\(\int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\) [137]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {1}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}+\frac {1}{a^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b x} \sqrt {a c-b c x}}{a \sqrt {c}}\right )}{a^5 c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {4 a^3-3 a b^2 x^2-6 \sqrt {a-b x} \sqrt {a+b x} \left (a^2-b^2 x^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{3 a^5 c (c (a-b x))^{3/2} (a+b x)^{3/2}} \] Input:

Integrate[1/(x*(a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

(4*a^3 - 3*a*b^2*x^2 - 6*Sqrt[a - b*x]*Sqrt[a + b*x]*(a^2 - b^2*x^2)*ArcTa 
nh[Sqrt[a + b*x]/Sqrt[a - b*x]])/(3*a^5*c*(c*(a - b*x))^(3/2)*(a + b*x)^(3 
/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {115, 27, 35, 115, 27, 35, 103, 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.

\(\displaystyle \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\)

\(\Big \downarrow \) 115

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 35

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

\(\Big \downarrow \) 115

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 35

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

\(\Big \downarrow \) 103

\(\displaystyle \frac {\frac {b \int \frac {1}{b (a+b x) (a c-b c x)-a^2 b c}d\left (\sqrt {a+b x} \sqrt {a c-b c x}\right )}{a^2 c}+\frac {1}{a^2 c \sqrt {a+b x} \sqrt {a c-b c x}}}{a^2 c}+\frac {1}{3 a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}\)

\(\Big \downarrow \) 221

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

Input:

Int[1/(x*(a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

1/(3*a^2*c*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)) + (1/(a^2*c*Sqrt[a + b*x]* 
Sqrt[a*c - b*c*x]) - ArcTanh[(Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/(a*Sqrt[c]) 
]/(a^3*c^(3/2)))/(a^2*c)
 

Defintions of rubi rules used

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 35

Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} 
, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] &&  !(IntegerQ[n] && SimplerQ[a + 
b*x, c + d*x])
 

rule 103

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ 
))), x_] :> Simp[b*f   Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq 
rt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[2*b*d 
*e - f*(b*c + a*d), 0]
 

rule 115

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2 
*n, 2*p]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(251\) vs. \(2(86)=172\).

Time = 0.29 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.42

method	result	size
default	\(-\frac {\sqrt {c \left (-b x +a \right )}\, \left (3 \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}{x}\right ) b^{4} c \,x^{4}-6 \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}{x}\right ) a^{2} b^{2} c \,x^{2}+3 \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}{x}\right ) a^{4} c +3 b^{2} x^{2} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {a^{2} c}-4 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {a^{2} c}\, a^{2}\right )}{3 c^{3} a^{4} \left (-b x +a \right )^{2} \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \left (b x +a \right )^{\frac {3}{2}}}\)	\(252\)

Input:

int(1/x/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*(c*(-b*x+a))^(1/2)/c^3/a^4*(3*ln(2*(a^2*c+(a^2*c)^(1/2)*(c*(-b^2*x^2+ 
a^2))^(1/2))/x)*b^4*c*x^4-6*ln(2*(a^2*c+(a^2*c)^(1/2)*(c*(-b^2*x^2+a^2))^( 
1/2))/x)*a^2*b^2*c*x^2+3*ln(2*(a^2*c+(a^2*c)^(1/2)*(c*(-b^2*x^2+a^2))^(1/2 
))/x)*a^4*c+3*b^2*x^2*(c*(-b^2*x^2+a^2))^(1/2)*(a^2*c)^(1/2)-4*(c*(-b^2*x^ 
2+a^2))^(1/2)*(a^2*c)^(1/2)*a^2)/(-b*x+a)^2/(a^2*c)^(1/2)/(c*(-b^2*x^2+a^2 
))^(1/2)/(b*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.83 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \sqrt {c} \log \left (-\frac {b^{2} c x^{2} - 2 \, a^{2} c + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {c}}{x^{2}}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} - 4 \, a^{3}\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{6 \, {\left (a^{5} b^{4} c^{3} x^{4} - 2 \, a^{7} b^{2} c^{3} x^{2} + a^{9} c^{3}\right )}}, -\frac {3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {-c}}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (3 \, a b^{2} x^{2} - 4 \, a^{3}\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{3 \, {\left (a^{5} b^{4} c^{3} x^{4} - 2 \, a^{7} b^{2} c^{3} x^{2} + a^{9} c^{3}\right )}}\right ] \] Input:

