\(\int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx\) [7]

Optimal result

Integrand size = 29, antiderivative size = 193 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=-\frac {2 A \sqrt {a x+b x^2}}{5 c x^3}-\frac {2 (A b c+5 a B c-5 a A d) \sqrt {a x+b x^2}}{15 a c^2 x^2}-\frac {2 \left (5 a B c (b c-3 a d)-A \left (2 b^2 c^2+5 a b c d-15 a^2 d^2\right )\right ) \sqrt {a x+b x^2}}{15 a^2 c^3 x}-\frac {2 d \sqrt {b c-a d} (B c-A d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{c^{7/2}} \] Output:

-2/5*A*(b*x^2+a*x)^(1/2)/c/x^3-2/15*(-5*A*a*d+A*b*c+5*B*a*c)*(b*x^2+a*x)^( 
1/2)/a/c^2/x^2-2/15*(5*a*B*c*(-3*a*d+b*c)-A*(-15*a^2*d^2+5*a*b*c*d+2*b^2*c 
^2))*(b*x^2+a*x)^(1/2)/a^2/c^3/x-2*d*(-a*d+b*c)^(1/2)*(-A*d+B*c)*arctanh(( 
-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a*x)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\frac {2 \sqrt {x (a+b x)} \left (-\frac {\sqrt {c} \left (-2 A b^2 c^2 x^2+a b c x (5 B c x+A (c-5 d x))+a^2 \left (5 B c x (c-3 d x)+A \left (3 c^2-5 c d x+15 d^2 x^2\right )\right )\right )}{a^2}+\frac {15 d \sqrt {-b c+a d} (-B c+A d) x^{5/2} \arctan \left (\frac {-d \sqrt {x} \sqrt {a+b x}+\sqrt {b} (c+d x)}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {a+b x}}\right )}{15 c^{7/2} x^3} \] Input:

Integrate[((A + B*x)*Sqrt[a*x + b*x^2])/(x^4*(c + d*x)),x]
 

Output:

(2*Sqrt[x*(a + b*x)]*(-((Sqrt[c]*(-2*A*b^2*c^2*x^2 + a*b*c*x*(5*B*c*x + A* 
(c - 5*d*x)) + a^2*(5*B*c*x*(c - 3*d*x) + A*(3*c^2 - 5*c*d*x + 15*d^2*x^2) 
)))/a^2) + (15*d*Sqrt[-(b*c) + a*d]*(-(B*c) + A*d)*x^(5/2)*ArcTan[(-(d*Sqr 
t[x]*Sqrt[a + b*x]) + Sqrt[b]*(c + d*x))/(Sqrt[c]*Sqrt[-(b*c) + a*d])])/Sq 
rt[a + b*x]))/(15*c^(7/2)*x^3)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2153, 2009}

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.

\(\displaystyle \int \frac {\sqrt {a x+b x^2} (A+B x)}{x^4 (c+d x)} \, dx\)

\(\Big \downarrow \) 2153

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 A b \left (a x+b x^2\right )^{3/2}}{15 a^2 c x^3}-\frac {d \sqrt {b c-a d} (B c-A d) \text {arctanh}\left (\frac {x (2 b c-a d)+a c}{2 \sqrt {c} \sqrt {a x+b x^2} \sqrt {b c-a d}}\right )}{c^{7/2}}+\frac {a d^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (B c-A d)}{\sqrt {b} c^4}+\frac {d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (2 b c-a d) (B c-A d)}{\sqrt {b} c^4}-\frac {2 \sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (B c-A d)}{c^3}+\frac {2 d \sqrt {a x+b x^2} (B c-A d)}{c^3 x}-\frac {2 \left (a x+b x^2\right )^{3/2} (B c-A d)}{3 a c^2 x^3}-\frac {2 A \left (a x+b x^2\right )^{3/2}}{5 a c x^4}\)

Input:

Int[((A + B*x)*Sqrt[a*x + b*x^2])/(x^4*(c + d*x)),x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2153

Int[(Px_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b 
_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e* 
x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, n, p}, x] && PolyQ[Px, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ 
[m] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] && IGtQ[n, 0])
 

