\(\int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx\) [26]

Optimal result

Integrand size = 41, antiderivative size = 175 \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\sqrt {2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d (d e-c f)}-\frac {2 (B e-A f) \text {arctanh}\left (\frac {\sqrt {f} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d e+c f} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {f} (d e-c f) \sqrt {d e+c f}} \] Output:

2^(1/2)*(-A*d+B*c)*arctanh(1/2*(-b*d^2*x^2+b*c^2)^(1/2)*2^(1/2)/b^(1/2)/c^ 
(1/2)/(d*x+c)^(1/2))/b^(1/2)/c^(1/2)/d/(-c*f+d*e)-2*(-A*f+B*e)*arctanh(f^( 
1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/(c*f+d*e)^(1/2)/(d*x+c)^(1/2))/b^(1/ 
2)/f^(1/2)/(-c*f+d*e)/(c*f+d*e)^(1/2)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (2 \sqrt {c} d (-B e+A f) \arctan \left (\frac {\sqrt {-d e-c f} \sqrt {c^2-d^2 x^2}}{\sqrt {f} (c-d x) \sqrt {c+d x}}\right )+\sqrt {2} (B c-A d) \sqrt {f} \sqrt {-d e-c f} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{\sqrt {c} d \sqrt {f} \sqrt {-d e-c f} (-d e+c f) \sqrt {b \left (c^2-d^2 x^2\right )}} \] Input:

Integrate[(A + B*x)/(Sqrt[c + d*x]*(e + f*x)*Sqrt[b*c^2 - b*d^2*x^2]),x]
 

Output:

-((Sqrt[c^2 - d^2*x^2]*(2*Sqrt[c]*d*(-(B*e) + A*f)*ArcTan[(Sqrt[-(d*e) - c 
*f]*Sqrt[c^2 - d^2*x^2])/(Sqrt[f]*(c - d*x)*Sqrt[c + d*x])] + Sqrt[2]*(B*c 
 - A*d)*Sqrt[f]*Sqrt[-(d*e) - c*f]*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x]) 
/Sqrt[c^2 - d^2*x^2]]))/(Sqrt[c]*d*Sqrt[f]*Sqrt[-(d*e) - c*f]*(-(d*e) + c* 
f)*Sqrt[b*(c^2 - d^2*x^2)]))
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.42, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2349, 27, 471, 221, 718, 97, 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.

\(\displaystyle \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx\)

\(\Big \downarrow \) 2349

\(\displaystyle \left (A-\frac {B e}{f}\right ) \int \frac {1}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}}dx+\int \frac {B}{f \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \left (A-\frac {B e}{f}\right ) \int \frac {1}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}}dx+\frac {B \int \frac {1}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx}{f}\)

\(\Big \downarrow \) 471

\(\displaystyle \left (A-\frac {B e}{f}\right ) \int \frac {1}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}}dx+\frac {2 B d \int \frac {1}{\frac {d^2 \left (b c^2-b d^2 x^2\right )}{c+d x}-2 b c d^2}d\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}}{f}\)

\(\Big \downarrow \) 221

\(\displaystyle \left (A-\frac {B e}{f}\right ) \int \frac {1}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}}dx-\frac {\sqrt {2} B \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d f}\)

\(\Big \downarrow \) 718

\(\displaystyle \frac {\sqrt {c+d x} \left (A-\frac {B e}{f}\right ) \sqrt {b c-b d x} \int \frac {1}{(c+d x) \sqrt {b c-b d x} (e+f x)}dx}{\sqrt {b c^2-b d^2 x^2}}-\frac {\sqrt {2} B \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d f}\)