integrate(1/x/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*sqrt(c)*log(-(b^2*c*x^2 - 2*a^2*c 
+ 2*sqrt(-b*c*x + a*c)*sqrt(b*x + a)*a*sqrt(c))/x^2) - 2*(3*a*b^2*x^2 - 4* 
a^3)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(a^5*b^4*c^3*x^4 - 2*a^7*b^2*c^3*x^ 
2 + a^9*c^3), -1/3*(3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*sqrt(-c)*arctan(sqrt 
(-b*c*x + a*c)*sqrt(b*x + a)*a*sqrt(-c)/(b^2*c*x^2 - a^2*c)) + (3*a*b^2*x^ 
2 - 4*a^3)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(a^5*b^4*c^3*x^4 - 2*a^7*b^2* 
c^3*x^2 + a^9*c^3)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(1/x/(b*x+a)**(5/2)/(-b*c*x+a*c)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {1}{3 \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{2} c} + \frac {1}{\sqrt {-b^{2} c x^{2} + a^{2} c} a^{4} c^{2}} - \frac {\log \left (\frac {2 \, a^{2} c}{{\left | x \right |}} + \frac {2 \, \sqrt {-b^{2} c x^{2} + a^{2} c} a \sqrt {c}}{{\left | x \right |}}\right )}{a^{5} c^{\frac {5}{2}}} \] Input:

integrate(1/x/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="maxima")
 

Output:

1/3/((-b^2*c*x^2 + a^2*c)^(3/2)*a^2*c) + 1/(sqrt(-b^2*c*x^2 + a^2*c)*a^4*c 
^2) - log(2*a^2*c/abs(x) + 2*sqrt(-b^2*c*x^2 + a^2*c)*a*sqrt(c)/abs(x))/(a 
^5*c^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (86) = 172\).

Time = 0.19 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.40 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=-\frac {\sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a} {\left (\frac {7 \, {\left (b x + a\right )}}{a^{5} c} - \frac {15}{a^{4} c}\right )}}{12 \, {\left ({\left (b x + a\right )} c - 2 \, a c\right )}^{2}} + \frac {2 \, \arctan \left (\frac {{\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}}{2 \, a c}\right )}{a^{5} \sqrt {-c} c^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} - 15 \, a {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} c + 14 \, a^{2} c^{2}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} - 2 \, a c\right )}^{3} a^{4} \sqrt {-c} c} \] Input:

integrate(1/x/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x, algorithm="giac")
 

Output:

-1/12*sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)*(7*(b*x + a)/(a^5*c) - 15/( 
a^4*c))/((b*x + a)*c - 2*a*c)^2 + 2*arctan(1/2*(sqrt(b*x + a)*sqrt(-c) - s 
qrt(-(b*x + a)*c + 2*a*c))^2/(a*c))/(a^5*sqrt(-c)*c^2) + 2/3*(3*(sqrt(b*x 
+ a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 - 15*a*(sqrt(b*x + a)*sqrt(- 
c) - sqrt(-(b*x + a)*c + 2*a*c))^2*c + 14*a^2*c^2)/(((sqrt(b*x + a)*sqrt(- 
c) - sqrt(-(b*x + a)*c + 2*a*c))^2 - 2*a*c)^3*a^4*sqrt(-c)*c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\int \frac {1}{x\,{\left (a\,c-b\,c\,x\right )}^{5/2}\,{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:

int(1/(x*(a*c - b*c*x)^(5/2)*(a + b*x)^(5/2)),x)
 

Output:

int(1/(x*(a*c - b*c*x)^(5/2)*(a + b*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 420, normalized size of antiderivative = 4.04 \[ \int \frac {1}{x (a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) a^{2}+3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) b^{2} x^{2}+3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) a^{2}-3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) b^{2} x^{2}-3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) a^{2}+3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) b^{2} x^{2}+3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) a^{2}-3 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) b^{2} x^{2}+4 a^{3}-3 a \,b^{2} x^{2}\right )}{3 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{5} c^{3} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:

int(1/x/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x)
 

Output:

(sqrt(c)*( - 3*sqrt(a + b*x)*sqrt(a - b*x)*log( - sqrt(2) + tan(asin(sqrt( 
a - b*x)/(sqrt(a)*sqrt(2)))/2) - 1)*a**2 + 3*sqrt(a + b*x)*sqrt(a - b*x)*l 
og( - sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2) - 1)*b**2*x** 
2 + 3*sqrt(a + b*x)*sqrt(a - b*x)*log( - sqrt(2) + tan(asin(sqrt(a - b*x)/ 
(sqrt(a)*sqrt(2)))/2) + 1)*a**2 - 3*sqrt(a + b*x)*sqrt(a - b*x)*log( - sqr 
t(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2) + 1)*b**2*x**2 - 3*sqr 
t(a + b*x)*sqrt(a - b*x)*log(sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqr 
t(2)))/2) - 1)*a**2 + 3*sqrt(a + b*x)*sqrt(a - b*x)*log(sqrt(2) + tan(asin 
(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2) - 1)*b**2*x**2 + 3*sqrt(a + b*x)*sqrt 
(a - b*x)*log(sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))/2) + 1)* 
a**2 - 3*sqrt(a + b*x)*sqrt(a - b*x)*log(sqrt(2) + tan(asin(sqrt(a - b*x)/ 
(sqrt(a)*sqrt(2)))/2) + 1)*b**2*x**2 + 4*a**3 - 3*a*b**2*x**2))/(3*sqrt(a 
+ b*x)*sqrt(a - b*x)*a**5*c**3*(a**2 - b**2*x**2))