3.379 \(\int \frac{A+B x^2}{\sqrt{x} (a+b x^2)^2} \, dx\)

Optimal. Leaf size=261 \[ -\frac{(a B+3 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a B+3 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}-\frac{(a B+3 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\sqrt{x} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

[Out]

((A*b - a*B)*Sqrt[x])/(2*a*b*(a + b*x^2)) - ((3*A*b + a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*S
qrt[2]*a^(7/4)*b^(5/4)) + ((3*A*b + a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(
5/4)) - ((3*A*b + a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*b^(5/4))
+ ((3*A*b + a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*b^(5/4))

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Rubi [A]  time = 0.180924, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {457, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{(a B+3 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a B+3 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}-\frac{(a B+3 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}+\frac{\sqrt{x} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*Sqrt[x])/(2*a*b*(a + b*x^2)) - ((3*A*b + a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*S
qrt[2]*a^(7/4)*b^(5/4)) + ((3*A*b + a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(7/4)*b^(
5/4)) - ((3*A*b + a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*b^(5/4))
+ ((3*A*b + a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(7/4)*b^(5/4))

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 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 (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}+\frac{\left (\frac{3 A b}{2}+\frac{a B}{2}\right ) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{2 a b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}+\frac{\left (\frac{3 A b}{2}+\frac{a B}{2}\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{a b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 a^{3/2} b}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 a^{3/2} b}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{8 a^{3/2} b^{3/2}}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{8 a^{3/2} b^{3/2}}-\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}-\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}-\frac{(3 A b+a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(3 A b+a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}-\frac{(3 A b+a B) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}\\ &=\frac{(A b-a B) \sqrt{x}}{2 a b \left (a+b x^2\right )}-\frac{(3 A b+a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(3 A b+a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{7/4} b^{5/4}}-\frac{(3 A b+a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}+\frac{(3 A b+a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{7/4} b^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.244844, size = 203, normalized size = 0.78 \[ \frac{\frac{(a B+3 A b) \left (8 a^{3/4} \sqrt [4]{b} \sqrt{x}-3 \sqrt{2} \left (a+b x^2\right ) \left (\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )-\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )\right )\right )}{a^{7/4} \sqrt [4]{b}}-32 B \sqrt{x}}{48 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*B*Sqrt[x] + ((3*A*b + a*B)*(8*a^(3/4)*b^(1/4)*Sqrt[x] - 3*Sqrt[2]*(a + b*x^2)*(2*ArcTan[1 - (Sqrt[2]*b^(1
/4)*Sqrt[x])/a^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)] + Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4
)*Sqrt[x] + Sqrt[b]*x] - Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])))/(a^(7/4)*b^(1/4)))/(48*
b*(a + b*x^2))

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Maple [A]  time = 0.012, size = 305, normalized size = 1.2 \begin{align*}{\frac{Ab-Ba}{2\,ab \left ( b{x}^{2}+a \right ) }\sqrt{x}}+{\frac{3\,\sqrt{2}A}{8\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}A}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}A}{8\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{\sqrt{2}B}{8\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{\sqrt{2}B}{16\,ab}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}B}{8\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(A*b-B*a)*x^(1/2)/a/b/(b*x^2+a)+3/8/a^2*(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+3/
16/a^2*(1/b*a)^(1/4)*2^(1/2)*A*ln((x+(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(
1/2)+(1/b*a)^(1/2)))+3/8/a^2*(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+1/8/a/b*(1/b*a)^(
1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+1/16/a/b*(1/b*a)^(1/4)*2^(1/2)*B*ln((x+(1/b*a)^(1/4)*x^
(1/2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))+1/8/a/b*(1/b*a)^(1/4)*2^(1/2)*B*
arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)

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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)/(b*x^2+a)^2/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.0147, size = 1559, normalized size = 5.97 \begin{align*} \frac{4 \,{\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{a^{4} b^{2} \sqrt{-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}} +{\left (B^{2} a^{2} + 6 \, A B a b + 9 \, A^{2} b^{2}\right )} x} a^{5} b^{4} \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{3}{4}} -{\left (B a^{6} b^{4} + 3 \, A a^{5} b^{5}\right )} \sqrt{x} \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{3}{4}}}{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}\right ) +{\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{1}{4}} \log \left (a^{2} b \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{1}{4}} +{\left (B a + 3 \, A b\right )} \sqrt{x}\right ) -{\left (a b^{2} x^{2} + a^{2} b\right )} \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{1}{4}} \log \left (-a^{2} b \left (-\frac{B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac{1}{4}} +{\left (B a + 3 \, A b\right )} \sqrt{x}\right ) - 4 \,{\left (B a - A b\right )} \sqrt{x}}{8 \,{\left (a b^{2} x^{2} + a^{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(a*b^2*x^2 + a^2*b)*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a
^7*b^5))^(1/4)*arctan((sqrt(a^4*b^2*sqrt(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 8
1*A^4*b^4)/(a^7*b^5)) + (B^2*a^2 + 6*A*B*a*b + 9*A^2*b^2)*x)*a^5*b^4*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*
a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(3/4) - (B*a^6*b^4 + 3*A*a^5*b^5)*sqrt(x)*(-(B^4*a^4 + 12*A
*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(3/4))/(B^4*a^4 + 12*A*B^3*a^3*b +
54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)) + (a*b^2*x^2 + a^2*b)*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2
*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4)*log(a^2*b*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B
^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4) + (B*a + 3*A*b)*sqrt(x)) - (a*b^2*x^2 + a^2*b)*(-(
B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4)*log(-a^2*b*(-(B
^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4) + (B*a + 3*A*b)*
sqrt(x)) - 4*(B*a - A*b)*sqrt(x))/(a*b^2*x^2 + a^2*b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21231, size = 369, normalized size = 1.41 \begin{align*} \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{8 \, a^{2} b^{2}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{8 \, a^{2} b^{2}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{16 \, a^{2} b^{2}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} A b\right )} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{16 \, a^{2} b^{2}} - \frac{B a \sqrt{x} - A b \sqrt{x}}{2 \,{\left (b x^{2} + a\right )} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/
b)^(1/4))/(a^2*b^2) + 1/8*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)
^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^2) + 1/16*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*log(sqrt(2
)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^2) - 1/16*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*log(
-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^2) - 1/2*(B*a*sqrt(x) - A*b*sqrt(x))/((b*x^2 + a)*a*b)