3.5.52 \(\int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx\) [452]

3.5.52.1 Optimal result
3.5.52.2 Mathematica [A] (verified)
3.5.52.3 Rubi [A] (verified)
3.5.52.4 Maple [A] (verified)
3.5.52.5 Fricas [A] (verification not implemented)
3.5.52.6 Sympy [F(-1)]
3.5.52.7 Maxima [A] (verification not implemented)
3.5.52.8 Giac [A] (verification not implemented)
3.5.52.9 Mupad [B] (verification not implemented)

3.5.52.1 Optimal result

Integrand size = 18, antiderivative size = 171 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=-\frac {35 b^2 (3 A b-2 a B)}{24 a^4 (a+b x)^{3/2}}-\frac {A}{3 a x^3 (a+b x)^{3/2}}+\frac {3 A b-2 a B}{4 a^2 x^2 (a+b x)^{3/2}}-\frac {7 b (3 A b-2 a B)}{8 a^3 x (a+b x)^{3/2}}-\frac {35 b^2 (3 A b-2 a B)}{8 a^5 \sqrt {a+b x}}+\frac {35 b^2 (3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}} \]

output
-35/24*b^2*(3*A*b-2*B*a)/a^4/(b*x+a)^(3/2)-1/3*A/a/x^3/(b*x+a)^(3/2)+1/4*( 
3*A*b-2*B*a)/a^2/x^2/(b*x+a)^(3/2)-7/8*b*(3*A*b-2*B*a)/a^3/x/(b*x+a)^(3/2) 
+35/8*b^2*(3*A*b-2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(11/2)-35/8*b^2*( 
3*A*b-2*B*a)/a^5/(b*x+a)^(1/2)
 
3.5.52.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {-315 A b^4 x^4+210 a b^3 x^3 (-2 A+B x)-4 a^4 (2 A+3 B x)+6 a^3 b x (3 A+7 B x)+7 a^2 b^2 x^2 (-9 A+40 B x)}{24 a^5 x^3 (a+b x)^{3/2}}+\frac {35 b^2 (3 A b-2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{11/2}} \]

input
Integrate[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]
 
output
(-315*A*b^4*x^4 + 210*a*b^3*x^3*(-2*A + B*x) - 4*a^4*(2*A + 3*B*x) + 6*a^3 
*b*x*(3*A + 7*B*x) + 7*a^2*b^2*x^2*(-9*A + 40*B*x))/(24*a^5*x^3*(a + b*x)^ 
(3/2)) + (35*b^2*(3*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(11/ 
2))
 
3.5.52.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {87, 52, 52, 61, 61, 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.

\(\displaystyle \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \int \frac {1}{x^2 (a+b x)^{5/2}}dx}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \left (-\frac {5 b \int \frac {1}{x (a+b x)^{5/2}}dx}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\right )}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \left (-\frac {5 b \left (\frac {\int \frac {1}{x (a+b x)^{3/2}}dx}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\right )}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \left (-\frac {5 b \left (\frac {\frac {\int \frac {1}{x \sqrt {a+b x}}dx}{a}+\frac {2}{a \sqrt {a+b x}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\right )}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \left (-\frac {5 b \left (\frac {\frac {2 \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a b}+\frac {2}{a \sqrt {a+b x}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\right )}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(3 A b-2 a B) \left (-\frac {7 b \left (-\frac {5 b \left (\frac {\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}}{a}+\frac {2}{3 a (a+b x)^{3/2}}\right )}{2 a}-\frac {1}{a x (a+b x)^{3/2}}\right )}{4 a}-\frac {1}{2 a x^2 (a+b x)^{3/2}}\right )}{2 a}-\frac {A}{3 a x^3 (a+b x)^{3/2}}\)

input
Int[(A + B*x)/(x^4*(a + b*x)^(5/2)),x]
 
output
-1/3*A/(a*x^3*(a + b*x)^(3/2)) - ((3*A*b - 2*a*B)*(-1/2*1/(a*x^2*(a + b*x) 
^(3/2)) - (7*b*(-(1/(a*x*(a + b*x)^(3/2))) - (5*b*(2/(3*a*(a + b*x)^(3/2)) 
 + (2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2))/a))/ 
(2*a)))/(4*a)))/(2*a)
 

3.5.52.3.1 Defintions of rubi rules used

rule 52
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61
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] - Simp[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] && 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 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 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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]
 