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \left (\left (\left (\frac {5 B x}{3}+A \right ) a -\frac {2 A b x}{3}\right ) \left (b x +a \right ) c^{2}-\frac {5 d \left (\left (3 B x +A \right ) a +A b x \right ) x a c}{3}+5 A \,a^{2} d^{2} x^{2}\right ) \sqrt {c \left (a d -b c \right )}\, \sqrt {x \left (b x +a \right )}}{5}+2 a^{2} d \,x^{3} \left (a d -b c \right ) \left (A d -B c \right ) \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )}{c^{3} x^{3} \sqrt {c \left (a d -b c \right )}\, a^{2}}\)	\(154\)
risch	\(-\frac {2 \left (b x +a \right ) \left (15 A \,a^{2} d^{2} x^{2}-5 A a b c d \,x^{2}-2 A \,b^{2} c^{2} x^{2}-15 B \,a^{2} c d \,x^{2}+5 B a b \,c^{2} x^{2}-5 A \,a^{2} c d x +A a b \,c^{2} x +5 B \,a^{2} c^{2} x +3 a^{2} A \,c^{2}\right )}{15 a^{2} c^{3} \sqrt {x \left (b x +a \right )}\, x^{2}}+\frac {\left (A a \,d^{2}-A b c d -B a c d +B b \,c^{2}\right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{c^{3} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\)	\(275\)
default	\(\frac {A \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{5 a \,x^{4}}+\frac {4 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{15 a^{2} x^{3}}\right )}{c}+\frac {2 \left (A d -B c \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3 c^{2} a \,x^{3}}+\frac {d^{2} \left (A d -B c \right ) \left (\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}+\frac {\left (a d -2 b c \right ) \ln \left (\frac {\frac {a d -2 b c}{2 d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}\right )}{2 d \sqrt {b}}+\frac {c \left (a d -b c \right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\right )}{c^{4}}+\frac {\left (A d -B c \right ) d \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{a \,x^{2}}+\frac {2 b \left (\sqrt {b \,x^{2}+a x}+\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b}}\right )}{a}\right )}{c^{3}}-\frac {\left (A d -B c \right ) d^{2} \left (\sqrt {b \,x^{2}+a x}+\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b}}\right )}{c^{4}}\)	\(510\)

Input:

int((B*x+A)*(b*x^2+a*x)^(1/2)/x^4/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

2/(c*(a*d-b*c))^(1/2)*(-1/5*(((5/3*B*x+A)*a-2/3*A*b*x)*(b*x+a)*c^2-5/3*d*( 
(3*B*x+A)*a+A*b*x)*x*a*c+5*A*a^2*d^2*x^2)*(c*(a*d-b*c))^(1/2)*(x*(b*x+a))^ 
(1/2)+a^2*d*x^3*(a*d-b*c)*(A*d-B*c)*arctan((x*(b*x+a))^(1/2)/x*c/(c*(a*d-b 
*c))^(1/2)))/c^3/x^3/a^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.99 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\left [-\frac {15 \, {\left (B a^{2} c d - A a^{2} d^{2}\right )} x^{3} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x + 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right ) + 2 \, {\left (3 \, A a^{2} c^{2} + {\left (15 \, A a^{2} d^{2} + {\left (5 \, B a b - 2 \, A b^{2}\right )} c^{2} - 5 \, {\left (3 \, B a^{2} + A a b\right )} c d\right )} x^{2} - {\left (5 \, A a^{2} c d - {\left (5 \, B a^{2} + A a b\right )} c^{2}\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, a^{2} c^{3} x^{3}}, -\frac {2 \, {\left (15 \, {\left (B a^{2} c d - A a^{2} d^{2}\right )} x^{3} \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) + {\left (3 \, A a^{2} c^{2} + {\left (15 \, A a^{2} d^{2} + {\left (5 \, B a b - 2 \, A b^{2}\right )} c^{2} - 5 \, {\left (3 \, B a^{2} + A a b\right )} c d\right )} x^{2} - {\left (5 \, A a^{2} c d - {\left (5 \, B a^{2} + A a b\right )} c^{2}\right )} x\right )} \sqrt {b x^{2} + a x}\right )}}{15 \, a^{2} c^{3} x^{3}}\right ] \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^4/(d*x+c),x, algorithm="fricas")
 

Output:

[-1/15*(15*(B*a^2*c*d - A*a^2*d^2)*x^3*sqrt((b*c - a*d)/c)*log((a*c + (2*b 
*c - a*d)*x + 2*sqrt(b*x^2 + a*x)*c*sqrt((b*c - a*d)/c))/(d*x + c)) + 2*(3 
*A*a^2*c^2 + (15*A*a^2*d^2 + (5*B*a*b - 2*A*b^2)*c^2 - 5*(3*B*a^2 + A*a*b) 
*c*d)*x^2 - (5*A*a^2*c*d - (5*B*a^2 + A*a*b)*c^2)*x)*sqrt(b*x^2 + a*x))/(a 
^2*c^3*x^3), -2/15*(15*(B*a^2*c*d - A*a^2*d^2)*x^3*sqrt(-(b*c - a*d)/c)*ar 
ctan(-sqrt(b*x^2 + a*x)*c*sqrt(-(b*c - a*d)/c)/((b*c - a*d)*x)) + (3*A*a^2 
*c^2 + (15*A*a^2*d^2 + (5*B*a*b - 2*A*b^2)*c^2 - 5*(3*B*a^2 + A*a*b)*c*d)* 
x^2 - (5*A*a^2*c*d - (5*B*a^2 + A*a*b)*c^2)*x)*sqrt(b*x^2 + a*x))/(a^2*c^3 
*x^3)]
 

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\int \frac {\sqrt {x \left (a + b x\right )} \left (A + B x\right )}{x^{4} \left (c + d x\right )}\, dx \] Input:

integrate((B*x+A)*(b*x**2+a*x)**(1/2)/x**4/(d*x+c),x)
 

Output:

Integral(sqrt(x*(a + b*x))*(A + B*x)/(x**4*(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\int { \frac {\sqrt {b x^{2} + a x} {\left (B x + A\right )}}{{\left (d x + c\right )} x^{4}} \,d x } \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^4/(d*x+c),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^2 + a*x)*(B*x + A)/((d*x + c)*x^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (171) = 342\).

