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

Optimal. Leaf size=293 \[ -\frac{3 (a B+7 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (a B+7 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}-\frac{3 (a B+7 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (a B+7 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}+\frac{\sqrt{x} (a B+7 A b)}{16 a^2 b \left (a+b x^2\right )}+\frac{\sqrt{x} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

((A*b - a*B)*Sqrt[x])/(4*a*b*(a + b*x^2)^2) + ((7*A*b + a*B)*Sqrt[x])/(16*a^2*b*(a + b*x^2)) - (3*(7*A*b + a*B
)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*b^(5/4)) + (3*(7*A*b + a*B)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*b^(5/4)) - (3*(7*A*b + a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*b^(5/4)) + (3*(7*A*b + a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*b^(5/4))

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Rubi [A]  time = 0.212456, antiderivative size = 293, 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, 211, 1165, 628, 1162, 617, 204} \[ -\frac{3 (a B+7 A b) \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (a B+7 A b) \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}-\frac{3 (a B+7 A b) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (a B+7 A b) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}+\frac{\sqrt{x} (a B+7 A b)}{16 a^2 b \left (a+b x^2\right )}+\frac{\sqrt{x} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*Sqrt[x])/(4*a*b*(a + b*x^2)^2) + ((7*A*b + a*B)*Sqrt[x])/(16*a^2*b*(a + b*x^2)) - (3*(7*A*b + a*B
)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*b^(5/4)) + (3*(7*A*b + a*B)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(11/4)*b^(5/4)) - (3*(7*A*b + a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*b^(5/4)) + (3*(7*A*b + a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/
4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(11/4)*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 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 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 )^3} \, dx &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{\left (\frac{7 A b}{2}+\frac{a B}{2}\right ) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}+\frac{(3 (7 A b+a B)) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{32 a^2 b}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}+\frac{(3 (7 A b+a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{16 a^2 b}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}+\frac{(3 (7 A b+a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^{5/2} b}+\frac{(3 (7 A b+a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^{5/2} b}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}+\frac{(3 (7 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 )}{64 a^{5/2} b^{3/2}}+\frac{(3 (7 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 )}{64 a^{5/2} b^{3/2}}-\frac{(3 (7 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 )}{64 \sqrt{2} a^{11/4} b^{5/4}}-\frac{(3 (7 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 )}{64 \sqrt{2} a^{11/4} b^{5/4}}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}-\frac{3 (7 A b+a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (7 A b+a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}+\frac{(3 (7 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 )}{32 \sqrt{2} a^{11/4} b^{5/4}}-\frac{(3 (7 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 )}{32 \sqrt{2} a^{11/4} b^{5/4}}\\ &=\frac{(A b-a B) \sqrt{x}}{4 a b \left (a+b x^2\right )^2}+\frac{(7 A b+a B) \sqrt{x}}{16 a^2 b \left (a+b x^2\right )}-\frac{3 (7 A b+a B) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (7 A b+a B) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{11/4} b^{5/4}}-\frac{3 (7 A b+a B) \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}+\frac{3 (7 A b+a B) \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{11/4} b^{5/4}}\\ \end{align*}

Mathematica [A]  time = 0.279884, size = 230, normalized size = 0.78 \[ \frac{\frac{(a B+7 A b) \left (7 \left (a+b x^2\right ) \left (8 a^{3/4} \sqrt [4]{b} \sqrt{x}-3 \sqrt{2} \left (a+b x^2\right ) \left (\log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )-\log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )\right )\right )+32 a^{7/4} \sqrt [4]{b} \sqrt{x}\right )}{a^{11/4} \sqrt [4]{b}}-256 B \sqrt{x}}{896 b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*B*Sqrt[x] + ((7*A*b + a*B)*(32*a^(7/4)*b^(1/4)*Sqrt[x] + 7*(a + b*x^2)*(8*a^(3/4)*b^(1/4)*Sqrt[x] - 3*Sq
rt[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)*Sq
rt[x] + Sqrt[b]*x]))))/(a^(11/4)*b^(1/4)))/(896*b*(a + b*x^2)^2)