3.5.52.4 Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.71

method result size
pseudoelliptic \(\frac {\frac {105 x^{3} \left (b x +a \right )^{\frac {3}{2}} b^{2} \left (A b -\frac {2 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8}-\frac {35 x^{3} \left (-\frac {B x}{2}+A \right ) b^{3} a^{\frac {3}{2}}}{2}-\frac {21 x^{2} \left (-\frac {40 B x}{9}+A \right ) b^{2} a^{\frac {5}{2}}}{8}+\frac {3 x \left (\frac {7 B x}{3}+A \right ) b \,a^{\frac {7}{2}}}{4}+\frac {3 \left (-\frac {2 B x}{3}-\frac {4 A}{9}\right ) a^{\frac {9}{2}}}{4}-\frac {105 A \sqrt {a}\, b^{4} x^{4}}{8}}{a^{\frac {11}{2}} \left (b x +a \right )^{\frac {3}{2}} x^{3}}\) \(122\)
risch \(-\frac {\sqrt {b x +a}\, \left (123 A \,b^{2} x^{2}-66 B a b \,x^{2}-34 a A b x +12 a^{2} B x +8 a^{2} A \right )}{24 a^{5} x^{3}}-\frac {b^{2} \left (-\frac {2 \left (-64 A b +48 B a \right )}{\sqrt {b x +a}}+\frac {32 a \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}-\frac {2 \left (105 A b -70 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{16 a^{5}}\) \(125\)
derivativedivides \(2 b^{2} \left (\frac {-\frac {\left (\frac {41 A b}{16}-\frac {11 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a b A +3 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {55}{16} a^{2} b A -\frac {13}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {35 \left (3 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{5}}-\frac {A b -B a}{3 a^{4} \left (b x +a \right )^{\frac {3}{2}}}-\frac {4 A b -3 B a}{a^{5} \sqrt {b x +a}}\right )\) \(147\)
default \(2 b^{2} \left (\frac {-\frac {\left (\frac {41 A b}{16}-\frac {11 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {35}{6} a b A +3 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {55}{16} a^{2} b A -\frac {13}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {35 \left (3 A b -2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{5}}-\frac {A b -B a}{3 a^{4} \left (b x +a \right )^{\frac {3}{2}}}-\frac {4 A b -3 B a}{a^{5} \sqrt {b x +a}}\right )\) \(147\)

input
int((B*x+A)/x^4/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 
output
3/4/a^(11/2)*(35/2*x^3*(b*x+a)^(3/2)*b^2*(A*b-2/3*B*a)*arctanh((b*x+a)^(1/ 
2)/a^(1/2))-70/3*x^3*(-1/2*B*x+A)*b^3*a^(3/2)-7/2*x^2*(-40/9*B*x+A)*b^2*a^ 
(5/2)+x*(7/3*B*x+A)*b*a^(7/2)+(-2/3*B*x-4/9*A)*a^(9/2)-35/2*A*a^(1/2)*b^4* 
x^4)/(b*x+a)^(3/2)/x^3
 
3.5.52.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.62 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\left [-\frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{48 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}, \frac {105 \, {\left ({\left (2 \, B a b^{4} - 3 \, A b^{5}\right )} x^{5} + 2 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} + {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (8 \, A a^{5} - 105 \, {\left (2 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{4} - 140 \, {\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{3} - 21 \, {\left (2 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 6 \, {\left (2 \, B a^{5} - 3 \, A a^{4} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}}\right ] \]

input
integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="fricas")
 
output
[-1/48*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + 
 (2*B*a^3*b^2 - 3*A*a^2*b^3)*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt( 
a) + 2*a)/x) + 2*(8*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 - 140*(2*B*a 
^3*b^2 - 3*A*a^2*b^3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 
- 3*A*a^4*b)*x)*sqrt(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3), 1/24 
*(105*((2*B*a*b^4 - 3*A*b^5)*x^5 + 2*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 + (2*B* 
a^3*b^2 - 3*A*a^2*b^3)*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8 
*A*a^5 - 105*(2*B*a^2*b^3 - 3*A*a*b^4)*x^4 - 140*(2*B*a^3*b^2 - 3*A*a^2*b^ 
3)*x^3 - 21*(2*B*a^4*b - 3*A*a^3*b^2)*x^2 + 6*(2*B*a^5 - 3*A*a^4*b)*x)*sqr 
t(b*x + a))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3)]
 
