\(\int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx\) [85]

Optimal result

Integrand size = 26, antiderivative size = 112 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=-\frac {\sqrt {a+b x} \sqrt {a c-b c x}}{4 x^4}+\frac {b^2 \sqrt {a+b x} \sqrt {a c-b c x}}{8 a^2 x^2}+\frac {b^4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {a+b x} \sqrt {a c-b c x}}{a \sqrt {c}}\right )}{8 a^3} \] Output:

-1/4*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)/x^4+1/8*b^2*(b*x+a)^(1/2)*(-b*c*x+a* 
c)^(1/2)/a^2/x^2+1/8*b^4*c^(1/2)*arctanh((b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)/ 
a/c^(1/2))/a^3
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\frac {\sqrt {c (a-b x)} \left (a \sqrt {a-b x} \sqrt {a+b x} \left (-2 a^2+b^2 x^2\right )+2 b^4 x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )\right )}{8 a^3 x^4 \sqrt {a-b x}} \] Input:

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

Output:

(Sqrt[c*(a - b*x)]*(a*Sqrt[a - b*x]*Sqrt[a + b*x]*(-2*a^2 + b^2*x^2) + 2*b 
^4*x^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a - b*x]]))/(8*a^3*x^4*Sqrt[a - b*x])
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 121, 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 = {108, 25, 27, 114, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 103

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/4*(Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/x^4 - (b^2*c*(-1/2*(Sqrt[a + b*x]*S 
qrt[a*c - b*c*x])/(a^2*c*x^2) - (b^2*ArcTanh[(Sqrt[a + b*x]*Sqrt[a*c - b*c 
*x])/(a*Sqrt[c])])/(2*a^3*Sqrt[c])))/4
 

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 114

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] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

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 = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {\left (-b x +a \right ) \sqrt {b x +a}\, \left (-b^{2} x^{2}+2 a^{2}\right ) c}{8 x^{4} a^{2} \sqrt {-c \left (b x -a \right )}}+\frac {b^{4} \ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {-b^{2} c \,x^{2}+a^{2} c}}{x}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}\, c}{8 a^{2} \sqrt {a^{2} c}\, \sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}}\)	\(140\)
default	\(\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, \left (\ln \left (\frac {2 a^{2} c +2 \sqrt {a^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}{x}\right ) b^{4} c \,x^{4}+\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {a^{2} c}\, b^{2} x^{2}-2 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, a^{2} \sqrt {a^{2} c}\right )}{8 a^{2} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, x^{4} \sqrt {a^{2} c}}\)	\(152\)

Input:

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

Output:

-1/8*(-b*x+a)*(b*x+a)^(1/2)*(-b^2*x^2+2*a^2)/x^4/a^2/(-c*(b*x-a))^(1/2)*c+ 
1/8*b^4/a^2/(a^2*c)^(1/2)*ln((2*a^2*c+2*(a^2*c)^(1/2)*(-b^2*c*x^2+a^2*c)^( 
1/2))/x)*(-(b*x+a)*c*(b*x-a))^(1/2)/(b*x+a)^(1/2)/(-c*(b*x-a))^(1/2)*c
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\left [\frac {b^{4} \sqrt {c} x^{4} \log \left (-\frac {b^{2} c x^{2} - 2 \, a^{2} c - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {c}}{x^{2}}\right ) + 2 \, {\left (a b^{2} x^{2} - 2 \, a^{3}\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{16 \, a^{3} x^{4}}, \frac {b^{4} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} a \sqrt {-c}}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (a b^{2} x^{2} - 2 \, a^{3}\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{8 \, a^{3} x^{4}}\right ] \] Input:

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

Output:

