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

Optimal. Leaf size=298 \[ \frac{(3 a B+5 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(3 a B+5 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(3 a B+5 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(3 a B+5 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{x^{3/2} (3 a B+5 A b)}{16 a^2 b \left (a+b x^2\right )}+\frac{x^{3/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

((A*b - a*B)*x^(3/2))/(4*a*b*(a + b*x^2)^2) + ((5*A*b + 3*a*B)*x^(3/2))/(16*a^2*b*(a + b*x^2)) - ((5*A*b + 3*a
*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4)) - ((5*A*b + 3*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4))

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Rubi [A]  time = 0.216198, antiderivative size = 298, 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, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{(3 a B+5 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(3 a B+5 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(3 a B+5 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(3 a B+5 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{x^{3/2} (3 a B+5 A b)}{16 a^2 b \left (a+b x^2\right )}+\frac{x^{3/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(3/2))/(4*a*b*(a + b*x^2)^2) + ((5*A*b + 3*a*B)*x^(3/2))/(16*a^2*b*(a + b*x^2)) - ((5*A*b + 3*a
*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(9/4)*b^(7/4)) + ((5*A*b + 3*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4)) - ((5*A*b + 3*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(9/4)*b^(7/4))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -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{\sqrt{x} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{\left (\frac{5 A b}{2}+\frac{3 a B}{2}\right ) \int \frac{\sqrt{x}}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac{(5 A b+3 a B) \int \frac{\sqrt{x}}{a+b x^2} \, dx}{32 a^2 b}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac{(5 A b+3 a B) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{16 a^2 b}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}-\frac{(5 A b+3 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^2 b^{3/2}}+\frac{(5 A b+3 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^2 b^{3/2}}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac{(5 A b+3 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 )}{64 a^2 b^2}+\frac{(5 A b+3 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 )}{64 a^2 b^2}+\frac{(5 A b+3 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 )}{64 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(5 A b+3 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 )}{64 \sqrt{2} a^{9/4} b^{7/4}}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}+\frac{(5 A b+3 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(5 A b+3 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(5 A b+3 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 )}{32 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(5 A b+3 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 )}{32 \sqrt{2} a^{9/4} b^{7/4}}\\ &=\frac{(A b-a B) x^{3/2}}{4 a b \left (a+b x^2\right )^2}+\frac{(5 A b+3 a B) x^{3/2}}{16 a^2 b \left (a+b x^2\right )}-\frac{(5 A b+3 a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(5 A b+3 a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{9/4} b^{7/4}}+\frac{(5 A b+3 a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}-\frac{(5 A b+3 a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{9/4} b^{7/4}}\\ \end{align*}

Mathematica [C]  time = 0.0527267, size = 62, normalized size = 0.21 \[ \frac{2 x^{3/2} \left ((A b-a B) \, _2F_1\left (\frac{3}{4},3;\frac{7}{4};-\frac{b x^2}{a}\right )+a B \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-\frac{b x^2}{a}\right )\right )}{3 a^3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 335, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 1/32\,{\frac{ \left ( 5\,Ab+3\,Ba \right ){x}^{7/2}}{{a}^{2}}}+1/32\,{\frac{ \left ( 9\,Ab-Ba \right ){x}^{3/2}}{ab}} \right ) }+{\frac{5\,\sqrt{2}A}{64\,{a}^{2}b}\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}{64\,{a}^{2}b}\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}{128\,{a}^{2}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{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{3\,\sqrt{2}B}{64\,a{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}B}{64\,a{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}B}{128\,a{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}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x)

[Out]

2*(1/32*(5*A*b+3*B*a)/a^2*x^(7/2)+1/32*(9*A*b-B*a)/a/b*x^(3/2))/(b*x^2+a)^2+5/64/a^2/b/(1/b*a)^(1/4)*2^(1/2)*A
*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+5/64/a^2/b/(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/
2)-1)+5/128/a^2/b/(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)))+3/64/a/b^2/(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+3/6
4/a/b^2/(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)-1)+3/128/a/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((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.0065, size = 2276, normalized size = 7.64 \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^(1/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/64*(4*(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A
^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*arctan((sqrt((729*B^6*a^6 + 7290*A*B^5*a^5*b + 30375*A^2*B^4*a^4*b^
2 + 67500*A^3*B^3*a^3*b^3 + 84375*A^4*B^2*a^2*b^4 + 56250*A^5*B*a*b^5 + 15625*A^6*b^6)*x - (81*B^4*a^9*b^3 + 5
40*A*B^3*a^8*b^4 + 1350*A^2*B^2*a^7*b^5 + 1500*A^3*B*a^6*b^6 + 625*A^4*a^5*b^7)*sqrt(-(81*B^4*a^4 + 540*A*B^3*
a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7)))*a^2*b^2*(-(81*B^4*a^4 + 540*A*B^3*a
^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4) - (27*B^3*a^5*b^2 + 135*A*B^2*a
^4*b^3 + 225*A^2*B*a^3*b^4 + 125*A^3*a^2*b^5)*sqrt(x)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 +
1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4))/(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500
*A^3*B*a*b^3 + 625*A^4*b^4)) - (a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b + 1350*A^
2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(a^7*b^5*(-(81*B^4*a^4 + 540*A*B^3*a^3*b +
1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135*A*B^2*a^2*b + 225
*A^2*B*a*b^2 + 125*A^3*b^3)*sqrt(x)) + (a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(81*B^4*a^4 + 540*A*B^3*a^3*b +
1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(1/4)*log(-a^7*b^5*(-(81*B^4*a^4 + 540*A*B^
3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 1500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^7))^(3/4) + (27*B^3*a^3 + 135*A*B^2*a^
2*b + 225*A^2*B*a*b^2 + 125*A^3*b^3)*sqrt(x)) - 4*((3*B*a*b + 5*A*b^2)*x^3 - (B*a^2 - 9*A*a*b)*x)*sqrt(x))/(a^
2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*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)*x**(1/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17936, size = 402, normalized size = 1.35 \begin{align*} \frac{3 \, B a b x^{\frac{7}{2}} + 5 \, A b^{2} x^{\frac{7}{2}} - B a^{2} x^{\frac{3}{2}} + 9 \, A a b x^{\frac{3}{2}}}{16 \,{\left (b x^{2} + a\right )}^{2} a^{2} b} + \frac{\sqrt{2}{\left (3 \, \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 )}{64 \, a^{3} b^{4}} + \frac{\sqrt{2}{\left (3 \, \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 )}{64 \, a^{3} b^{4}} - \frac{\sqrt{2}{\left (3 \, \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 )}{128 \, a^{3} b^{4}} + \frac{\sqrt{2}{\left (3 \, \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 )}{128 \, a^{3} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*x^(1/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/16*(3*B*a*b*x^(7/2) + 5*A*b^2*x^(7/2) - B*a^2*x^(3/2) + 9*A*a*b*x^(3/2))/((b*x^2 + a)^2*a^2*b) + 1/64*sqrt(2
)*(3*(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^4) + 1/64*sqrt(2)*(3*(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^4) - 1/128*sqrt(2)*(3*(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^4) + 1/128*sqrt(2)*(3*(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^4)