\(\int \frac {A+B x^2}{x^3 (a+b x^2)^{5/2}} \, dx\) [594]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-5 A b+2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b-2 a B}{2 a^3 \sqrt {a+b x^2}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \]

[Out]

1/6*(-5*A*b+2*B*a)/a^2/(b*x^2+a)^(3/2)-1/2*A/a/x^2/(b*x^2+a)^(3/2)+1/2*(5*A*b-2*B*a)*arctanh((b*x^2+a)^(1/2)/a
^(1/2))/a^(7/2)+1/2*(-5*A*b+2*B*a)/a^3/(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 79, 53, 65, 214} \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}}-\frac {5 A b-2 a B}{2 a^3 \sqrt {a+b x^2}}-\frac {5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}} \]

[In]

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

[Out]

-1/6*(5*A*b - 2*a*B)/(a^2*(a + b*x^2)^(3/2)) - A/(2*a*x^2*(a + b*x^2)^(3/2)) - (5*A*b - 2*a*B)/(2*a^3*Sqrt[a +
b*x^2]) + ((5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(7/2))

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x^2 (a+b x)^{5/2}} \, dx,x,x^2\right ) \\ & = -\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {\left (-\frac {5 A b}{2}+a B\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,x^2\right )}{2 a} \\ & = -\frac {5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {(5 A b-2 a B) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,x^2\right )}{4 a^2} \\ & = -\frac {5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b-2 a B}{2 a^3 \sqrt {a+b x^2}}-\frac {(5 A b-2 a B) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{4 a^3} \\ & = -\frac {5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b-2 a B}{2 a^3 \sqrt {a+b x^2}}-\frac {(5 A b-2 a B) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 a^3 b} \\ & = -\frac {5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac {A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b-2 a B}{2 a^3 \sqrt {a+b x^2}}+\frac {(5 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 a^2 A-20 a A b x^2+8 a^2 B x^2-15 A b^2 x^4+6 a b B x^4}{6 a^3 x^2 \left (a+b x^2\right )^{3/2}}+\frac {(5 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \]

[In]

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

[Out]

(-3*a^2*A - 20*a*A*b*x^2 + 8*a^2*B*x^2 - 15*A*b^2*x^4 + 6*a*b*B*x^4)/(6*a^3*x^2*(a + b*x^2)^(3/2)) + ((5*A*b -
2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(7/2))

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(-\frac {-5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} x^{2} \left (A b -\frac {2 B a}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\frac {20 x^{2} \left (-\frac {3 x^{2} B}{10}+A \right ) b \,a^{\frac {3}{2}}}{3}+\left (-\frac {8 x^{2} B}{3}+A \right ) a^{\frac {5}{2}}+5 A \sqrt {a}\, b^{2} x^{4}}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{2}}\)	\(96\)
default	\(B \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )+A \left (-\frac {1}{2 a \,x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 b \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )}{2 a}\right )\)	\(152\)
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{2 a^{3} x^{2}}+\frac {5 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A b}{2 a^{\frac {7}{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{a^{\frac {5}{2}}}+\frac {13 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {7 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {13 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {7 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} b \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} b \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}\)	\(595\)

[In]

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

[Out]

-1/2*(-5*(b*x^2+a)^(3/2)*x^2*(A*b-2/5*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+20/3*x^2*(-3/10*x^2*B+A)*b*a^(3/2)
+(-8/3*x^2*B+A)*a^(5/2)+5*A*a^(1/2)*b^2*x^4)/(b*x^2+a)^(3/2)/a^(7/2)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.09 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/12*(3*((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^4 + (2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(a)*log(
-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*(2*B*a^2*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*
A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), 1/6*(3*((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2
*B*a^2*b - 5*A*a*b^2)*x^4 + (2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(2*B*a^2
*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6
*x^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1608 vs. \(2 (99) = 198\).

Time = 21.73 (sec) , antiderivative size = 1608, normalized size of antiderivative = 14.23 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

A*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2
)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2
*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*
a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70*a**15*b**2*x**4*sqrt(1 + b*x**2
/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b**
2*x**4*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**
8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b*
*2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*
x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14*b**3*x**6*log(sqrt(1 + b*x
**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a
**13*b**4*x**8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*
b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(
35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8)) + B*(8*a**7*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 +
18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**7*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 1
8*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**7*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2
)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 14*a**6*b*x**2*sqrt(1 + b*x**2/a)/(6*a**(19/2) +
18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 9*a**6*b*x**2*log(b*x**2/a)/(6*a**(19/
2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 18*a**6*b*x**2*log(sqrt(1 + b*x**
2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 6*a**5*b**2*x
**4*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) +
9*a**5*b**2*x**4*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*
x**6) - 18*a**5*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x
**4 + 6*a**(13/2)*b**3*x**6) + 3*a**4*b**3*x**6*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2
)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**4*b**3*x**6*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2
)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {5 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a^{2}} + \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {5 \, A b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, A b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {A}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} \]

[In]

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

[Out]

-B*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 5/2*A*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) + B/(sqrt(b*x^2 + a)*
a^2) + 1/3*B/((b*x^2 + a)^(3/2)*a) - 5/2*A*b/(sqrt(b*x^2 + a)*a^3) - 5/6*A*b/((b*x^2 + a)^(3/2)*a^2) - 1/2*A/(
(b*x^2 + a)^(3/2)*a*x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{3}} + \frac {3 \, {\left (b x^{2} + a\right )} B a + B a^{2} - 6 \, {\left (b x^{2} + a\right )} A b - A a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x^{2} + a} A}{2 \, a^{3} x^{2}} \]

[In]

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

[Out]

1/2*(2*B*a - 5*A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 1/3*(3*(b*x^2 + a)*B*a + B*a^2 - 6*(b*x^
2 + a)*A*b - A*a*b)/((b*x^2 + a)^(3/2)*a^3) - 1/2*sqrt(b*x^2 + a)*A/(a^3*x^2)

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {\frac {B}{3\,a}+\frac {B\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {10\,A\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {5\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}-\frac {5\,A\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \]

[In]

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

[Out]

(B/(3*a) + (B*(a + b*x^2))/a^2)/(a + b*x^2)^(3/2) - (B*atanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(5/2) - (10*A*b)/(3
*a^2*(a + b*x^2)^(3/2)) - A/(2*a*x^2*(a + b*x^2)^(3/2)) + (5*A*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(7/2))
- (5*A*b^2*x^2)/(2*a^3*(a + b*x^2)^(3/2))