\(\int \frac {A+B x}{x^3 (a+b x^2)^{5/2}} \, dx\) [1265]

Optimal result

Integrand size = 20, antiderivative size = 123 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac {b (6 A+5 B x)}{3 a^3 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {B \sqrt {a+b x^2}}{a^3 x}+\frac {5 A b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 a^2 (A+2 B x)-4 a b x^2 (5 A+6 B x)-b^2 x^4 (15 A+16 B x)}{6 a^3 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:

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

Output:

(-3*a^2*(A + 2*B*x) - 4*a*b*x^2*(5*A + 6*B*x) - b^2*x^4*(15*A + 16*B*x))/( 
6*a^3*x^2*(a + b*x^2)^(3/2)) - (5*A*b*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2] 
)/Sqrt[a]])/a^(7/2)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {532, 25, 2336, 27, 2338, 25, 534, 243, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 532

\(\displaystyle -\frac {\int -\frac {-\frac {2 b B x^3}{a}-\frac {3 A b x^2}{a}+3 B x+3 A}{x^3 \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {-\frac {2 b B x^3}{a}-\frac {3 A b x^2}{a}+3 B x+3 A}{x^3 \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {3 \left (-\frac {2 A b x^2}{a}+B x+A\right )}{x^3 \sqrt {b x^2+a}}dx}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {-\frac {2 A b x^2}{a}+B x+A}{x^3 \sqrt {b x^2+a}}dx}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\frac {3 \left (-\frac {\int -\frac {2 a B-5 A b x}{x^2 \sqrt {b x^2+a}}dx}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {2 a B-5 A b x}{x^2 \sqrt {b x^2+a}}dx}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {3 \left (\frac {-5 A b \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {2 B \sqrt {a+b x^2}}{x}}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {3 \left (\frac {-\frac {5}{2} A b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {2 B \sqrt {a+b x^2}}{x}}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {3 \left (\frac {-5 A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {2 B \sqrt {a+b x^2}}{x}}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {5 A b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 B \sqrt {a+b x^2}}{x}}{2 a}-\frac {A \sqrt {a+b x^2}}{2 a x^2}\right )}{a}-\frac {b (6 A+5 B x)}{a^2 \sqrt {a+b x^2}}}{3 a}-\frac {b (A+B x)}{3 a^2 \left (a+b x^2\right )^{3/2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.19

method	result	size
default	\(A \left (-\frac {1}{2 a \,x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 b \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )}{2 a}\right )+B \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{a}\right )\)	\(146\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 a^{3} x^{2}}+\frac {5 b A \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {7}{2}}}-\frac {13 b \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {5 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{6 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {13 b \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {5 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{6 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}\)	\(565\)

Input:

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

Output:

A*(-1/2/a/x^2/(b*x^2+a)^(3/2)-5/2*b/a*(1/3/a/(b*x^2+a)^(3/2)+1/a*(1/a/(b*x 
^2+a)^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))))+B*(-1/a/x/( 
b*x^2+a)^(3/2)-4*b/a*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.52 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{4} + A a^{2} b x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (16 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} + 24 \, B a^{2} b x^{3} + 20 \, A a^{2} b x^{2} + 6 \, B a^{3} x + 3 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, -\frac {15 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{4} + A a^{2} b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (16 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} + 24 \, B a^{2} b x^{3} + 20 \, A a^{2} b x^{2} + 6 \, B a^{3} x + 3 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/12*(15*(A*b^3*x^6 + 2*A*a*b^2*x^4 + 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*(16*B*a*b^2*x^5 + 15*A*a*b^2*x^4 
 + 24*B*a^2*b*x^3 + 20*A*a^2*b*x^2 + 6*B*a^3*x + 3*A*a^3)*sqrt(b*x^2 + a)) 
/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), -1/6*(15*(A*b^3*x^6 + 2*A*a*b^2*x^ 
4 + A*a^2*b*x^2)*sqrt(-a)*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (16*B*a*b^2 
*x^5 + 15*A*a*b^2*x^4 + 24*B*a^2*b*x^3 + 20*A*a^2*b*x^2 + 6*B*a^3*x + 3*A* 
a^3)*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. 1034 vs. \(2 (112) = 224\).

Time = 8.20 (sec) , antiderivative size = 1034, normalized size of antiderivative = 8.41 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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 + 3 
6*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**(3 
7/2)*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*...
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {8 \, B b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, B b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {5 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} - \frac {5 \, A b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, A b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {B}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} - \frac {A}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} \] Input:

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

Output:

-8/3*B*b*x/(sqrt(b*x^2 + a)*a^3) - 4/3*B*b*x/((b*x^2 + a)^(3/2)*a^2) + 5/2 
*A*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 5/2*A*b/(sqrt(b*x^2 + a)*a^3) 
 - 5/6*A*b/((b*x^2 + a)^(3/2)*a^2) - B/((b*x^2 + a)^(3/2)*a*x) - 1/2*A/((b 
*x^2 + a)^(3/2)*a*x^2)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.60 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left ({\left (\frac {5 \, B b^{2} x}{a^{3}} + \frac {6 \, A b^{2}}{a^{3}}\right )} x + \frac {6 \, B b}{a^{2}}\right )} x + \frac {7 \, A b}{a^{2}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {5 \, A b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {B\,a^2-8\,B\,{\left (b\,x^2+a\right )}^2+4\,B\,a\,\left (b\,x^2+a\right )}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {10\,A\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {5\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}-\frac {5\,A\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \] Input:

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

Output:

(B*a^2 - 8*B*(a + b*x^2)^2 + 4*B*a*(a + b*x^2))/(3*a^3*x*(a + b*x^2)^(3/2) 
) - (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))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.86 \[ \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{3}-20 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}-6 \sqrt {b \,x^{2}+a}\, a^{2} b x -15 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}-24 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}-16 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,x^{2}-30 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x^{4}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,x^{2}+30 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}+16 \sqrt {b}\, a^{2} b \,x^{2}+32 \sqrt {b}\, a \,b^{2} x^{4}+16 \sqrt {b}\, b^{3} x^{6}}{6 a^{3} x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

( - 3*sqrt(a + b*x**2)*a**3 - 20*sqrt(a + b*x**2)*a**2*b*x**2 - 6*sqrt(a + 
 b*x**2)*a**2*b*x - 15*sqrt(a + b*x**2)*a*b**2*x**4 - 24*sqrt(a + b*x**2)* 
a*b**2*x**3 - 16*sqrt(a + b*x**2)*b**3*x**5 - 15*sqrt(a)*log((sqrt(a + b*x 
**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*x**2 - 30*sqrt(a)*log((sqrt(a 
+ b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*x**4 - 15*sqrt(a)*log((sq 
rt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**6 + 15*sqrt(a)*log( 
(sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*x**2 + 30*sqrt(a) 
*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*x**4 + 15*sq 
rt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**6 + 16 
*sqrt(b)*a**2*b*x**2 + 32*sqrt(b)*a*b**2*x**4 + 16*sqrt(b)*b**3*x**6)/(6*a 
**3*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4))