3.5.52.6 Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\text {Timed out} \]

input
integrate((B*x+A)/x**4/(b*x+a)**(5/2),x)
 
output
Timed out
 
3.5.52.7 Maxima [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=-\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (16 \, B a^{5} - 16 \, A a^{4} b - 105 \, {\left (2 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{4} + 280 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} {\left (b x + a\right )}^{3} - 231 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{2} + 48 \, {\left (2 \, B a^{4} - 3 \, A a^{3} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {9}{2}} a^{5} b - 3 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{6} b + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{7} b - {\left (b x + a\right )}^{\frac {3}{2}} a^{8} b} - \frac {105 \, {\left (2 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {11}{2}} b}\right )} \]

input
integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="maxima")
 
output
-1/48*b^3*(2*(16*B*a^5 - 16*A*a^4*b - 105*(2*B*a - 3*A*b)*(b*x + a)^4 + 28 
0*(2*B*a^2 - 3*A*a*b)*(b*x + a)^3 - 231*(2*B*a^3 - 3*A*a^2*b)*(b*x + a)^2 
+ 48*(2*B*a^4 - 3*A*a^3*b)*(b*x + a))/((b*x + a)^(9/2)*a^5*b - 3*(b*x + a) 
^(7/2)*a^6*b + 3*(b*x + a)^(5/2)*a^7*b - (b*x + a)^(3/2)*a^8*b) - 105*(2*B 
*a - 3*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(1 
1/2)*b))
 
3.5.52.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {35 \, {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{5}} + \frac {210 \, {\left (b x + a\right )}^{4} B a b^{2} - 560 \, {\left (b x + a\right )}^{3} B a^{2} b^{2} + 462 \, {\left (b x + a\right )}^{2} B a^{3} b^{2} - 96 \, {\left (b x + a\right )} B a^{4} b^{2} - 16 \, B a^{5} b^{2} - 315 \, {\left (b x + a\right )}^{4} A b^{3} + 840 \, {\left (b x + a\right )}^{3} A a b^{3} - 693 \, {\left (b x + a\right )}^{2} A a^{2} b^{3} + 144 \, {\left (b x + a\right )} A a^{3} b^{3} + 16 \, A a^{4} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )}^{3} a^{5}} \]

input
integrate((B*x+A)/x^4/(b*x+a)^(5/2),x, algorithm="giac")
 
output
35/8*(2*B*a*b^2 - 3*A*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^5) + 
 1/24*(210*(b*x + a)^4*B*a*b^2 - 560*(b*x + a)^3*B*a^2*b^2 + 462*(b*x + a) 
^2*B*a^3*b^2 - 96*(b*x + a)*B*a^4*b^2 - 16*B*a^5*b^2 - 315*(b*x + a)^4*A*b 
^3 + 840*(b*x + a)^3*A*a*b^3 - 693*(b*x + a)^2*A*a^2*b^3 + 144*(b*x + a)*A 
*a^3*b^3 + 16*A*a^4*b^3)/(((b*x + a)^(3/2) - sqrt(b*x + a)*a)^3*a^5)
 
3.5.52.9 Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{x^4 (a+b x)^{5/2}} \, dx=\frac {35\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-2\,B\,a\right )}{8\,a^{11/2}}-\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{3\,a}+\frac {2\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,\left (a+b\,x\right )}{a^2}-\frac {77\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^2}{8\,a^3}+\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^3}{3\,a^4}-\frac {35\,\left (3\,A\,b^3-2\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^4}{8\,a^5}}{3\,a\,{\left (a+b\,x\right )}^{7/2}-{\left (a+b\,x\right )}^{9/2}+a^3\,{\left (a+b\,x\right )}^{3/2}-3\,a^2\,{\left (a+b\,x\right )}^{5/2}} \]

input
int((A + B*x)/(x^4*(a + b*x)^(5/2)),x)
 
output
(35*b^2*atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b - 2*B*a))/(8*a^(11/2)) - ((2 
*(A*b^3 - B*a*b^2))/(3*a) + (2*(3*A*b^3 - 2*B*a*b^2)*(a + b*x))/a^2 - (77* 
(3*A*b^3 - 2*B*a*b^2)*(a + b*x)^2)/(8*a^3) + (35*(3*A*b^3 - 2*B*a*b^2)*(a 
+ b*x)^3)/(3*a^4) - (35*(3*A*b^3 - 2*B*a*b^2)*(a + b*x)^4)/(8*a^5))/(3*a*( 
a + b*x)^(7/2) - (a + b*x)^(9/2) + a^3*(a + b*x)^(3/2) - 3*a^2*(a + b*x)^( 
5/2))