\(\Big \downarrow \) 97

\(\displaystyle \frac {\sqrt {c+d x} \left (A-\frac {B e}{f}\right ) \sqrt {b c-b d x} \left (\frac {d \int \frac {1}{(c+d x) \sqrt {b c-b d x}}dx}{d e-c f}-\frac {f \int \frac {1}{\sqrt {b c-b d x} (e+f x)}dx}{d e-c f}\right )}{\sqrt {b c^2-b d^2 x^2}}-\frac {\sqrt {2} B \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {c+d x} \left (A-\frac {B e}{f}\right ) \sqrt {b c-b d x} \left (\frac {2 f \int \frac {1}{e+\frac {c f}{d}-\frac {f (b c-b d x)}{b d}}d\sqrt {b c-b d x}}{b d (d e-c f)}-\frac {2 \int \frac {1}{2 c-\frac {b c-b d x}{b}}d\sqrt {b c-b d x}}{b (d e-c f)}\right )}{\sqrt {b c^2-b d^2 x^2}}-\frac {\sqrt {2} B \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {c+d x} \left (A-\frac {B e}{f}\right ) \sqrt {b c-b d x} \left (\frac {2 \sqrt {f} \text {arctanh}\left (\frac {\sqrt {f} \sqrt {b c-b d x}}{\sqrt {b} \sqrt {c f+d e}}\right )}{\sqrt {b} (d e-c f) \sqrt {c f+d e}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {b c-b d x}}{\sqrt {2} \sqrt {b} \sqrt {c}}\right )}{\sqrt {b} \sqrt {c} (d e-c f)}\right )}{\sqrt {b c^2-b d^2 x^2}}-\frac {\sqrt {2} B \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {c} d f}\)

Input:

Int[(A + B*x)/(Sqrt[c + d*x]*(e + f*x)*Sqrt[b*c^2 - b*d^2*x^2]),x]
 

Output:

((A - (B*e)/f)*Sqrt[c + d*x]*Sqrt[b*c - b*d*x]*(-((Sqrt[2]*ArcTanh[Sqrt[b* 
c - b*d*x]/(Sqrt[2]*Sqrt[b]*Sqrt[c])])/(Sqrt[b]*Sqrt[c]*(d*e - c*f))) + (2 
*Sqrt[f]*ArcTanh[(Sqrt[f]*Sqrt[b*c - b*d*x])/(Sqrt[b]*Sqrt[d*e + c*f])])/( 
Sqrt[b]*(d*e - c*f)*Sqrt[d*e + c*f])))/Sqrt[b*c^2 - b*d^2*x^2] - (Sqrt[2]* 
B*ArcTanh[Sqrt[b*c^2 - b*d^2*x^2]/(Sqrt[2]*Sqrt[b]*Sqrt[c]*Sqrt[c + d*x])] 
)/(Sqrt[b]*Sqrt[c]*d*f)
 

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 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 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[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]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 718

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_) 
^2)^(p_), x_Symbol] :> Simp[(a + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]* 
(a/d + (c*x)/e)^FracPart[p])   Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c/ 
e)*x)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && EqQ[c*d^2 + a*e^2, 
 0]
 

rule 2349

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) \sqrt {b \left (c f +d e \right ) f}\, \sqrt {2}\, d -2 A \,\operatorname {arctanh}\left (\frac {f \sqrt {\left (-d x +c \right ) b}}{\sqrt {b \left (c f +d e \right ) f}}\right ) \sqrt {b c}\, d f -B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) \sqrt {b \left (c f +d e \right ) f}\, \sqrt {2}\, c +2 B \,\operatorname {arctanh}\left (\frac {f \sqrt {\left (-d x +c \right ) b}}{\sqrt {b \left (c f +d e \right ) f}}\right ) \sqrt {b c}\, d e \right )}{\sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b}\, d \left (c f -d e \right ) \sqrt {b \left (c f +d e \right ) f}\, \sqrt {b c}}\)	\(215\)

Input:

int((B*x+A)/(d*x+c)^(1/2)/(f*x+e)/(-b*d^2*x^2+b*c^2)^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

