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

Optimal. Leaf size=284 \[ -\frac{(A b-5 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}-\frac{(A b-5 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}-\frac{\sqrt{x} (A b-5 a B)}{2 a b^2}+\frac{x^{5/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.208611, antiderivative size = 284, 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, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{(A b-5 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}-\frac{(A b-5 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}-\frac{\sqrt{x} (A b-5 a B)}{2 a b^2}+\frac{x^{5/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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{x^{3/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac{\left (-\frac{A b}{2}+\frac{5 a B}{2}\right ) \int \frac{x^{3/2}}{a+b x^2} \, dx}{2 a b}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac{(A b-5 a B) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{4 b^2}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac{(A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{2 b^2}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac{(A b-5 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 \sqrt{a} b^2}+\frac{(A b-5 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 \sqrt{a} b^2}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac{(A b-5 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 \sqrt{a} b^{5/2}}+\frac{(A b-5 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 \sqrt{a} b^{5/2}}-\frac{(A b-5 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^{3/4} b^{9/4}}-\frac{(A b-5 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^{3/4} b^{9/4}}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}-\frac{(A b-5 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 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^{3/4} b^{9/4}}-\frac{(A b-5 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^{3/4} b^{9/4}}\\ &=-\frac{(A b-5 a B) \sqrt{x}}{2 a b^2}+\frac{(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}-\frac{(A b-5 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} a^{3/4} b^{9/4}}-\frac{(A b-5 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}+\frac{(A b-5 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} a^{3/4} b^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.388784, size = 353, normalized size = 1.24 \[ \frac{\frac{2 \sqrt{2} (5 a B-A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac{2 \sqrt{2} (A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac{\sqrt{2} A b \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}+\frac{\sqrt{2} A b \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}-\frac{8 A b^{5/4} \sqrt{x}}{a+b x^2}+\frac{8 a \sqrt [4]{b} B \sqrt{x}}{a+b x^2}+5 \sqrt{2} \sqrt [4]{a} B \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )-5 \sqrt{2} \sqrt [4]{a} B \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )+32 \sqrt [4]{b} B \sqrt{x}}{16 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^(1/4)*B*Sqrt[x] - (8*A*b^(5/4)*Sqrt[x])/(a + b*x^2) + (8*a*b^(1/4)*B*Sqrt[x])/(a + b*x^2) + (2*Sqrt[2]*(
-(A*b) + 5*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) + (2*Sqrt[2]*(A*b - 5*a*B)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) - (Sqrt[2]*A*b*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt
[b]*x])/a^(3/4) + 5*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + (Sqrt[2]*A*
b*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) - 5*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] + Sqrt
[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(16*b^(9/4))

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Maple [A]  time = 0.012, size = 323, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{{b}^{2}}}-{\frac{A}{2\,b \left ( b{x}^{2}+a \right ) }\sqrt{x}}+{\frac{Ba}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }\sqrt{x}}+{\frac{\sqrt{2}A}{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}A}{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}A}{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{5\,\sqrt{2}B}{8\,{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{5\,\sqrt{2}B}{8\,{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{5\,\sqrt{2}B}{16\,{b}^{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 ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.888488, size = 1594, normalized size = 5.61 \begin{align*} \frac{4 \,{\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{a^{2} b^{4} \sqrt{-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}} +{\left (25 \, B^{2} a^{2} - 10 \, A B a b + A^{2} b^{2}\right )} x} a^{2} b^{7} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{3}{4}} +{\left (5 \, B a^{3} b^{7} - A a^{2} b^{8}\right )} \sqrt{x} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{3}{4}}}{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}\right ) +{\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (a b^{2} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{1}{4}} -{\left (5 \, B a - A b\right )} \sqrt{x}\right ) -{\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (-a b^{2} \left (-\frac{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac{1}{4}} -{\left (5 \, B a - A b\right )} \sqrt{x}\right ) + 4 \,{\left (4 \, B b x^{2} + 5 \, B a - A b\right )} \sqrt{x}}{8 \,{\left (b^{3} x^{2} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(b^3*x^2 + a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a
^3*b^9))^(1/4)*arctan((sqrt(a^2*b^4*sqrt(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^
3 + A^4*b^4)/(a^3*b^9)) + (25*B^2*a^2 - 10*A*B*a*b + A^2*b^2)*x)*a^2*b^7*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 15
0*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(3/4) + (5*B*a^3*b^7 - A*a^2*b^8)*sqrt(x)*(-(625*B^4*
a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(3/4))/(625*B^4*a^4 - 500*A
*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)) + (b^3*x^2 + a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a
^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log(a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a
^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) - (5*B*a - A*b)*sqrt(x)) - (b^3*x^2 +
a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log
(-a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) -
(5*B*a - A*b)*sqrt(x)) + 4*(4*B*b*x^2 + 5*B*a - A*b)*sqrt(x))/(b^3*x^2 + a*b^2)

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

[Out]

Timed out

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Giac [A]  time = 1.15135, size = 382, normalized size = 1.35 \begin{align*} \frac{2 \, B \sqrt{x}}{b^{2}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \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 b^{3}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \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 b^{3}} - \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \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 b^{3}} + \frac{\sqrt{2}{\left (5 \, \left (a b^{3}\right )^{\frac{1}{4}} B a - \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 b^{3}} + \frac{B a \sqrt{x} - A b \sqrt{x}}{2 \,{\left (b x^{2} + a\right )} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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