}+147 \sqrt {2} b B c^{3/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-147 \sqrt {2} b B c^{3/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{336 b^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-96*A*b^(7/4))/x^(7/2) - (224*b^(7/4)*B)/x^(3/2) + (448*A*b^(3/4)*c)/x^(3/2) - (168*b^(7/4)*B*c*Sqrt[x])/(b
+ c*x^2) + (168*A*b^(3/4)*c^2*Sqrt[x])/(b + c*x^2) + 42*Sqrt[2]*c^(3/4)*(7*b*B - 11*A*c)*ArcTan[1 - (Sqrt[2]*c
^(1/4)*Sqrt[x])/b^(1/4)] + 42*Sqrt[2]*c^(3/4)*(-7*b*B + 11*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]
+ 147*Sqrt[2]*b*B*c^(3/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 231*Sqrt[2]*A*c^(7/4)*L
og[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 147*Sqrt[2]*b*B*c^(3/4)*Log[Sqrt[b] + Sqrt[2]*b^(1
/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 231*Sqrt[2]*A*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[
c]*x])/(336*b^(15/4))

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IntegrateAlgebraic [A]  time = 0.70, size = 200, normalized size = 0.65 \begin {gather*} \frac {\left (7 b B c^{3/4}-11 A c^{7/4}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{4 \sqrt {2} b^{15/4}}-\frac {\left (7 b B c^{3/4}-11 A c^{7/4}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{15/4}}+\frac {-12 A b^2+44 A b c x^2+77 A c^2 x^4-28 b^2 B x^2-49 b B c x^4}{42 b^3 x^{7/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*A*b^2 - 28*b^2*B*x^2 + 44*A*b*c*x^2 - 49*b*B*c*x^4 + 77*A*c^2*x^4)/(42*b^3*x^(7/2)*(b + c*x^2)) + ((7*b*B
*c^(3/4) - 11*A*c^(7/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(4*Sqrt[2]*b^(15/4))
- ((7*b*B*c^(3/4) - 11*A*c^(7/4))*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4*Sqrt[2
]*b^(15/4))

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fricas [B]  time = 0.45, size = 795, normalized size = 2.56 \begin {gather*} \frac {84 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{8} \sqrt {-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}} + {\left (49 \, B^{2} b^{2} c^{2} - 154 \, A B b c^{3} + 121 \, A^{2} c^{4}\right )} x} b^{11} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {3}{4}} + {\left (7 \, B b^{12} c - 11 \, A b^{11} c^{2}\right )} \sqrt {x} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {3}{4}}}{2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}\right ) + 21 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (b^{4} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} - {\left (7 \, B b c - 11 \, A c^{2}\right )} \sqrt {x}\right ) - 21 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-b^{4} \left (-\frac {2401 \, B^{4} b^{4} c^{3} - 15092 \, A B^{3} b^{3} c^{4} + 35574 \, A^{2} B^{2} b^{2} c^{5} - 37268 \, A^{3} B b c^{6} + 14641 \, A^{4} c^{7}}{b^{15}}\right )^{\frac {1}{4}} - {\left (7 \, B b c - 11 \, A c^{2}\right )} \sqrt {x}\right ) - 4 \, {\left (7 \, {\left (7 \, B b c - 11 \, A c^{2}\right )} x^{4} + 12 \, A b^{2} + 4 \, {\left (7 \, B b^{2} - 11 \, A b c\right )} x^{2}\right )} \sqrt {x}}{168 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(84*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*
B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*arctan((sqrt(b^8*sqrt(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^
2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15) + (49*B^2*b^2*c^2 - 154*A*B*b*c^3 + 121*A^2*c^4)*x)*b
^11*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^1
5)^(3/4) + (7*B*b^12*c - 11*A*b^11*c^2)*sqrt(x)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*
c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(3/4))/(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*
b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)) + 21*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3
*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*log(b^4*(-(2401*B^4*b^4*c^3 - 15
092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4) - (7*B*b*c - 11*A*c
^2)*sqrt(x)) - 21*(b^3*c*x^6 + b^4*x^4)*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^2*b^2*c^5 - 37
268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4)*log(-b^4*(-(2401*B^4*b^4*c^3 - 15092*A*B^3*b^3*c^4 + 35574*A^2*B^
2*b^2*c^5 - 37268*A^3*B*b*c^6 + 14641*A^4*c^7)/b^15)^(1/4) - (7*B*b*c - 11*A*c^2)*sqrt(x)) - 4*(7*(7*B*b*c - 1
1*A*c^2)*x^4 + 12*A*b^2 + 4*(7*B*b^2 - 11*A*b*c)*x^2)*sqrt(x))/(b^3*c*x^6 + b^4*x^4)

