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

Optimal. Leaf size=365 \[ -\frac {15 c^{7/4} (11 b B-19 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{23/4}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2} \]

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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, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {15 c^{7/4} (11 b B-19 A c) \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{23/4}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(11*b*B - 19*A*c))/(176*b^3*c*x^(11/2)) - (15*(11*b*B - 19*A*c))/(112*b^4*x^(7/2)) + (5*c*(11*b*B - 19*A*c
))/(16*b^5*x^(3/2)) - (b*B - A*c)/(4*b*c*x^(11/2)*(b + c*x^2)^2) - (11*b*B - 19*A*c)/(16*b^2*c*x^(11/2)*(b + c
*x^2)) - (15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) +
(15*c^(7/4)*(11*b*B - 19*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) - (15*c^(7/
4)*(11*b*B - 19*A*c)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4)) + (15*c
^(7/4)*(11*b*B - 19*A*c)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4))

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 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 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 )^3} \, dx &=\int \frac {A+B x^2}{x^{13/2} \left (b+c x^2\right )^3} \, dx\\ &=-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}+\frac {\left (-\frac {11 b B}{2}+\frac {19 A c}{2}\right ) \int \frac {1}{x^{13/2} \left (b+c x^2\right )^2} \, dx}{4 b c}\\ &=-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac {(15 (11 b B-19 A c)) \int \frac {1}{x^{13/2} \left (b+c x^2\right )} \, dx}{32 b^2 c}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {(15 (11 b B-19 A c)) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac {(15 c (11 b B-19 A c)) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {\left (15 c^2 (11 b B-19 A c)\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^5}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {\left (15 c^2 (11 b B-19 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^5}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {\left (15 c^2 (11 b B-19 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^{11/2}}+\frac {\left (15 c^2 (11 b B-19 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^{11/2}}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}+\frac {\left (15 c^{3/2} (11 b B-19 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 )}{64 b^{11/2}}+\frac {\left (15 c^{3/2} (11 b B-19 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 )}{64 b^{11/2}}-\frac {\left (15 c^{7/4} (11 b B-19 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^{23/4}}-\frac {\left (15 c^{7/4} (11 b B-19 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^{23/4}}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {\left (15 c^{7/4} (11 b B-19 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^{23/4}}-\frac {\left (15 c^{7/4} (11 b B-19 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^{23/4}}\\ &=\frac {15 (11 b B-19 A c)}{176 b^3 c x^{11/2}}-\frac {15 (11 b B-19 A c)}{112 b^4 x^{7/2}}+\frac {5 c (11 b B-19 A c)}{16 b^5 x^{3/2}}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}-\frac {11 b B-19 A c}{16 b^2 c x^{11/2} \left (b+c x^2\right )}-\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 467, normalized size = 1.28 \begin {gather*} \frac {-\frac {19096 A b^{3/4} c^3 \sqrt {x}}{b+c x^2}-\frac {2464 A b^{7/4} c^3 \sqrt {x}}{\left (b+c x^2\right )^2}-\frac {39424 A b^{3/4} c^2}{x^{3/2}}+\frac {8448 A b^{7/4} c}{x^{7/2}}-\frac {1792 A b^{11/4}}{x^{11/2}}+2310 \sqrt {2} c^{7/4} (19 A c-11 b B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )+2310 \sqrt {2} c^{7/4} (11 b B-19 A c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )+21945 \sqrt {2} A c^{11/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-21945 \sqrt {2} A c^{11/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+\frac {14168 b^{7/4} B c^2 \sqrt {x}}{b+c x^2}+\frac {2464 b^{11/4} B c^2 \sqrt {x}}{\left (b+c x^2\right )^2}+\frac {19712 b^{7/4} B c}{x^{3/2}}-\frac {2816 b^{11/4} B}{x^{7/2}}-12705 \sqrt {2} b B c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+12705 \sqrt {2} b B c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{9856 b^{23/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1792*A*b^(11/4))/x^(11/2) - (2816*b^(11/4)*B)/x^(7/2) + (8448*A*b^(7/4)*c)/x^(7/2) + (19712*b^(7/4)*B*c)/x^
(3/2) - (39424*A*b^(3/4)*c^2)/x^(3/2) + (2464*b^(11/4)*B*c^2*Sqrt[x])/(b + c*x^2)^2 - (2464*A*b^(7/4)*c^3*Sqrt
[x])/(b + c*x^2)^2 + (14168*b^(7/4)*B*c^2*Sqrt[x])/(b + c*x^2) - (19096*A*b^(3/4)*c^3*Sqrt[x])/(b + c*x^2) + 2
310*Sqrt[2]*c^(7/4)*(-11*b*B + 19*A*c)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 2310*Sqrt[2]*c^(7/4)*(1
1*b*B - 19*A*c)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 12705*Sqrt[2]*b*B*c^(7/4)*Log[Sqrt[b] - Sqrt[2
]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 21945*Sqrt[2]*A*c^(11/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x
] + Sqrt[c]*x] + 12705*Sqrt[2]*b*B*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 21945*
Sqrt[2]*A*c^(11/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(9856*b^(23/4))