Time = 0.33 (sec) , antiderivative size = 448, normalized size of antiderivative = 2.32 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=-\frac {2 \, {\left (B b c^{2} d - B a c d^{2} - A b c d^{2} + A a d^{3}\right )} \arctan \left (-\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} d + \sqrt {b} c}{\sqrt {-b c^{2} + a c d}}\right )}{\sqrt {-b c^{2} + a c d} c^{3}} + \frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} B b c^{2} - 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} B a c d - 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} A b c d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} A a d^{2} + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} B a \sqrt {b} c^{2} + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} A b^{\frac {3}{2}} c^{2} - 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} A a \sqrt {b} c d + 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} B a^{2} c^{2} + 25 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} A a b c^{2} - 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} A a^{2} c d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} A a^{2} \sqrt {b} c^{2} + 3 \, A a^{3} c^{2}\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5} c^{3}} \] Input:

integrate((B*x+A)*(b*x^2+a*x)^(1/2)/x^4/(d*x+c),x, algorithm="giac")
 

Output:

-2*(B*b*c^2*d - B*a*c*d^2 - A*b*c*d^2 + A*a*d^3)*arctan(-((sqrt(b)*x - sqr 
t(b*x^2 + a*x))*d + sqrt(b)*c)/sqrt(-b*c^2 + a*c*d))/(sqrt(-b*c^2 + a*c*d) 
*c^3) + 2/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*B*b*c^2 - 15*(sqrt(b)*x 
 - sqrt(b*x^2 + a*x))^4*B*a*c*d - 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*A*b 
*c*d + 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*A*a*d^2 + 15*(sqrt(b)*x - sqrt 
(b*x^2 + a*x))^3*B*a*sqrt(b)*c^2 + 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*A* 
b^(3/2)*c^2 - 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*A*a*sqrt(b)*c*d + 5*(sq 
rt(b)*x - sqrt(b*x^2 + a*x))^2*B*a^2*c^2 + 25*(sqrt(b)*x - sqrt(b*x^2 + a* 
x))^2*A*a*b*c^2 - 5*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*A*a^2*c*d + 15*(sqrt 
(b)*x - sqrt(b*x^2 + a*x))*A*a^2*sqrt(b)*c^2 + 3*A*a^3*c^2)/((sqrt(b)*x - 
sqrt(b*x^2 + a*x))^5*c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\int \frac {\sqrt {b\,x^2+a\,x}\,\left (A+B\,x\right )}{x^4\,\left (c+d\,x\right )} \,d x \] Input:

int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x^4*(c + d*x)),x)
 

Output:

int(((a*x + b*x^2)^(1/2)*(A + B*x))/(x^4*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.03 \[ \int \frac {(A+B x) \sqrt {a x+b x^2}}{x^4 (c+d x)} \, dx=\frac {2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x^{3}-2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d \,x^{3}+2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x^{3}-2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d \,x^{3}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c^{3}}{5}+\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c^{2} d x}{3}-2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c \,d^{2} x^{2}-\frac {4 \sqrt {x}\, \sqrt {b x +a}\, a b \,c^{3} x}{5}+\frac {8 \sqrt {x}\, \sqrt {b x +a}\, a b \,c^{2} d \,x^{2}}{3}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c^{3} x^{2}}{5}+\frac {6 \sqrt {b}\, a^{2} c \,d^{2} x^{3}}{5}-\frac {16 \sqrt {b}\, a b \,c^{2} d \,x^{3}}{15}-\frac {2 \sqrt {b}\, b^{2} c^{3} x^{3}}{5}}{a \,c^{4} x^{3}} \] Input:

int((B*x+A)*(b*x^2+a*x)^(1/2)/x^4/(d*x+c),x)
 

Output:

(2*(15*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b* 
x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a**2*d**2*x**3 - 15*sqrt( 
c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x) - sqrt(x) 
*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*a*b*c*d*x**3 + 15*sqrt(c)*sqrt(a*d - 
b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt( 
b))/(sqrt(c)*sqrt(b)))*a**2*d**2*x**3 - 15*sqrt(c)*sqrt(a*d - b*c)*atan((s 
qrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c) 
*sqrt(b)))*a*b*c*d*x**3 - 3*sqrt(x)*sqrt(a + b*x)*a**2*c**3 + 5*sqrt(x)*sq 
rt(a + b*x)*a**2*c**2*d*x - 15*sqrt(x)*sqrt(a + b*x)*a**2*c*d**2*x**2 - 6* 
sqrt(x)*sqrt(a + b*x)*a*b*c**3*x + 20*sqrt(x)*sqrt(a + b*x)*a*b*c**2*d*x** 
2 - 3*sqrt(x)*sqrt(a + b*x)*b**2*c**3*x**2 + 9*sqrt(b)*a**2*c*d**2*x**3 - 
8*sqrt(b)*a*b*c**2*d*x**3 - 3*sqrt(b)*b**2*c**3*x**3))/(15*a*c**4*x**3)