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Maple [A]  time = 0.015, size = 325, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 1/32\,{\frac{ \left ( 7\,Ab+Ba \right ){x}^{5/2}}{{a}^{2}}}+1/32\,{\frac{ \left ( 11\,Ab-3\,Ba \right ) \sqrt{x}}{ab}} \right ) }+{\frac{21\,\sqrt{2}A}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{21\,\sqrt{2}A}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{21\,\sqrt{2}A}{128\,{a}^{3}}\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}B}{64\,{a}^{2}b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{3\,\sqrt{2}B}{64\,{a}^{2}b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}B}{128\,{a}^{2}b}\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((B*x^2+A)/(b*x^2+a)^3/x^(1/2),x)

[Out]

2*(1/32*(7*A*b+B*a)/a^2*x^(5/2)+1/32*(11*A*b-3*B*a)/a/b*x^(1/2))/(b*x^2+a)^2+21/64/a^3*(1/b*a)^(1/4)*2^(1/2)*A
*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+21/64/a^3*(1/b*a)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2
)-1)+21/128/a^3*(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^2/b*(1/b*a)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+3/64/
a^2/b*(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^2/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)))

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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)^3/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.950624, size = 1782, normalized size = 6.08 \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)/(b*x^2+a)^3/x^(1/2),x, algorithm="fricas")

[Out]

1/64*(12*(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*
a*b^3 + 2401*A^4*b^4)/(a^11*b^5))^(1/4)*arctan((sqrt(a^6*b^2*sqrt(-(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2
*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)/(a^11*b^5)) + (B^2*a^2 + 14*A*B*a*b + 49*A^2*b^2)*x)*a^8*b^4*(-(B^4*a^
4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)/(a^11*b^5))^(3/4) - (B*a^9*b^4 + 7
*A*a^8*b^5)*sqrt(x)*(-(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)/(a^11
*b^5))^(3/4))/(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)) + 3*(a^2*b^3
*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*
b^4)/(a^11*b^5))^(1/4)*log(3*a^3*b*(-(B^4*a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401
*A^4*b^4)/(a^11*b^5))^(1/4) + 3*(B*a + 7*A*b)*sqrt(x)) - 3*(a^2*b^3*x^4 + 2*a^3*b^2*x^2 + a^4*b)*(-(B^4*a^4 +
28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)/(a^11*b^5))^(1/4)*log(-3*a^3*b*(-(B^4*
a^4 + 28*A*B^3*a^3*b + 294*A^2*B^2*a^2*b^2 + 1372*A^3*B*a*b^3 + 2401*A^4*b^4)/(a^11*b^5))^(1/4) + 3*(B*a + 7*A
*b)*sqrt(x)) - 4*(3*B*a^2 - 11*A*a*b - (B*a*b + 7*A*b^2)*x^2)*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)/(b*x**2+a)**3/x**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.16115, size = 396, normalized size = 1.35 \begin{align*} \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 7 \, \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 )}{64 \, a^{3} b^{2}} + \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 7 \, \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 )}{64 \, a^{3} b^{2}} + \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 7 \, \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 )}{128 \, a^{3} b^{2}} - \frac{3 \, \sqrt{2}{\left (\left (a b^{3}\right )^{\frac{1}{4}} B a + 7 \, \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 )}{128 \, a^{3} b^{2}} + \frac{B a b x^{\frac{5}{2}} + 7 \, A b^{2} x^{\frac{5}{2}} - 3 \, B a^{2} \sqrt{x} + 11 \, A a b \sqrt{x}}{16 \,{\left (b x^{2} + a\right )}^{2} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*sqrt(2)*((a*b^3)^(1/4)*B*a + 7*(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^3*b^2) + 3/64*sqrt(2)*((a*b^3)^(1/4)*B*a + 7*(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^3*b^2) + 3/128*sqrt(2)*((a*b^3)^(1/4)*B*a + 7*(a*b^3)^(1/4)*A*b)*log(sqr
t(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^2) - 3/128*sqrt(2)*((a*b^3)^(1/4)*B*a + 7*(a*b^3)^(1/4)*A*b)*
log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^3*b^2) + 1/16*(B*a*b*x^(5/2) + 7*A*b^2*x^(5/2) - 3*B*a^2*
sqrt(x) + 11*A*a*b*sqrt(x))/((b*x^2 + a)^2*a^2*b)