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giac [A]  time = 0.25, size = 292, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{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 )}{8 \, b^{4}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{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 )}{8 \, b^{4}} - \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{4}} + \frac {\sqrt {2} {\left (7 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 11 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{4}} - \frac {B b c \sqrt {x} - A c^{2} \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b^{3}} - \frac {2 \, {\left (7 \, B b x^{2} - 14 \, A c x^{2} + 3 \, A b\right )}}{21 \, b^{3} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))
/(b/c)^(1/4))/b^4 - 1/8*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c
)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^4 - 1/16*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*
sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 + 1/16*sqrt(2)*(7*(b*c^3)^(1/4)*B*b - 11*(b*c^3)^(1/4)*A*c)*log(-sqrt
(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 1/2*(B*b*c*sqrt(x) - A*c^2*sqrt(x))/((c*x^2 + b)*b^3) - 2/21*(7
*B*b*x^2 - 14*A*c*x^2 + 3*A*b)/(b^3*x^(7/2))

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maple [A]  time = 0.06, size = 348, normalized size = 1.12 \begin {gather*} \frac {A \,c^{2} \sqrt {x}}{2 \left (c \,x^{2}+b \right ) b^{3}}-\frac {B c \sqrt {x}}{2 \left (c \,x^{2}+b \right ) b^{2}}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{4}}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{4}}+\frac {11 \left (\frac {b}{c}\right )^{\frac {1}{4}} \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 )}{16 b^{4}}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{3}}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{3}}-\frac {7 \left (\frac {b}{c}\right )^{\frac {1}{4}} \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 )}{16 b^{3}}+\frac {4 A c}{3 b^{3} x^{\frac {3}{2}}}-\frac {2 B}{3 b^{2} x^{\frac {3}{2}}}-\frac {2 A}{7 b^{2} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b^3*c^2*x^(1/2)/(c*x^2+b)*A-1/2/b^2*c*x^(1/2)/(c*x^2+b)*B+11/8/b^4*c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2
)/(b/c)^(1/4)*x^(1/2)-1)+11/8/b^4*c^2*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+11/16/b^4*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)))-7/8/b^3*c*(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-7/8/b^3*c*(b/c)^(1/4)*2^(1/2)*B*ar
ctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-7/16/b^3*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)))-2/7*A/b^2/x^(7/2)+4/3/b^3/x^(3/2)*A*c-2/3/b^2/x^(3/2)*B

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maxima [A]  time = 3.12, size = 286, normalized size = 0.92 \begin {gather*} -\frac {7 \, {\left (7 \, B b c - 11 \, A c^{2}\right )} x^{4} + 12 \, A b^{2} + 4 \, {\left (7 \, B b^{2} - 11 \, A b c\right )} x^{2}}{42 \, {\left (b^{3} c x^{\frac {11}{2}} + b^{4} x^{\frac {7}{2}}\right )}} - \frac {\frac {2 \, \sqrt {2} {\left (7 \, B b c - 11 \, A c^{2}\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 (7 \, B b c - 11 \, A c^{2}\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 (7 \, B b c - 11 \, A c^{2}\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 (7 \, B b c - 11 \, A c^{2}\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}}}}{16 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/42*(7*(7*B*b*c - 11*A*c^2)*x^4 + 12*A*b^2 + 4*(7*B*b^2 - 11*A*b*c)*x^2)/(b^3*c*x^(11/2) + b^4*x^(7/2)) - 1/
16*(2*sqrt(2)*(7*B*b*c - 11*A*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(7*B*b*c - 11*A*c^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(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(7*B*b*c -
11*A*c^2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(7*B*b*c - 11
*A*c^2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^3