(b*(-d^2*x^2+c^2))^(1/2)*(A*arctanh(1/2*((-d*x+c)*b)^(1/2)*2^(1/2)/(b*c)^( 
1/2))*(b*(c*f+d*e)*f)^(1/2)*2^(1/2)*d-2*A*arctanh(f*((-d*x+c)*b)^(1/2)/(b* 
(c*f+d*e)*f)^(1/2))*(b*c)^(1/2)*d*f-B*arctanh(1/2*((-d*x+c)*b)^(1/2)*2^(1/ 
2)/(b*c)^(1/2))*(b*(c*f+d*e)*f)^(1/2)*2^(1/2)*c+2*B*arctanh(f*((-d*x+c)*b) 
^(1/2)/(b*(c*f+d*e)*f)^(1/2))*(b*c)^(1/2)*d*e)/(d*x+c)^(1/2)/((-d*x+c)*b)^ 
(1/2)/d/(c*f-d*e)/(b*(c*f+d*e)*f)^(1/2)/(b*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1006, normalized size of antiderivative = 5.75 \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)/(d*x+c)^(1/2)/(f*x+e)/(-b*d^2*x^2+b*c^2)^(1/2),x, algori 
thm="fricas")
 

Output:

[1/2*(sqrt(2)*((B*b*c*d - A*b*d^2)*e*f + (B*b*c^2 - A*b*c*d)*f^2)*sqrt(1/( 
b*c))*log(-(d^2*x^2 - 2*c*d*x - 2*sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d* 
x + c)*c*sqrt(1/(b*c)) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 2*sqrt(b*d*e* 
f + b*c*f^2)*(B*d*e - A*d*f)*log(-(b*d^2*f*x^2 - b*c*d*e - 2*b*c^2*f - (b* 
d^2*e + b*c*d*f)*x + 2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(b*d*e*f + b*c*f^2)*sq 
rt(d*x + c))/(d*f*x^2 + c*e + (d*e + c*f)*x)))/(b*d^3*e^2*f - b*c^2*d*f^3) 
, (sqrt(2)*((B*b*c*d - A*b*d^2)*e*f + (B*b*c^2 - A*b*c*d)*f^2)*sqrt(-1/(b* 
c))*arctan(sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c*sqrt(-1/(b*c)) 
/(d^2*x^2 - c^2)) + sqrt(b*d*e*f + b*c*f^2)*(B*d*e - A*d*f)*log(-(b*d^2*f* 
x^2 - b*c*d*e - 2*b*c^2*f - (b*d^2*e + b*c*d*f)*x + 2*sqrt(-b*d^2*x^2 + b* 
c^2)*sqrt(b*d*e*f + b*c*f^2)*sqrt(d*x + c))/(d*f*x^2 + c*e + (d*e + c*f)*x 
)))/(b*d^3*e^2*f - b*c^2*d*f^3), 1/2*(sqrt(2)*((B*b*c*d - A*b*d^2)*e*f + ( 
B*b*c^2 - A*b*c*d)*f^2)*sqrt(1/(b*c))*log(-(d^2*x^2 - 2*c*d*x - 2*sqrt(2)* 
sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c*sqrt(1/(b*c)) - 3*c^2)/(d^2*x^2 + 
 2*c*d*x + c^2)) + 4*sqrt(-b*d*e*f - b*c*f^2)*(B*d*e - A*d*f)*arctan(sqrt( 
-b*d^2*x^2 + b*c^2)*sqrt(-b*d*e*f - b*c*f^2)*sqrt(d*x + c)/(b*c*d*e + b*c^ 
2*f + (b*d^2*e + b*c*d*f)*x)))/(b*d^3*e^2*f - b*c^2*d*f^3), (sqrt(2)*((B*b 
*c*d - A*b*d^2)*e*f + (B*b*c^2 - A*b*c*d)*f^2)*sqrt(-1/(b*c))*arctan(sqrt( 
2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c*sqrt(-1/(b*c))/(d^2*x^2 - c^2) 
) + 2*sqrt(-b*d*e*f - b*c*f^2)*(B*d*e - A*d*f)*arctan(sqrt(-b*d^2*x^2 +...
 

