\(\int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx\) [274]

Optimal result

Integrand size = 18, antiderivative size = 107 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {2 b (A b-a B)}{a^3 \sqrt {a+b x}}-\frac {A \sqrt {a+b x}}{2 a^2 x^2}+\frac {(7 A b-4 a B) \sqrt {a+b x}}{4 a^3 x}-\frac {3 b (5 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 A b^2 x^2+a b x (5 A-12 B x)-2 a^2 (A+2 B x)}{4 a^3 x^2 \sqrt {a+b x}}+\frac {3 b (-5 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \] Input:

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

Output:

(15*A*b^2*x^2 + a*b*x*(5*A - 12*B*x) - 2*a^2*(A + 2*B*x))/(4*a^3*x^2*Sqrt[ 
a + b*x]) + (3*b*(-5*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(7/ 
2))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {87, 52, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {b \left (-\frac {\sqrt {b x +a}\, \left (-7 A b x +4 B a x +2 A a \right )}{4 b \,x^{2}}-\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {2 A b -2 B a}{\sqrt {b x +a}}\right )}{a^{3}}\)	\(80\)
risch	\(-\frac {\sqrt {b x +a}\, \left (-7 A b x +4 B a x +2 A a \right )}{4 a^{3} x^{2}}+\frac {b \left (-\frac {2 \left (-8 A b +8 B a \right )}{\sqrt {b x +a}}-\frac {2 \left (15 A b -12 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{8 a^{3}}\)	\(83\)
derivativedivides	\(2 b \left (-\frac {-A b +B a}{a^{3} \sqrt {b x +a}}-\frac {\frac {\left (-\frac {7 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {9}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\)	\(102\)
default	\(2 b \left (-\frac {-A b +B a}{a^{3} \sqrt {b x +a}}-\frac {\frac {\left (-\frac {7 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {9}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\)	\(102\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.54 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, A a^{3} + 3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, -\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (2 \, A a^{3} + 3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \] Input:

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

Output:

[-1/8*(3*((4*B*a*b^2 - 5*A*b^3)*x^3 + (4*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(a) 
*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(2*A*a^3 + 3*(4*B*a^2*b 
- 5*A*a*b^2)*x^2 + (4*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x + a))/(a^4*b*x^3 + a^ 
5*x^2), -1/4*(3*((4*B*a*b^2 - 5*A*b^3)*x^3 + (4*B*a^2*b - 5*A*a*b^2)*x^2)* 
sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x + a)) + (2*A*a^3 + 3*(4*B*a^2*b - 5*A*a* 
b^2)*x^2 + (4*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2)]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 51.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.73 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=A \left (- \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}}\right ) + B \left (- \frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}}\right ) \] Input:

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

Output:

A*(-1/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2 
)*sqrt(a/(b*x) + 1)) + 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) - 15 
*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2))) + B*(-1/(a*sqrt(b)*x* 
*(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3 
*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(5/2))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=-\frac {1}{8} \, b^{2} {\left (\frac {2 \, {\left (8 \, B a^{3} - 8 \, A a^{2} b + 3 \, {\left (4 \, B a - 5 \, A b\right )} {\left (b x + a\right )}^{2} - 5 \, {\left (4 \, B a^{2} - 5 \, A a b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {5}{2}} a^{3} b - 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b + \sqrt {b x + a} a^{5} b} + \frac {3 \, {\left (4 \, B a - 5 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} - \frac {2 \, {\left (B a b - A b^{2}\right )}}{\sqrt {b x + a} a^{3}} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x + a} B a^{2} b - 7 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{2} + 9 \, \sqrt {b x + a} A a b^{2}}{4 \, a^{3} b^{2} x^{2}} \] Input:

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

Output:

-3/4*(4*B*a*b - 5*A*b^2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) - 2 
*(B*a*b - A*b^2)/(sqrt(b*x + a)*a^3) - 1/4*(4*(b*x + a)^(3/2)*B*a*b - 4*sq 
rt(b*x + a)*B*a^2*b - 7*(b*x + a)^(3/2)*A*b^2 + 9*sqrt(b*x + a)*A*a*b^2)/( 
a^3*b^2*x^2)
 

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {\frac {2\,\left (A\,b^2-B\,a\,b\right )}{a}-\frac {5\,\left (5\,A\,b^2-4\,B\,a\,b\right )\,\left (a+b\,x\right )}{4\,a^2}+\frac {3\,\left (5\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^2}{4\,a^3}}{{\left (a+b\,x\right )}^{5/2}-2\,a\,{\left (a+b\,x\right )}^{3/2}+a^2\,\sqrt {a+b\,x}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (5\,A\,b-4\,B\,a\right )}{4\,a^{7/2}} \] Input:

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

Output:

((2*(A*b^2 - B*a*b))/a - (5*(5*A*b^2 - 4*B*a*b)*(a + b*x))/(4*a^2) + (3*(5 
*A*b^2 - 4*B*a*b)*(a + b*x)^2)/(4*a^3))/((a + b*x)^(5/2) - 2*a*(a + b*x)^( 
3/2) + a^2*(a + b*x)^(1/2)) - (3*b*atanh((a + b*x)^(1/2)/a^(1/2))*(5*A*b - 
 4*B*a))/(4*a^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2}+6 \sqrt {b x +a}\, a b x +3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} x^{2}}{8 a^{3} x^{2}} \] Input:

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

Output:

( - 4*sqrt(a + b*x)*a**2 + 6*sqrt(a + b*x)*a*b*x + 3*sqrt(a)*log(sqrt(a + 
b*x) - sqrt(a))*b**2*x**2 - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*x* 
*2)/(8*a**3*x**2)