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

Optimal. Leaf size=237 \[ -\frac{(b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{7/4} \sqrt [4]{c}}+\frac{(b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{7/4} \sqrt [4]{c}}-\frac{(b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{7/4} \sqrt [4]{c}}+\frac{(b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{7/4} \sqrt [4]{c}}-\frac{2 A}{3 b x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.188393, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {1584, 453, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{(b B-A c) \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{7/4} \sqrt [4]{c}}+\frac{(b B-A c) \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} b^{7/4} \sqrt [4]{c}}-\frac{(b B-A c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} b^{7/4} \sqrt [4]{c}}+\frac{(b B-A c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} b^{7/4} \sqrt [4]{c}}-\frac{2 A}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.135793, size = 168, normalized size = 0.71 \[ -\frac{(b B-A c) \left (\log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-\log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )\right )}{2 \sqrt{2} b^{7/4} \sqrt [4]{c}}-\frac{2 A}{3 b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 280, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{3\,b}{x}^{-{\frac{3}{2}}}}-{\frac{\sqrt{2}Ac}{2\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{\sqrt{2}Ac}{2\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }-{\frac{\sqrt{2}Ac}{4\,{b}^{2}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}B}{2\,b}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }+{\frac{\sqrt{2}B}{2\,b}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) }+{\frac{\sqrt{2}B}{4\,b}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.29509, size = 1347, normalized size = 5.68 \begin{align*} -\frac{12 \, b x^{2} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{b^{4} \sqrt{-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}} +{\left (B^{2} b^{2} - 2 \, A B b c + A^{2} c^{2}\right )} x} b^{5} c \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{3}{4}} +{\left (B b^{6} c - A b^{5} c^{2}\right )} \sqrt{x} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{3}{4}}}{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}\right ) + 3 \, b x^{2} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{1}{4}} \log \left (b^{2} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{1}{4}} -{\left (B b - A c\right )} \sqrt{x}\right ) - 3 \, b x^{2} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{1}{4}} \log \left (-b^{2} \left (-\frac{B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{7} c}\right )^{\frac{1}{4}} -{\left (B b - A c\right )} \sqrt{x}\right ) + 4 \, A \sqrt{x}}{6 \, b x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*b*x^2*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c))^(1/4)*arctan
((sqrt(b^4*sqrt(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c)) + (B^2*b^2 -
2*A*B*b*c + A^2*c^2)*x)*b^5*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*
c))^(3/4) + (B*b^6*c - A*b^5*c^2)*sqrt(x)*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4
*c^4)/(b^7*c))^(3/4))/(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)) + 3*b*x^2*(-(B^
4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c))^(1/4)*log(b^2*(-(B^4*b^4 - 4*A*B
^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c))^(1/4) - (B*b - A*c)*sqrt(x)) - 3*b*x^2*(-(B^4
*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c))^(1/4)*log(-b^2*(-(B^4*b^4 - 4*A*B
^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^7*c))^(1/4) - (B*b - A*c)*sqrt(x)) + 4*A*sqrt(x))/(
b*x^2)

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Sympy [A]  time = 160.822, size = 374, normalized size = 1.58 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}\right ) & \text{for}\: b = 0 \wedge c = 0 \\\frac{- \frac{2 A}{7 x^{\frac{7}{2}}} - \frac{2 B}{3 x^{\frac{3}{2}}}}{c} & \text{for}\: b = 0 \\\frac{- \frac{2 A}{3 x^{\frac{3}{2}}} + 2 B \sqrt{x}}{b} & \text{for}\: c = 0 \\- \frac{2 A}{3 b x^{\frac{3}{2}}} + \frac{\sqrt [4]{-1} A \log{\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{7}{4}} c \left (\frac{1}{c}\right )^{\frac{7}{4}}} - \frac{\sqrt [4]{-1} A \log{\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{7}{4}} c \left (\frac{1}{c}\right )^{\frac{7}{4}}} + \frac{\sqrt [4]{-1} A \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} \sqrt{x}}{\sqrt [4]{b} \sqrt [4]{\frac{1}{c}}} \right )}}{b^{\frac{7}{4}} c \left (\frac{1}{c}\right )^{\frac{7}{4}}} - \frac{\sqrt [4]{-1} B \log{\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{3}{4}} c^{2} \left (\frac{1}{c}\right )^{\frac{7}{4}}} + \frac{\sqrt [4]{-1} B \log{\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 b^{\frac{3}{4}} c^{2} \left (\frac{1}{c}\right )^{\frac{7}{4}}} - \frac{\sqrt [4]{-1} B \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} \sqrt{x}}{\sqrt [4]{b} \sqrt [4]{\frac{1}{c}}} \right )}}{b^{\frac{3}{4}} c^{2} \left (\frac{1}{c}\right )^{\frac{7}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(3*x**(3/2))), Eq(b, 0) & Eq(c, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(3*x**(
3/2)))/c, Eq(b, 0)), ((-2*A/(3*x**(3/2)) + 2*B*sqrt(x))/b, Eq(c, 0)), (-2*A/(3*b*x**(3/2)) + (-1)**(1/4)*A*log
(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(7/4)*c*(1/c)**(7/4)) - (-1)**(1/4)*A*log((-1)**(1/4)*b**
(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(7/4)*c*(1/c)**(7/4)) + (-1)**(1/4)*A*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(
1/c)**(1/4)))/(b**(7/4)*c*(1/c)**(7/4)) - (-1)**(1/4)*B*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b
**(3/4)*c**2*(1/c)**(7/4)) + (-1)**(1/4)*B*log((-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(3/4)*c**2*(
1/c)**(7/4)) - (-1)**(1/4)*B*atan((-1)**(3/4)*sqrt(x)/(b**(1/4)*(1/c)**(1/4)))/(b**(3/4)*c**2*(1/c)**(7/4)), T
rue))

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Giac [A]  time = 1.18459, size = 339, normalized size = 1.43 \begin{align*} \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{2 \, b^{2} c} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{2 \, b^{2} c} + \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{4 \, b^{2} c} - \frac{\sqrt{2}{\left (\left (b c^{3}\right )^{\frac{1}{4}} B b - \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 )}{4 \, b^{2} c} - \frac{2 \, A}{3 \, b x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*((b*c^3)^(1/4)*B*b - (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^2*c) + 1/2*sqrt(2)*((b*c^3)^(1/4)*B*b - (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^2*c) + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(
b/c)^(1/4) + x + sqrt(b/c))/(b^2*c) - 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(-sqrt(2)*sqrt(x)
*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c) - 2/3*A/(b*x^(3/2))