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mupad [B]  time = 0.40, size = 595, normalized size = 1.92 \begin {gather*} \frac {\frac {2\,x^2\,\left (11\,A\,c-7\,B\,b\right )}{21\,b^2}-\frac {2\,A}{7\,b}+\frac {c\,x^4\,\left (11\,A\,c-7\,B\,b\right )}{6\,b^3}}{b\,x^{7/2}+c\,x^{11/2}}+\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )}{8\,b^{15/4}}+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )}{8\,b^{15/4}}}{\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )\,1{}\mathrm {i}}{8\,b^{15/4}}-\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (\sqrt {x}\,\left (3872\,A^2\,b^9\,c^7-4928\,A\,B\,b^{10}\,c^6+1568\,B^2\,b^{11}\,c^5\right )+\frac {{\left (-c\right )}^{3/4}\,\left (11\,A\,c-7\,B\,b\right )\,\left (2816\,A\,b^{13}\,c^5-1792\,B\,b^{14}\,c^4\right )\,1{}\mathrm {i}}{8\,b^{15/4}}\right )\,1{}\mathrm {i}}{8\,b^{15/4}}}\right )\,\left (11\,A\,c-7\,B\,b\right )}{4\,b^{15/4}}-\frac {{\left (-c\right )}^{3/4}\,\mathrm {atan}\left (\frac {A^3\,c^8\,\sqrt {x}\,1331{}\mathrm {i}-B^3\,b^3\,c^5\,\sqrt {x}\,343{}\mathrm {i}-A^2\,B\,b\,c^7\,\sqrt {x}\,2541{}\mathrm {i}+A\,B^2\,b^2\,c^6\,\sqrt {x}\,1617{}\mathrm {i}}{b^{1/4}\,{\left (-c\right )}^{19/4}\,\left (c\,\left (c\,\left (1331\,A^3\,c-2541\,A^2\,B\,b\right )+1617\,A\,B^2\,b^2\right )-343\,B^3\,b^3\right )}\right )\,\left (11\,A\,c-7\,B\,b\right )\,1{}\mathrm {i}}{4\,b^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x^2*(11*A*c - 7*B*b))/(21*b^2) - (2*A)/(7*b) + (c*x^4*(11*A*c - 7*B*b))/(6*b^3))/(b*x^(7/2) + c*x^(11/2))
+ ((-c)^(3/4)*atan((((-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^1
0*c^6) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4))))/(8*b^(15/4)) + ((
-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) + ((-c)^(3/4)*(
11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4))))/(8*b^(15/4)))/(((-c)^(3/4)*(11*A*c - 7*
B*b)*(x^(1/2)*(3872*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*
A*b^13*c^5 - 1792*B*b^14*c^4)*1i)/(8*b^(15/4)))*1i)/(8*b^(15/4)) - ((-c)^(3/4)*(11*A*c - 7*B*b)*(x^(1/2)*(3872
*A^2*b^9*c^7 + 1568*B^2*b^11*c^5 - 4928*A*B*b^10*c^6) + ((-c)^(3/4)*(11*A*c - 7*B*b)*(2816*A*b^13*c^5 - 1792*B
*b^14*c^4)*1i)/(8*b^(15/4)))*1i)/(8*b^(15/4))))*(11*A*c - 7*B*b))/(4*b^(15/4)) - ((-c)^(3/4)*atan((A^3*c^8*x^(
1/2)*1331i - B^3*b^3*c^5*x^(1/2)*343i - A^2*B*b*c^7*x^(1/2)*2541i + A*B^2*b^2*c^6*x^(1/2)*1617i)/(b^(1/4)*(-c)
^(19/4)*(c*(c*(1331*A^3*c - 2541*A^2*B*b) + 1617*A*B^2*b^2) - 343*B^3*b^3)))*(11*A*c - 7*B*b)*1i)/(4*b^(15/4))

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

[Out]

Timed out

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