Sympy [F]

\[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=\int \frac {A + B x}{\sqrt {- b \left (- c + d x\right ) \left (c + d x\right )} \sqrt {c + d x} \left (e + f x\right )}\, dx \] Input:

integrate((B*x+A)/(d*x+c)**(1/2)/(f*x+e)/(-b*d**2*x**2+b*c**2)**(1/2),x)
 

Output:

Integral((A + B*x)/(sqrt(-b*(-c + d*x)*(c + d*x))*sqrt(c + d*x)*(e + f*x)) 
, x)
 

Maxima [F]

\[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=\int { \frac {B x + A}{\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} {\left (f x + e\right )}} \,d x } \] Input:

integrate((B*x+A)/(d*x+c)^(1/2)/(f*x+e)/(-b*d^2*x^2+b*c^2)^(1/2),x, algori 
thm="maxima")
 

Output:

integrate((B*x + A)/(sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*(f*x + e)), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=-\frac {\frac {\sqrt {2} {\left (B c - A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{2 \, \sqrt {-b c}}\right )}{\sqrt {-b c} {\left (d e - c f\right )}} - \frac {2 \, {\left (B d e - A d f\right )} \arctan \left (\frac {\sqrt {-{\left (d x + c\right )} b + 2 \, b c} f}{\sqrt {-b d e f - b c f^{2}}}\right )}{\sqrt {-b d e f - b c f^{2}} {\left (d e - c f\right )}}}{d} \] Input:

integrate((B*x+A)/(d*x+c)^(1/2)/(f*x+e)/(-b*d^2*x^2+b*c^2)^(1/2),x, algori 
thm="giac")
 

Output:

-(sqrt(2)*(B*c - A*d)*arctan(1/2*sqrt(2)*sqrt(-(d*x + c)*b + 2*b*c)/sqrt(- 
b*c))/(sqrt(-b*c)*(d*e - c*f)) - 2*(B*d*e - A*d*f)*arctan(sqrt(-(d*x + c)* 
b + 2*b*c)*f/sqrt(-b*d*e*f - b*c*f^2))/(sqrt(-b*d*e*f - b*c*f^2)*(d*e - c* 
f)))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=\int \frac {A+B\,x}{\left (e+f\,x\right )\,\sqrt {b\,c^2-b\,d^2\,x^2}\,\sqrt {c+d\,x}} \,d x \] Input:

int((A + B*x)/((e + f*x)*(b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2)),x)
 

Output:

int((A + B*x)/((e + f*x)*(b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.87 \[ \int \frac {A+B x}{\sqrt {c+d x} (e+f x) \sqrt {b c^2-b d^2 x^2}} \, dx=\frac {\sqrt {b}\, \left (4 \sqrt {f}\, \sqrt {c f +d e}\, \mathit {atan} \left (\frac {\sqrt {-d x +c}\, f i}{\sqrt {f}\, \sqrt {c f +d e}}\right ) a c d f i -4 \sqrt {f}\, \sqrt {c f +d e}\, \mathit {atan} \left (\frac {\sqrt {-d x +c}\, f i}{\sqrt {f}\, \sqrt {c f +d e}}\right ) b c d e i -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c d \,f^{2}-\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} e f +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} f^{2}+\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c d e f +\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c d \,f^{2}+\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{2} e f -\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} f^{2}-\sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c d e f \right )}{2 b c d f \left (c^{2} f^{2}-d^{2} e^{2}\right )} \] Input:

int((B*x+A)/(d*x+c)^(1/2)/(f*x+e)/(-b*d^2*x^2+b*c^2)^(1/2),x)
 

Output:

(sqrt(b)*(4*sqrt(f)*sqrt(c*f + d*e)*atan((sqrt(c - d*x)*f*i)/(sqrt(f)*sqrt 
(c*f + d*e)))*a*c*d*f*i - 4*sqrt(f)*sqrt(c*f + d*e)*atan((sqrt(c - d*x)*f* 
i)/(sqrt(f)*sqrt(c*f + d*e)))*b*c*d*e*i - sqrt(c)*sqrt(2)*log(sqrt(c - d*x 
) - sqrt(c)*sqrt(2))*a*c*d*f**2 - sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt 
(c)*sqrt(2))*a*d**2*e*f + sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt 
(2))*b*c**2*f**2 + sqrt(c)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b* 
c*d*e*f + sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d*f**2 
+ sqrt(c)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**2*e*f - sqrt(c 
)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**2*f**2 - sqrt(c)*sqrt( 
2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d*e*f))/(2*b*c*d*f*(c**2*f**2 
- d**2*e**2))