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IntegrateAlgebraic [A]  time = 0.69, size = 248, normalized size = 0.68 \begin {gather*} -\frac {15 \left (11 b B c^{7/4}-19 A c^{11/4}\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^{23/4}}+\frac {15 \left (11 b B c^{7/4}-19 A c^{11/4}\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^{23/4}}+\frac {-224 A b^4+608 A b^3 c x^2-3040 A b^2 c^2 x^4-11495 A b c^3 x^6-7315 A c^4 x^8-352 b^4 B x^2+1760 b^3 B c x^4+6655 b^2 B c^2 x^6+4235 b B c^3 x^8}{1232 b^5 x^{11/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-224*A*b^4 - 352*b^4*B*x^2 + 608*A*b^3*c*x^2 + 1760*b^3*B*c*x^4 - 3040*A*b^2*c^2*x^4 + 6655*b^2*B*c^2*x^6 - 1
1495*A*b*c^3*x^6 + 4235*b*B*c^3*x^8 - 7315*A*c^4*x^8)/(1232*b^5*x^(11/2)*(b + c*x^2)^2) - (15*(11*b*B*c^(7/4)
- 19*A*c^(11/4))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(32*Sqrt[2]*b^(23/4)) + (15*
(11*b*B*c^(7/4) - 19*A*c^(11/4))*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]
*b^(23/4))

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fricas [B]  time = 0.44, size = 894, normalized size = 2.45 \begin {gather*} -\frac {4620 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{12} \sqrt {-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}} + {\left (121 \, B^{2} b^{2} c^{4} - 418 \, A B b c^{5} + 361 \, A^{2} c^{6}\right )} x} b^{17} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {3}{4}} + {\left (11 \, B b^{18} c^{2} - 19 \, A b^{17} c^{3}\right )} \sqrt {x} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {3}{4}}}{14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}\right ) + 1155 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (15 \, b^{6} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {1}{4}} - 15 \, {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \sqrt {x}\right ) - 1155 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (-15 \, b^{6} \left (-\frac {14641 \, B^{4} b^{4} c^{7} - 101156 \, A B^{3} b^{3} c^{8} + 262086 \, A^{2} B^{2} b^{2} c^{9} - 301796 \, A^{3} B b c^{10} + 130321 \, A^{4} c^{11}}{b^{23}}\right )^{\frac {1}{4}} - 15 \, {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (385 \, {\left (11 \, B b c^{3} - 19 \, A c^{4}\right )} x^{8} + 605 \, {\left (11 \, B b^{2} c^{2} - 19 \, A b c^{3}\right )} x^{6} - 224 \, A b^{4} + 160 \, {\left (11 \, B b^{3} c - 19 \, A b^{2} c^{2}\right )} x^{4} - 32 \, {\left (11 \, B b^{4} - 19 \, A b^{3} c\right )} x^{2}\right )} \sqrt {x}}{4928 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\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)^3/x^(1/2),x, algorithm="fricas")

