3.380 \(\int \frac{A+B x^2}{x^{3/2} (a+b x^2)^2} \, dx\)

Optimal. Leaf size=289 \[ -\frac{(5 A b-a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{3/4}}+\frac{(5 A b-a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{3/4}}+\frac{(5 A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{(5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )} \]

[Out]

-(5*A*b - a*B)/(2*a^2*b*Sqrt[x]) + (A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x^2)) + ((5*A*b - a*B)*ArcTan[1 - (Sqrt[2
]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(9/4)*b^(3/4)) - ((5*A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])
/a^(1/4)])/(4*Sqrt[2]*a^(9/4)*b^(3/4)) - ((5*A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b
]*x])/(8*Sqrt[2]*a^(9/4)*b^(3/4)) + ((5*A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])
/(8*Sqrt[2]*a^(9/4)*b^(3/4))

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Rubi [A]  time = 0.211744, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {457, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{(5 A b-a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{3/4}}+\frac{(5 A b-a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{9/4} b^{3/4}}+\frac{(5 A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{(5 A b-a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*A*b - a*B)/(2*a^2*b*Sqrt[x]) + (A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x^2)) + ((5*A*b - a*B)*ArcTan[1 - (Sqrt[2
]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(9/4)*b^(3/4)) - ((5*A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])
/a^(1/4)])/(4*Sqrt[2]*a^(9/4)*b^(3/4)) - ((5*A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b
]*x])/(8*Sqrt[2]*a^(9/4)*b^(3/4)) + ((5*A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])
/(8*Sqrt[2]*a^(9/4)*b^(3/4))

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 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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 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])

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]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^{3/2} \left (a+b x^2\right )^2} \, dx &=\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}+\frac{\left (\frac{5 A b}{2}-\frac{a B}{2}\right ) \int \frac{1}{x^{3/2} \left (a+b x^2\right )} \, dx}{2 a b}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{(5 A b-a B) \int \frac{\sqrt{x}}{a+b x^2} \, dx}{4 a^2}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{(5 A b-a B) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{2 a^2}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}+\frac{(5 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^2 \sqrt{b}}-\frac{(5 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^2 \sqrt{b}}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{(5 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^2 b}-\frac{(5 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^2 b}-\frac{(5 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^{9/4} b^{3/4}}-\frac{(5 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^{9/4} b^{3/4}}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}-\frac{(5 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^{9/4} b^{3/4}}+\frac{(5 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^{9/4} b^{3/4}}-\frac{(5 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^{9/4} b^{3/4}}+\frac{(5 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^{9/4} b^{3/4}}\\ &=-\frac{5 A b-a B}{2 a^2 b \sqrt{x}}+\frac{A b-a B}{2 a b \sqrt{x} \left (a+b x^2\right )}+\frac{(5 A b-a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{(5 A b-a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{9/4} b^{3/4}}-\frac{(5 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^{9/4} b^{3/4}}+\frac{(5 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^{9/4} b^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.212999, size = 117, normalized size = 0.4 \[ \frac{2 x^{3/2} (a B-A b) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{b x^2}{a}\right )+3 A \left ((-a)^{3/4} \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )-(-a)^{3/4} \sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )-\frac{2 a}{\sqrt{x}}\right )}{3 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*((-2*a)/Sqrt[x] + (-a)^(3/4)*b^(1/4)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)] - (-a)^(3/4)*b^(1/4)*ArcTanh[(b
^(1/4)*Sqrt[x])/(-a)^(1/4)]) + 2*(-(A*b) + a*B)*x^(3/2)*Hypergeometric2F1[3/4, 2, 7/4, -((b*x^2)/a)])/(3*a^3)

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Maple [A]  time = 0.013, size = 323, normalized size = 1.1 \begin{align*} -2\,{\frac{A}{{a}^{2}\sqrt{x}}}-{\frac{Ab}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{B}{2\,a \left ( b{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}-{\frac{5\,\sqrt{2}A}{16\,{a}^{2}}\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{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{5\,\sqrt{2}A}{8\,{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{5\,\sqrt{2}A}{8\,{a}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{\sqrt{2}B}{16\,ab}\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{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{\sqrt{2}B}{8\,ab}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{\sqrt{2}B}{8\,ab}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^(3/2)/(b*x^2+a)^2,x)

[Out]

-2*A/a^2/x^(1/2)-1/2/a^2*x^(3/2)/(b*x^2+a)*A*b+1/2/a*x^(3/2)/(b*x^2+a)*B-5/16/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)))-5/8/a^2/(1/b*a
)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-5/8/a^2/(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*
a)^(1/4)*x^(1/2)-1)+1/16/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)+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)/x^(3/2)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.07271, size = 2026, normalized size = 7.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(a^2*b*x^3 + a^3*x)*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/
(a^9*b^3))^(1/4)*arctan((sqrt((B^6*a^6 - 30*A*B^5*a^5*b + 375*A^2*B^4*a^4*b^2 - 2500*A^3*B^3*a^3*b^3 + 9375*A^
4*B^2*a^2*b^4 - 18750*A^5*B*a*b^5 + 15625*A^6*b^6)*x - (B^4*a^9*b - 20*A*B^3*a^8*b^2 + 150*A^2*B^2*a^7*b^3 - 5
00*A^3*B*a^6*b^4 + 625*A^4*a^5*b^5)*sqrt(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 +
625*A^4*b^4)/(a^9*b^3)))*a^2*b*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b
^4)/(a^9*b^3))^(1/4) + (B^3*a^5*b - 15*A*B^2*a^4*b^2 + 75*A^2*B*a^3*b^3 - 125*A^3*a^2*b^4)*sqrt(x)*(-(B^4*a^4
- 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(1/4))/(B^4*a^4 - 20*A*B^3*
a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)) - (a^2*b*x^3 + a^3*x)*(-(B^4*a^4 - 20*A*B^3*a^3*
b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(1/4)*log(a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^
3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A*B^2*a^2*b + 75*A
^2*B*a*b^2 - 125*A^3*b^3)*sqrt(x)) + (a^2*b*x^3 + a^3*x)*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 5
00*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(1/4)*log(-a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2
- 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A*B^2*a^2*b + 75*A^2*B*a*b^2 - 125*A^3*b^3)*
sqrt(x)) + 4*((B*a - 5*A*b)*x^2 - 4*A*a)*sqrt(x))/(a^2*b*x^3 + a^3*x)

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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)/x**(3/2)/(b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21587, size = 375, normalized size = 1.3 \begin{align*} \frac{B a x^{2} - 5 \, A b x^{2} - 4 \, A a}{2 \,{\left (b x^{\frac{5}{2}} + a \sqrt{x}\right )} a^{2}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac{3}{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^{3} b^{3}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac{3}{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^{3} b^{3}} - \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac{3}{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^{3} b^{3}} + \frac{\sqrt{2}{\left (\left (a b^{3}\right )^{\frac{3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac{3}{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^{3} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*a*x^2 - 5*A*b*x^2 - 4*A*a)/((b*x^(5/2) + a*sqrt(x))*a^2) + 1/8*sqrt(2)*((a*b^3)^(3/4)*B*a - 5*(a*b^3)^(
3/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^3) + 1/8*sqrt(2)*((a*b^3)^(
3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^3*b^3) -
1/16*sqrt(2)*((a*b^3)^(3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*
b^3) + 1/16*sqrt(2)*((a*b^3)^(3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b)
)/(a^3*b^3)