\(\int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx\) [99]

Optimal result

Integrand size = 26, antiderivative size = 105 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=-\frac {3}{2} b^2 c \sqrt {a+b x} \sqrt {a c-b c x}-\frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{2 x^2}+\frac {3}{2} a b^2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x} \sqrt {a c-b c x}}{a \sqrt {c}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\frac {1}{2} c \sqrt {c (a-b x)} \left (-\frac {\sqrt {a+b x} \left (a^2+2 b^2 x^2\right )}{x^2}+\frac {6 a b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{\sqrt {a-b x}}\right ) \] Input:

Integrate[((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/x^3,x]
 

Output:

(c*Sqrt[c*(a - b*x)]*(-((Sqrt[a + b*x]*(a^2 + 2*b^2*x^2))/x^2) + (6*a*b^2* 
ArcTanh[Sqrt[a + b*x]/Sqrt[a - b*x]])/Sqrt[a - b*x]))/2
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {108, 27, 112, 25, 27, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 112

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 103

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2))/x^3,x]
 

Output:

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

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 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 108

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 112

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + 
p + 1))), x] - Simp[1/(f*(m + n + p + 1))   Int[(a + b*x)^(m - 1)*(c + d*x) 
^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a 
*f) + b*n*(d*e - c*f))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && 
GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (IntegersQ[2*m, 2*n, 2*p 
] || (IntegersQ[m, n + p] || IntegersQ[p, m + 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 = 155, normalized size of antiderivative = 1.48

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, c \left (3 \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}{x}\right ) a^{2} b^{2} c \,x^{2}-2 b^{2} x^{2} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {a^{2} c}-\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {a^{2} c}\, a^{2}\right )}{2 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, x^{2} \sqrt {a^{2} c}}\)	\(155\)
risch	\(-\frac {a^{2} \left (-b x +a \right ) \sqrt {b x +a}\, c^{2}}{2 x^{2} \sqrt {-c \left (b x -a \right )}}+\frac {\left (-\frac {b^{2} \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}}{c}+\frac {3 b^{2} a^{2} \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {-b^{2} c \,x^{2}+a^{2} c}}{x}\right )}{2 \sqrt {a^{2} c}}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}\, c^{2}}{\sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}}\)	\(157\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.89 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\left [\frac {3 \, a b^{2} c^{\frac {3}{2}} x^{2} \log \left (-\frac {b^{2} c x^{2} - 2 \, a^{2} c - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {c}}{x^{2}}\right ) - 2 \, {\left (2 \, b^{2} c x^{2} + a^{2} c\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{4 \, x^{2}}, \frac {3 \, a b^{2} \sqrt {-c} c x^{2} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {-c}}{b^{2} c x^{2} - a^{2} c}\right ) - {\left (2 \, b^{2} c x^{2} + a^{2} c\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{2 \, x^{2}}\right ] \] Input:

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

Output:

[1/4*(3*a*b^2*c^(3/2)*x^2*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*(2*b^2*c*x^2 + a^2*c)*sqrt(-b*c*x + a*c 
)*sqrt(b*x + a))/x^2, 1/2*(3*a*b^2*sqrt(-c)*c*x^2*arctan(sqrt(-b*c*x + a*c 
)*sqrt(b*x + a)*a*sqrt(-c)/(b^2*c*x^2 - a^2*c)) - (2*b^2*c*x^2 + a^2*c)*sq 
rt(-b*c*x + a*c)*sqrt(b*x + a))/x^2]
 

Sympy [F]

\[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\int \frac {\left (- c \left (- a + b x\right )\right )^{\frac {3}{2}} \left (a + b x\right )^{\frac {3}{2}}}{x^{3}}\, dx \] Input:

integrate((b*x+a)**(3/2)*(-b*c*x+a*c)**(3/2)/x**3,x)
 

Output:

Integral((-c*(-a + b*x))**(3/2)*(a + b*x)**(3/2)/x**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\frac {3}{2} \, a b^{2} c^{\frac {3}{2}} \log \left (\frac {2 \, a^{2} c}{{\left | x \right |}} + \frac {2 \, \sqrt {-b^{2} c x^{2} + a^{2} c} a \sqrt {c}}{{\left | x \right |}}\right ) - \frac {3}{2} \, \sqrt {-b^{2} c x^{2} + a^{2} c} b^{2} c - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2}}{2 \, a^{2}} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {5}{2}}}{2 \, a^{2} c x^{2}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (83) = 166\).

Time = 0.22 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.20 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\frac {3 \, a b^{3} \sqrt {-c} c \arctan \left (\frac {{\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}}{2 \, a c}\right ) - \sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a} b^{3} c - \frac {2 \, {\left (a^{2} b^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{6} \sqrt {-c} c^{2} - 4 \, a^{4} b^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} \sqrt {-c} c^{4}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} + 4 \, a^{2} c^{2}\right )}^{2}}}{b} \] Input:

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

Output:

(3*a*b^3*sqrt(-c)*c*arctan(1/2*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c 
 + 2*a*c))^2/(a*c)) - sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)*b^3*c - 2*( 
a^2*b^3*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^6*sqrt(-c)*c 
^2 - 4*a^4*b^3*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^2*sqr 
t(-c)*c^4)/((sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 + 4*a^ 
2*c^2)^2)/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\int \frac {{\left (a\,c-b\,c\,x\right )}^{3/2}\,{\left (a+b\,x\right )}^{3/2}}{x^3} \,d x \] Input:

int(((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2))/x^3,x)
 

Output:

int(((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2))/x^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.79 \[ \int \frac {(a+b x)^{3/2} (a c-b c x)^{3/2}}{x^3} \, dx=\frac {\sqrt {c}\, c \left (-\sqrt {b x +a}\, \sqrt {-b x +a}\, a^{2}-2 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{2} x^{2}+3 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) a \,b^{2} x^{2}-3 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) a \,b^{2} x^{2}+3 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) a \,b^{2} x^{2}-3 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) a \,b^{2} x^{2}\right )}{2 x^{2}} \] Input:

int((b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)/x^3,x)
 

Output:

(sqrt(c)*c*( - sqrt(a + b*x)*sqrt(a - b*x)*a**2 - 2*sqrt(a + b*x)*sqrt(a - 
 b*x)*b**2*x**2 + 3*log( - sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt( 
2)))/2) - 1)*a*b**2*x**2 - 3*log( - sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt 
(a)*sqrt(2)))/2) + 1)*a*b**2*x**2 + 3*log(sqrt(2) + tan(asin(sqrt(a - b*x) 
/(sqrt(a)*sqrt(2)))/2) - 1)*a*b**2*x**2 - 3*log(sqrt(2) + tan(asin(sqrt(a 
- b*x)/(sqrt(a)*sqrt(2)))/2) + 1)*a*b**2*x**2))/(2*x**2)