[Out]

-1/4928*(4620*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*
B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*arctan((sqrt(b^12*sqrt(-(14641*B^4*b^4*c^7 -
101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23) + (121*B^2*b^2*c^
4 - 418*A*B*b*c^5 + 361*A^2*c^6)*x)*b^17*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9
- 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(3/4) + (11*B*b^18*c^2 - 19*A*b^17*c^3)*sqrt(x)*(-(14641*B^4*b^
4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(3/4))/(1
4641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)) + 1
155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^
9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*log(15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8
+ 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B*b*c^2 - 19*A*c^3)*sqr
t(x)) - 1155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B
^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*log(-15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^
3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B*b*c^2 - 19*
A*c^3)*sqrt(x)) - 4*(385*(11*B*b*c^3 - 19*A*c^4)*x^8 + 605*(11*B*b^2*c^2 - 19*A*b*c^3)*x^6 - 224*A*b^4 + 160*(
11*B*b^3*c - 19*A*b^2*c^2)*x^4 - 32*(11*B*b^4 - 19*A*b^3*c)*x^2)*sqrt(x))/(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^
6)

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giac [A]  time = 0.28, size = 351, normalized size = 0.96 \begin {gather*} \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\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}} + \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\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}} + \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} - \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} + \frac {23 \, B b c^{3} x^{\frac {5}{2}} - 31 \, A c^{4} x^{\frac {5}{2}} + 27 \, B b^{2} c^{2} \sqrt {x} - 35 \, A b c^{3} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{5}} + \frac {2 \, {\left (77 \, B b c x^{4} - 154 \, A c^{2} x^{4} - 11 \, B b^{2} x^{2} + 33 \, A b c x^{2} - 7 \, A b^{2}\right )}}{77 \, b^{5} x^{\frac {11}{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)^3/x^(1/2),x, algorithm="giac")

[Out]

15/64*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sq
rt(x))/(b/c)^(1/4))/b^6 + 15/64*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(-1/2*sqrt(2)*
(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^6 + 15/128*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)
*A*c^2)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 - 15/128*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*
c^3)^(1/4)*A*c^2)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 + 1/16*(23*B*b*c^3*x^(5/2) - 31*A*c^4*
x^(5/2) + 27*B*b^2*c^2*sqrt(x) - 35*A*b*c^3*sqrt(x))/((c*x^2 + b)^2*b^5) + 2/77*(77*B*b*c*x^4 - 154*A*c^2*x^4
- 11*B*b^2*x^2 + 33*A*b*c*x^2 - 7*A*b^2)/(b^5*x^(11/2))

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maple [A]  time = 0.07, size = 420, normalized size = 1.15 \begin {gather*} -\frac {31 A \,c^{4} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{5}}+\frac {23 B \,c^{3} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {35 A \,c^{3} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}+\frac {27 B \,c^{2} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{6}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{6}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, A \,c^{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 )}{128 b^{6}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, B \,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 b^{5}}-\frac {4 A \,c^{2}}{b^{5} x^{\frac {3}{2}}}+\frac {2 B c}{b^{4} x^{\frac {3}{2}}}+\frac {6 A c}{7 b^{4} x^{\frac {7}{2}}}-\frac {2 B}{7 b^{3} x^{\frac {7}{2}}}-\frac {2 A}{11 b^{3} x^{\frac {11}{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)^3/x^(1/2),x)

[Out]

-31/16/b^5*c^4/(c*x^2+b)^2*x^(5/2)*A+23/16/b^4*c^3/(c*x^2+b)^2*x^(5/2)*B-35/16/b^4*c^3/(c*x^2+b)^2*A*x^(1/2)+2
7/16/b^3*c^2/(c*x^2+b)^2*B*x^(1/2)-285/64/b^6*c^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-
285/128/b^6*c^3*(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)))-285/64/b^6*c^3*(b/c)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+165/64/b^5*c^2*
(b/c)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)+165/128/b^5*c^2*(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)))+165/64/b^5*c^2*(b/c)^(1/4)*2^(
1/2)*B*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-2/11*A/b^3/x^(11/2)+6/7/b^4/x^(7/2)*A*c-2/7/b^3/x^(7/2)*B-4*c^2/b
^5/x^(3/2)*A+2*c/b^4/x^(3/2)*B

