3.21 \(\int \frac{(A+B x) (a+b x^2)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=141 \[ -\frac{5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{15}{8} a^2 \sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2} (2 A-B x)}{4 x^2}-\frac{5 \left (a+b x^2\right )^{3/2} (3 a B-2 A b x)}{12 x}+\frac{5}{8} a b \sqrt{a+b x^2} (4 A+3 B x) \]

[Out]

(5*a*b*(4*A + 3*B*x)*Sqrt[a + b*x^2])/8 - (5*(3*a*B - 2*A*b*x)*(a + b*x^2)^(3/2))/(12*x) - ((2*A - B*x)*(a + b
*x^2)^(5/2))/(4*x^2) + (15*a^2*Sqrt[b]*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/8 - (5*a^(3/2)*A*b*ArcTanh[Sqrt
[a + b*x^2]/Sqrt[a]])/2

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Rubi [A]  time = 0.117082, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {813, 815, 844, 217, 206, 266, 63, 208} \[ -\frac{5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{15}{8} a^2 \sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2} (2 A-B x)}{4 x^2}-\frac{5 \left (a+b x^2\right )^{3/2} (3 a B-2 A b x)}{12 x}+\frac{5}{8} a b \sqrt{a+b x^2} (4 A+3 B x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x^2)^(5/2))/x^3,x]

[Out]

(5*a*b*(4*A + 3*B*x)*Sqrt[a + b*x^2])/8 - (5*(3*a*B - 2*A*b*x)*(a + b*x^2)^(3/2))/(12*x) - ((2*A - B*x)*(a + b
*x^2)^(5/2))/(4*x^2) + (15*a^2*Sqrt[b]*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/8 - (5*a^(3/2)*A*b*ArcTanh[Sqrt
[a + b*x^2]/Sqrt[a]])/2

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x^2\right )^{5/2}}{x^3} \, dx &=-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}-\frac{5}{16} \int \frac{(-4 a B-8 A b x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx\\ &=-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{5}{32} \int \frac{(16 a A b+24 a b B x) \sqrt{a+b x^2}}{x} \, dx\\ &=\frac{5}{8} a b (4 A+3 B x) \sqrt{a+b x^2}-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{5 \int \frac{32 a^2 A b^2+24 a^2 b^2 B x}{x \sqrt{a+b x^2}} \, dx}{64 b}\\ &=\frac{5}{8} a b (4 A+3 B x) \sqrt{a+b x^2}-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{1}{2} \left (5 a^2 A b\right ) \int \frac{1}{x \sqrt{a+b x^2}} \, dx+\frac{1}{8} \left (15 a^2 b B\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{5}{8} a b (4 A+3 B x) \sqrt{a+b x^2}-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{1}{4} \left (5 a^2 A b\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )+\frac{1}{8} \left (15 a^2 b B\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{5}{8} a b (4 A+3 B x) \sqrt{a+b x^2}-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{15}{8} a^2 \sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{1}{2} \left (5 a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{5}{8} a b (4 A+3 B x) \sqrt{a+b x^2}-\frac{5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac{(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac{15}{8} a^2 \sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0259002, size = 92, normalized size = 0.65 \[ \frac{A b \left (a+b x^2\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )}{7 a^2}-\frac{a^2 B \sqrt{a+b x^2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x^2)^(5/2))/x^3,x]

[Out]

-((a^2*B*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, -((b*x^2)/a)])/(x*Sqrt[1 + (b*x^2)/a])) + (A*b*(a
+ b*x^2)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, 1 + (b*x^2)/a])/(7*a^2)

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Maple [A]  time = 0.007, size = 181, normalized size = 1.3 \begin{align*} -{\frac{A}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Ab}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,Ab}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Ab}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{5\,Aab}{2}\sqrt{b{x}^{2}+a}}-{\frac{B}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{bBx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,bBx}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{15\,Bbax}{8}\sqrt{b{x}^{2}+a}}+{\frac{15\,B{a}^{2}}{8}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(5/2)/x^3,x)

[Out]

