3.376 \(\int \frac{x^{5/2} (A+B x^2)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=289 \[ -\frac{x^{3/2} (3 A b-7 a B)}{6 a b^2}+\frac{(3 A b-7 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{x^{7/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

[Out]

-((3*A*b - 7*a*B)*x^(3/2))/(6*a*b^2) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x^2)) - ((3*A*b - 7*a*B)*ArcTan[1 -
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/
4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(11/4)) - ((3*A*b - 7*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt
[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(11/4))

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Rubi [A]  time = 0.21859, 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, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{x^{3/2} (3 A b-7 a B)}{6 a b^2}+\frac{(3 A b-7 a B) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{x^{7/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

-((3*A*b - 7*a*B)*x^(3/2))/(6*a*b^2) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x^2)) - ((3*A*b - 7*a*B)*ArcTan[1 -
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/
4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(11/4)) + ((3*A*b - 7*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(11/4)) - ((3*A*b - 7*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt
[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(11/4))

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 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{x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac{\left (-\frac{3 A b}{2}+\frac{7 a B}{2}\right ) \int \frac{x^{5/2}}{a+b x^2} \, dx}{2 a b}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b-7 a B) \int \frac{\sqrt{x}}{a+b x^2} \, dx}{4 b^2}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b-7 a B) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{2 b^2}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}-\frac{(3 A b-7 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 b^{5/2}}+\frac{(3 A b-7 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{4 b^{5/2}}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b-7 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 b^3}+\frac{(3 A b-7 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 b^3}+\frac{(3 A b-7 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} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 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} \sqrt [4]{a} b^{11/4}}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}+\frac{(3 A b-7 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 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} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 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} \sqrt [4]{a} b^{11/4}}\\ &=-\frac{(3 A b-7 a B) x^{3/2}}{6 a b^2}+\frac{(A b-a B) x^{7/2}}{2 a b \left (a+b x^2\right )}-\frac{(3 A b-7 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{4 \sqrt{2} \sqrt [4]{a} b^{11/4}}+\frac{(3 A b-7 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}-\frac{(3 A b-7 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{8 \sqrt{2} \sqrt [4]{a} b^{11/4}}\\ \end{align*}

Mathematica [C]  time = 0.192071, size = 136, normalized size = 0.47 \[ \frac{2 x^{3/2} (a B-A b) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{b x^2}{a}\right )}{3 a b^2}+\frac{3 (A b-2 a B) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )+(6 a B-3 A b) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{-a}}\right )+2 \sqrt [4]{-a} b^{3/4} B x^{3/2}}{3 \sqrt [4]{-a} b^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

(2*(-a)^(1/4)*b^(3/4)*B*x^(3/2) + 3*(A*b - 2*a*B)*ArcTan[(b^(1/4)*Sqrt[x])/(-a)^(1/4)] + (-3*A*b + 6*a*B)*ArcT
anh[(b^(1/4)*Sqrt[x])/(-a)^(1/4)])/(3*(-a)^(1/4)*b^(11/4)) + (2*(-(A*b) + a*B)*x^(3/2)*Hypergeometric2F1[3/4,
2, 7/4, -((b*x^2)/a)])/(3*a*b^2)

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Maple [A]  time = 0.013, size = 317, normalized size = 1.1 \begin{align*}{\frac{2\,B}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}-{\frac{A}{2\,b \left ( b{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{Ba}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }{x}^{{\frac{3}{2}}}}-{\frac{7\,\sqrt{2}Ba}{16\,{b}^{3}}\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{7\,\sqrt{2}Ba}{8\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{7\,\sqrt{2}Ba}{8\,{b}^{3}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{3\,\sqrt{2}A}{16\,{b}^{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{3\,\sqrt{2}A}{8\,{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{3\,\sqrt{2}A}{8\,{b}^{2}}\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(x^(5/2)*(B*x^2+A)/(b*x^2+a)^2,x)

[Out]

2/3*B*x^(3/2)/b^2-1/2/b*x^(3/2)/(b*x^2+a)*A+1/2/b^2*x^(3/2)/(b*x^2+a)*B*a-7/16/b^3/(1/b*a)^(1/4)*2^(1/2)*B*a*l
n((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)))-7/8/b^3/(1/
b*a)^(1/4)*2^(1/2)*B*a*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-7/8/b^3/(1/b*a)^(1/4)*2^(1/2)*B*a*arctan(2^(1/2
)/(1/b*a)^(1/4)*x^(1/2)-1)+3/16/b^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/b^2/(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)
*x^(1/2)+1)+3/8/b^2/(1/b*a)^(1/4)*2^(1/2)*A*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(x^(5/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.999223, size = 2141, normalized size = 7.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/24*(12*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A
^4*b^4)/(a*b^11))^(1/4)*arctan((sqrt((117649*B^6*a^6 - 302526*A*B^5*a^5*b + 324135*A^2*B^4*a^4*b^2 - 185220*A^
3*B^3*a^3*b^3 + 59535*A^4*B^2*a^2*b^4 - 10206*A^5*B*a*b^5 + 729*A^6*b^6)*x - (2401*B^4*a^5*b^5 - 4116*A*B^3*a^
4*b^6 + 2646*A^2*B^2*a^3*b^7 - 756*A^3*B*a^2*b^8 + 81*A^4*a*b^9)*sqrt(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646
*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11)))*b^3*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*
B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4) + (343*B^3*a^3*b^3 - 441*A*B^2*a^2*b^4 + 189*A^2*B
*a*b^5 - 27*A^3*b^6)*sqrt(x)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*
A^4*b^4)/(a*b^11))^(1/4))/(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b
^4)) - 3*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^
4*b^4)/(a*b^11))^(1/4)*log(a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 +
81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) + 3*(b^
3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^
11))^(1/4)*log(-a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4
)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) - 4*(4*B*b*x^3 + (
7*B*a - 3*A*b)*x)*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**(5/2)*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1643, size = 382, normalized size = 1.32 \begin{align*} \frac{2 \, B x^{\frac{3}{2}}}{3 \, b^{2}} + \frac{B a x^{\frac{3}{2}} - A b x^{\frac{3}{2}}}{2 \,{\left (b x^{2} + a\right )} b^{2}} - \frac{\sqrt{2}{\left (7 \, \left (a b^{3}\right )^{\frac{3}{4}} B a - 3 \, \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 b^{5}} - \frac{\sqrt{2}{\left (7 \, \left (a b^{3}\right )^{\frac{3}{4}} B a - 3 \, \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 b^{5}} + \frac{\sqrt{2}{\left (7 \, \left (a b^{3}\right )^{\frac{3}{4}} B a - 3 \, \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 b^{5}} - \frac{\sqrt{2}{\left (7 \, \left (a b^{3}\right )^{\frac{3}{4}} B a - 3 \, \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 b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

2/3*B*x^(3/2)/b^2 + 1/2*(B*a*x^(3/2) - A*b*x^(3/2))/((b*x^2 + a)*b^2) - 1/8*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(
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*b^5) - 1/8*sqrt(2)*(7*(
a*b^3)^(3/4)*B*a - 3*(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*
b^5) + 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b
))/(a*b^5) - 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + s
qrt(a/b))/(a*b^5)