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maxima [A]  time = 3.10, size = 353, normalized size = 0.97 \begin {gather*} \frac {385 \, {\left (11 \, B b c^{3} - 19 \, A c^{4}\right )} x^{8} + 605 \, {\left (11 \, B b^{2} c^{2} - 19 \, A b c^{3}\right )} x^{6} - 224 \, A b^{4} + 160 \, {\left (11 \, B b^{3} c - 19 \, A b^{2} c^{2}\right )} x^{4} - 32 \, {\left (11 \, B b^{4} - 19 \, A b^{3} c\right )} x^{2}}{1232 \, {\left (b^{5} c^{2} x^{\frac {19}{2}} + 2 \, b^{6} c x^{\frac {15}{2}} + b^{7} x^{\frac {11}{2}}\right )}} + \frac {15 \, {\left (\frac {2 \, \sqrt {2} {\left (11 \, B b c^{2} - 19 \, A c^{3}\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 (11 \, B b c^{2} - 19 \, A c^{3}\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 (11 \, B b c^{2} - 19 \, A c^{3}\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 (11 \, B b c^{2} - 19 \, A c^{3}\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1232*(385*(11*B*b*c^3 - 19*A*c^4)*x^8 + 605*(11*B*b^2*c^2 - 19*A*b*c^3)*x^6 - 224*A*b^4 + 160*(11*B*b^3*c -
19*A*b^2*c^2)*x^4 - 32*(11*B*b^4 - 19*A*b^3*c)*x^2)/(b^5*c^2*x^(19/2) + 2*b^6*c*x^(15/2) + b^7*x^(11/2)) + 15/
128*(2*sqrt(2)*(11*B*b*c^2 - 19*A*c^3)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(s
qrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*(11*B*b*c^2 - 19*A*c^3)*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)*(11*
B*b*c^2 - 19*A*c^3)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(11
*B*b*c^2 - 19*A*c^3)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^5

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mupad [B]  time = 0.51, size = 639, normalized size = 1.75 \begin {gather*} \frac {15\,{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {6859\,A^3\,c^{10}\,\sqrt {x}-1331\,B^3\,b^3\,c^7\,\sqrt {x}-11913\,A^2\,B\,b\,c^9\,\sqrt {x}+6897\,A\,B^2\,b^2\,c^8\,\sqrt {x}}{b^{1/4}\,{\left (-c\right )}^{27/4}\,\left (c\,\left (c\,\left (6859\,A^3\,c-11913\,A^2\,B\,b\right )+6897\,A\,B^2\,b^2\right )-1331\,B^3\,b^3\right )}\right )\,\left (19\,A\,c-11\,B\,b\right )}{32\,b^{23/4}}-\frac {\frac {2\,A}{11\,b}-\frac {2\,x^2\,\left (19\,A\,c-11\,B\,b\right )}{77\,b^2}+\frac {55\,c^2\,x^6\,\left (19\,A\,c-11\,B\,b\right )}{112\,b^4}+\frac {5\,c^3\,x^8\,\left (19\,A\,c-11\,B\,b\right )}{16\,b^5}+\frac {10\,c\,x^4\,\left (19\,A\,c-11\,B\,b\right )}{77\,b^3}}{b^2\,x^{11/2}+c^2\,x^{19/2}+2\,b\,c\,x^{15/2}}-\frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (\sqrt {x}\,\left (1330790400\,A^2\,b^{15}\,c^9-1540915200\,A\,B\,b^{16}\,c^8+446054400\,B^2\,b^{17}\,c^7\right )-\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (298844160\,A\,b^{21}\,c^6-173015040\,B\,b^{22}\,c^5\right )}{64\,b^{23/4}}\right )\,15{}\mathrm {i}}{64\,b^{23/4}}+\frac {{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (\sqrt {x}\,\left (1330790400\,A^2\,b^{15}\,c^9-1540915200\,A\,B\,b^{16}\,c^8+446054400\,B^2\,b^{17}\,c^7\right )+\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (298844160\,A\,b^{21}\,c^6-173015040\,B\,b^{22}\,c^5\right )}{64\,b^{23/4}}\right )\,15{}\mathrm {i}}{64\,b^{23/4}}}{\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (\sqrt {x}\,\left (1330790400\,A^2\,b^{15}\,c^9-1540915200\,A\,B\,b^{16}\,c^8+446054400\,B^2\,b^{17}\,c^7\right )-\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (298844160\,A\,b^{21}\,c^6-173015040\,B\,b^{22}\,c^5\right )}{64\,b^{23/4}}\right )}{64\,b^{23/4}}-\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (\sqrt {x}\,\left (1330790400\,A^2\,b^{15}\,c^9-1540915200\,A\,B\,b^{16}\,c^8+446054400\,B^2\,b^{17}\,c^7\right )+\frac {15\,{\left (-c\right )}^{7/4}\,\left (19\,A\,c-11\,B\,b\right )\,\left (298844160\,A\,b^{21}\,c^6-173015040\,B\,b^{22}\,c^5\right )}{64\,b^{23/4}}\right )}{64\,b^{23/4}}}\right )\,\left (19\,A\,c-11\,B\,b\right )\,15{}\mathrm {i}}{32\,b^{23/4}} \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)^3),x)