[1/16*(b^4*sqrt(c)*x^4*log(-(b^2*c*x^2 - 2*a^2*c - 2*sqrt(-b*c*x + a*c)*sq 
rt(b*x + a)*a*sqrt(c))/x^2) + 2*(a*b^2*x^2 - 2*a^3)*sqrt(-b*c*x + a*c)*sqr 
t(b*x + a))/(a^3*x^4), 1/8*(b^4*sqrt(-c)*x^4*arctan(sqrt(-b*c*x + a*c)*sqr 
t(b*x + a)*a*sqrt(-c)/(b^2*c*x^2 - a^2*c)) + (a*b^2*x^2 - 2*a^3)*sqrt(-b*c 
*x + a*c)*sqrt(b*x + a))/(a^3*x^4)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\int \frac {\sqrt {- c \left (- a + b x\right )} \sqrt {a + b x}}{x^{5}}\, dx \] Input:

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

Output:

Integral(sqrt(-c*(-a + b*x))*sqrt(a + b*x)/x**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\frac {b^{4} \sqrt {c} \log \left (\frac {2 \, a^{2} c}{{\left | x \right |}} + \frac {2 \, \sqrt {-b^{2} c x^{2} + a^{2} c} a \sqrt {c}}{{\left | x \right |}}\right )}{8 \, a^{3}} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} b^{4}}{8 \, a^{4}} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2}}{8 \, a^{4} c x^{2}} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}}}{4 \, a^{2} c x^{4}} \] Input:

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

Output:

1/8*b^4*sqrt(c)*log(2*a^2*c/abs(x) + 2*sqrt(-b^2*c*x^2 + a^2*c)*a*sqrt(c)/ 
abs(x))/a^3 - 1/8*sqrt(-b^2*c*x^2 + a^2*c)*b^4/a^4 - 1/8*(-b^2*c*x^2 + a^2 
*c)^(3/2)*b^2/(a^4*c*x^2) - 1/4*(-b^2*c*x^2 + a^2*c)^(3/2)/(a^2*c*x^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (90) = 180\).

Time = 0.17 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.68 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\frac {\frac {b^{5} \sqrt {-c} \arctan \left (\frac {{\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}}{2 \, a c}\right )}{a^{3}} + \frac {2 \, {\left (b^{5} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{14} \sqrt {-c} c - 28 \, a^{2} b^{5} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{10} \sqrt {-c} c^{3} + 112 \, a^{4} b^{5} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{6} \sqrt {-c} c^{5} - 64 \, a^{6} b^{5} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} \sqrt {-c} c^{7}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} + 4 \, a^{2} c^{2}\right )}^{4} a^{2}}}{4 \, b} \] Input:

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

Output:

1/4*(b^5*sqrt(-c)*arctan(1/2*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c + 
 2*a*c))^2/(a*c))/a^3 + 2*(b^5*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x + a)*c 
 + 2*a*c))^14*sqrt(-c)*c - 28*a^2*b^5*(sqrt(b*x + a)*sqrt(-c) - sqrt(-(b*x 
 + a)*c + 2*a*c))^10*sqrt(-c)*c^3 + 112*a^4*b^5*(sqrt(b*x + a)*sqrt(-c) - 
sqrt(-(b*x + a)*c + 2*a*c))^6*sqrt(-c)*c^5 - 64*a^6*b^5*(sqrt(b*x + a)*sqr 
t(-c) - sqrt(-(b*x + a)*c + 2*a*c))^2*sqrt(-c)*c^7)/(((sqrt(b*x + a)*sqrt( 
-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 + 4*a^2*c^2)^4*a^2))/b
 

Mupad [B] (verification not implemented)

Time = 6.73 (sec) , antiderivative size = 685, normalized size of antiderivative = 6.12 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\frac {b^4\,\ln \left (\frac {\sqrt {c\,\left (a-b\,x\right )}-\sqrt {a\,c}}{\sqrt {a+b\,x}-\sqrt {a}}\right )\,\sqrt {a\,c}}{8\,a^{7/2}}-\frac {b^4\,\ln \left (\frac {{\left (\sqrt {c\,\left (a-b\,x\right )}-\sqrt {a\,c}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}-c\right )\,\sqrt {a\,c}}{8\,a^{7/2}}-\frac {\frac {b^4\,c\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^{10}}{256\,a^{7/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}-\frac {b^4\,c^6\,\sqrt {a\,c}}{1024\,a^{7/2}}+\frac {b^4\,c^5\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{128\,a^{7/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}-\frac {11\,b^4\,c^4\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{512\,a^{7/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}+\frac {7\,b^4\,c^3\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{256\,a^{7/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}+\frac {239\,b^4\,c^2\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{1024\,a^{7/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}}{\frac {{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^{12}}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}-\frac {4\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^{10}}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}+\frac {c^4\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}-\frac {4\,c^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}+\frac {6\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}}-\frac {b^4\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{256\,a^{7/2}\,c\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {b^4\,\sqrt {a\,c}\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{1024\,a^{7/2}\,c^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4} \] Input:

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

Output:

(b^4*log(((c*(a - b*x))^(1/2) - (a*c)^(1/2))/((a + b*x)^(1/2) - a^(1/2)))* 
(a*c)^(1/2))/(8*a^(7/2)) - (b^4*log(((c*(a - b*x))^(1/2) - (a*c)^(1/2))^2/ 
((a + b*x)^(1/2) - a^(1/2))^2 - c)*(a*c)^(1/2))/(8*a^(7/2)) - ((b^4*c*(a*c 
)^(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^10)/(256*a^(7/2)*((a + b*x)^(1 
/2) - a^(1/2))^10) - (b^4*c^6*(a*c)^(1/2))/(1024*a^(7/2)) + (b^4*c^5*(a*c) 
^(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/(128*a^(7/2)*((a + b*x)^(1/2 
) - a^(1/2))^2) - (11*b^4*c^4*(a*c)^(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/ 
2))^4)/(512*a^(7/2)*((a + b*x)^(1/2) - a^(1/2))^4) + (7*b^4*c^3*(a*c)^(1/2 
)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^6)/(256*a^(7/2)*((a + b*x)^(1/2) - a 
^(1/2))^6) + (239*b^4*c^2*(a*c)^(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^ 
8)/(1024*a^(7/2)*((a + b*x)^(1/2) - a^(1/2))^8))/(((a*c - b*c*x)^(1/2) - ( 
a*c)^(1/2))^12/((a + b*x)^(1/2) - a^(1/2))^12 - (4*c*((a*c - b*c*x)^(1/2) 
- (a*c)^(1/2))^10)/((a + b*x)^(1/2) - a^(1/2))^10 + (c^4*((a*c - b*c*x)^(1 
/2) - (a*c)^(1/2))^4)/((a + b*x)^(1/2) - a^(1/2))^4 - (4*c^3*((a*c - b*c*x 
)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2) - a^(1/2))^6 + (6*c^2*((a*c - b 
*c*x)^(1/2) - (a*c)^(1/2))^8)/((a + b*x)^(1/2) - a^(1/2))^8) - (b^4*(a*c)^ 
(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/(256*a^(7/2)*c*((a + b*x)^(1/ 
2) - a^(1/2))^2) + (b^4*(a*c)^(1/2)*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^4) 
/(1024*a^(7/2)*c^2*((a + b*x)^(1/2) - a^(1/2))^4)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {a+b x} \sqrt {a c-b c x}}{x^5} \, dx=\frac {\sqrt {c}\, \left (-2 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{3}+\sqrt {b x +a}\, \sqrt {-b x +a}\, a \,b^{2} x^{2}+\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) b^{4} x^{4}-\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) b^{4} x^{4}+\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )-1\right ) b^{4} x^{4}-\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right )}{2}\right )+1\right ) b^{4} x^{4}\right )}{8 a^{3} x^{4}} \] Input:

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

Output:

(sqrt(c)*( - 2*sqrt(a + b*x)*sqrt(a - b*x)*a**3 + sqrt(a + b*x)*sqrt(a - b 
*x)*a*b**2*x**2 + log( - sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2) 
))/2) - 1)*b**4*x**4 - log( - sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)*sq 
rt(2)))/2) + 1)*b**4*x**4 + log(sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a)* 
sqrt(2)))/2) - 1)*b**4*x**4 - log(sqrt(2) + tan(asin(sqrt(a - b*x)/(sqrt(a 
)*sqrt(2)))/2) + 1)*b**4*x**4))/(8*a**3*x**4)