-1/2*A*(b*x^2+a)^(7/2)/a/x^2+1/2*A*b/a*(b*x^2+a)^(5/2)+5/6*A*b*(b*x^2+a)^(3/2)-5/2*A*b*a^(3/2)*ln((2*a+2*a^(1/
2)*(b*x^2+a)^(1/2))/x)+5/2*A*b*a*(b*x^2+a)^(1/2)-B/a/x*(b*x^2+a)^(7/2)+B*b/a*x*(b*x^2+a)^(5/2)+5/4*B*b*x*(b*x^
2+a)^(3/2)+15/8*B*b*a*x*(b*x^2+a)^(1/2)+15/8*B*b^(1/2)*a^2*ln(x*b^(1/2)+(b*x^2+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+A)*(b*x^2+a)^(5/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70185, size = 1316, normalized size = 9.33 \begin{align*} \left [\frac{45 \, B a^{2} \sqrt{b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 60 \, A a^{\frac{3}{2}} b x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{48 \, x^{2}}, -\frac{45 \, B a^{2} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - 30 \, A a^{\frac{3}{2}} b x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) -{\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{24 \, x^{2}}, \frac{120 \, A \sqrt{-a} a b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + 45 \, B a^{2} \sqrt{b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{48 \, x^{2}}, -\frac{45 \, B a^{2} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - 60 \, A \sqrt{-a} a b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt{b x^{2} + a}}{24 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/48*(45*B*a^2*sqrt(b)*x^2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 60*A*a^(3/2)*b*x^2*log(-(b*x^2 -
2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B*b^2*x^5 + 8*A*b^2*x^4 + 27*B*a*b*x^3 + 56*A*a*b*x^2 - 24*B*a^2
*x - 12*A*a^2)*sqrt(b*x^2 + a))/x^2, -1/24*(45*B*a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 30*A*a^
(3/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - (6*B*b^2*x^5 + 8*A*b^2*x^4 + 27*B*a*b*x^3 +
56*A*a*b*x^2 - 24*B*a^2*x - 12*A*a^2)*sqrt(b*x^2 + a))/x^2, 1/48*(120*A*sqrt(-a)*a*b*x^2*arctan(sqrt(-a)/sqrt(
b*x^2 + a)) + 45*B*a^2*sqrt(b)*x^2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(6*B*b^2*x^5 + 8*A*b^2*
x^4 + 27*B*a*b*x^3 + 56*A*a*b*x^2 - 24*B*a^2*x - 12*A*a^2)*sqrt(b*x^2 + a))/x^2, -1/24*(45*B*a^2*sqrt(-b)*x^2*
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 60*A*sqrt(-a)*a*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (6*B*b^2*x^5 + 8
*A*b^2*x^4 + 27*B*a*b*x^3 + 56*A*a*b*x^2 - 24*B*a^2*x - 12*A*a^2)*sqrt(b*x^2 + a))/x^2]

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Sympy [A]  time = 11.3184, size = 279, normalized size = 1.98 \begin{align*} - \frac{5 A a^{\frac{3}{2}} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2} - \frac{A a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} + \frac{2 A a^{2} \sqrt{b}}{x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{2 A a b^{\frac{3}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + A b^{2} \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) - \frac{B a^{\frac{5}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + B a^{\frac{3}{2}} b x \sqrt{1 + \frac{b x^{2}}{a}} - \frac{7 B a^{\frac{3}{2}} b x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B \sqrt{a} b^{2} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{15 B a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8} + \frac{B b^{3} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x**2+a)**(5/2)/x**3,x)

[Out]

-5*A*a**(3/2)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) + 2*A*a**2*sqrt(b)/(x
*sqrt(a/(b*x**2) + 1)) + 2*A*a*b**(3/2)*x/sqrt(a/(b*x**2) + 1) + A*b**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)),
((a + b*x**2)**(3/2)/(3*b), True)) - B*a**(5/2)/(x*sqrt(1 + b*x**2/a)) + B*a**(3/2)*b*x*sqrt(1 + b*x**2/a) - 7
*B*a**(3/2)*b*x/(8*sqrt(1 + b*x**2/a)) + 3*B*sqrt(a)*b**2*x**3/(8*sqrt(1 + b*x**2/a)) + 15*B*a**2*sqrt(b)*asin
h(sqrt(b)*x/sqrt(a))/8 + B*b**3*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.18631, size = 296, normalized size = 2.1 \begin{align*} \frac{5 \, A a^{2} b \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{15}{8} \, B a^{2} \sqrt{b} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right ) + \frac{1}{24} \,{\left (56 \, A a b +{\left (27 \, B a b + 2 \,{\left (3 \, B b^{2} x + 4 \, A b^{2}\right )} x\right )} x\right )} \sqrt{b x^{2} + a} + \frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{3} A a^{2} b + 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{3} \sqrt{b} +{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )} A a^{3} b - 2 \, B a^{4} \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^3,x, algorithm="giac")

[Out]

5*A*a^2*b*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 15/8*B*a^2*sqrt(b)*log(abs(-sqrt(b)*x + s
qrt(b*x^2 + a))) + 1/24*(56*A*a*b + (27*B*a*b + 2*(3*B*b^2*x + 4*A*b^2)*x)*x)*sqrt(b*x^2 + a) + ((sqrt(b)*x -
sqrt(b*x^2 + a))^3*A*a^2*b + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^3*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A
*a^3*b - 2*B*a^4*sqrt(b))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^2