[Out]

(15*(-c)^(7/4)*atan((6859*A^3*c^10*x^(1/2) - 1331*B^3*b^3*c^7*x^(1/2) - 11913*A^2*B*b*c^9*x^(1/2) + 6897*A*B^2
*b^2*c^8*x^(1/2))/(b^(1/4)*(-c)^(27/4)*(c*(c*(6859*A^3*c - 11913*A^2*B*b) + 6897*A*B^2*b^2) - 1331*B^3*b^3)))*
(19*A*c - 11*B*b))/(32*b^(23/4)) - ((2*A)/(11*b) - (2*x^2*(19*A*c - 11*B*b))/(77*b^2) + (55*c^2*x^6*(19*A*c -
11*B*b))/(112*b^4) + (5*c^3*x^8*(19*A*c - 11*B*b))/(16*b^5) + (10*c*x^4*(19*A*c - 11*B*b))/(77*b^3))/(b^2*x^(1
1/2) + c^2*x^(19/2) + 2*b*c*x^(15/2)) - ((-c)^(7/4)*atan((((-c)^(7/4)*(19*A*c - 11*B*b)*(x^(1/2)*(1330790400*A
^2*b^15*c^9 + 446054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) - (15*(-c)^(7/4)*(19*A*c - 11*B*b)*(298844160*
A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4)))*15i)/(64*b^(23/4)) + ((-c)^(7/4)*(19*A*c - 11*B*b)*(x^(1/2)
*(1330790400*A^2*b^15*c^9 + 446054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) + (15*(-c)^(7/4)*(19*A*c - 11*B*
b)*(298844160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4)))*15i)/(64*b^(23/4)))/((15*(-c)^(7/4)*(19*A*c -
11*B*b)*(x^(1/2)*(1330790400*A^2*b^15*c^9 + 446054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) - (15*(-c)^(7/4
)*(19*A*c - 11*B*b)*(298844160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4))))/(64*b^(23/4)) - (15*(-c)^(7
/4)*(19*A*c - 11*B*b)*(x^(1/2)*(1330790400*A^2*b^15*c^9 + 446054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) +
(15*(-c)^(7/4)*(19*A*c - 11*B*b)*(298844160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4))))/(64*b^(23/4)))
)*(19*A*c - 11*B*b)*15i)/(32*b^(23/4))

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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)**3/x**(1/2),x)

[Out]

Timed out

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