\(\int \frac {(a x+b x^2)^{5/2}}{x^6 (c+d x)} \, dx\) [122]

Optimal result

Integrand size = 24, antiderivative size = 138 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=-\frac {2 (b c-a d)^2 \sqrt {a x+b x^2}}{c^3 x}-\frac {2 (b c-a d) \left (a x+b x^2\right )^{3/2}}{3 c^2 x^3}-\frac {2 \left (a x+b x^2\right )^{5/2}}{5 c x^5}+\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{c^{7/2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.03 (sec) , antiderivative size = 421, normalized size of antiderivative = 3.05 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {2 (x (a+b x))^{5/2} \left (-b c^{3/2} \sqrt {a+b x} \left (23 b^2 c^2 x^2+a b c x (11 c-35 d x)+a^2 \left (3 c^2-5 c d x+15 d^2 x^2\right )\right )-15 (b c-a d)^2 \left (b c-a d-i \sqrt {a} \sqrt {d} \sqrt {b c-a d}\right ) \sqrt {-b c+2 a d-2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} x^{5/2} \arctan \left (\frac {\sqrt {-b c+2 a d-2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x}}{\sqrt {c} \left (-\sqrt {a}+\sqrt {a+b x}\right )}\right )-15 (b c-a d)^2 \left (b c-a d+i \sqrt {a} \sqrt {d} \sqrt {b c-a d}\right ) \sqrt {-b c+2 a d+2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} x^{5/2} \arctan \left (\frac {\sqrt {-b c+2 a d+2 i \sqrt {a} \sqrt {d} \sqrt {b c-a d}} \sqrt {x}}{\sqrt {c} \left (-\sqrt {a}+\sqrt {a+b x}\right )}\right )\right )}{15 b c^{9/2} x^5 (a+b x)^{5/2}} \] Input:

Integrate[(a*x + b*x^2)^(5/2)/(x^6*(c + d*x)),x]
 

Output:

(2*(x*(a + b*x))^(5/2)*(-(b*c^(3/2)*Sqrt[a + b*x]*(23*b^2*c^2*x^2 + a*b*c* 
x*(11*c - 35*d*x) + a^2*(3*c^2 - 5*c*d*x + 15*d^2*x^2))) - 15*(b*c - a*d)^ 
2*(b*c - a*d - I*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d])*Sqrt[-(b*c) + 2*a*d - (2 
*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*x^(5/2)*ArcTan[(Sqrt[-(b*c) + 2*a*d - 
 (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*Sqrt[x])/(Sqrt[c]*(-Sqrt[a] + Sqrt 
[a + b*x]))] - 15*(b*c - a*d)^2*(b*c - a*d + I*Sqrt[a]*Sqrt[d]*Sqrt[b*c - 
a*d])*Sqrt[-(b*c) + 2*a*d + (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*x^(5/2) 
*ArcTan[(Sqrt[-(b*c) + 2*a*d + (2*I)*Sqrt[a]*Sqrt[d]*Sqrt[b*c - a*d]]*Sqrt 
[x])/(Sqrt[c]*(-Sqrt[a] + Sqrt[a + b*x]))]))/(15*b*c^(9/2)*x^5*(a + b*x)^( 
5/2))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1261, 105, 105, 105, 104, 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.

\(\displaystyle \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx\)

\(\Big \downarrow \) 1261

\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2} \int \frac {(a+b x)^{5/2}}{x^{7/2} (c+d x)}dx}{x^{5/2} (a+b x)^{5/2}}\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[(a*x + b*x^2)^(5/2)/(x^6*(c + d*x)),x]
 

Output:

((a*x + b*x^2)^(5/2)*((-2*(a + b*x)^(5/2))/(5*c*x^(5/2)) + ((b*c - a*d)*(( 
-2*(a + b*x)^(3/2))/(3*c*x^(3/2)) + ((b*c - a*d)*((-2*Sqrt[a + b*x])/(c*Sq 
rt[x]) + (2*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b*c - a*d]*Sqrt[x])/(Sqrt[c]*Sqr 
t[a + b*x])])/c^(3/2)))/c))/c))/(x^(5/2)*(a + b*x)^(5/2))
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

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 + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

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 1261

Int[((e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((b_.)*(x_) + (c_.)*(x_)^2) 
^(p_), x_Symbol] :> Simp[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p)) 
  Int[x^(m + p)*(f + g*x)^n*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m, 
 n}, x] &&  !IGtQ[n, 0]
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \sqrt {c \left (a d -b c \right )}\, \left (\left (\frac {23}{3} b^{2} x^{2}+\frac {11}{3} a b x +a^{2}\right ) c^{2}-\frac {5 a d x \left (7 b x +a \right ) c}{3}+5 a^{2} d^{2} x^{2}\right ) \sqrt {x \left (b x +a \right )}}{5}+2 x^{3} \left (a d -b c \right )^{3} \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )}{c^{3} x^{3} \sqrt {c \left (a d -b c \right )}}\)	\(132\)
risch	\(-\frac {2 \left (b x +a \right ) \left (15 a^{2} d^{2} x^{2}-35 a b c d \,x^{2}+23 b^{2} c^{2} x^{2}-5 a^{2} c d x +11 a b \,c^{2} x +3 a^{2} c^{2}\right )}{15 c^{3} \sqrt {x \left (b x +a \right )}\, x^{2}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d \,c^{3} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}\)	\(249\)
default	\(\text {Expression too large to display}\)	\(1978\)

Input:

int((b*x^2+a*x)^(5/2)/x^6/(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

2/5/(c*(a*d-b*c))^(1/2)*(-(c*(a*d-b*c))^(1/2)*((23/3*b^2*x^2+11/3*a*b*x+a^ 
2)*c^2-5/3*a*d*x*(7*b*x+a)*c+5*a^2*d^2*x^2)*(x*(b*x+a))^(1/2)+5*x^3*(a*d-b 
*c)^3*arctan((x*(b*x+a))^(1/2)/x*c/(c*(a*d-b*c))^(1/2)))/c^3/x^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.37 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x + 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right ) - 2 \, {\left (3 \, a^{2} c^{2} + {\left (23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} + {\left (11 \, a b c^{2} - 5 \, a^{2} c d\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, c^{3} x^{3}}, \frac {2 \, {\left (15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{3} \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) - {\left (3 \, a^{2} c^{2} + {\left (23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2}\right )} x^{2} + {\left (11 \, a b c^{2} - 5 \, a^{2} c d\right )} x\right )} \sqrt {b x^{2} + a x}\right )}}{15 \, c^{3} x^{3}}\right ] \] Input:

integrate((b*x^2+a*x)^(5/2)/x^6/(d*x+c),x, algorithm="fricas")
 

Output:

[1/15*(15*(b^2*c^2 - 2*a*b*c*d + 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^2*c^2 + (23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2)*x^2 + (11*a*b*c^2 
 - 5*a^2*c*d)*x)*sqrt(b*x^2 + a*x))/(c^3*x^3), 2/15*(15*(b^2*c^2 - 2*a*b*c 
*d + a^2*d^2)*x^3*sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sqrt(-( 
b*c - a*d)/c)/((b*c - a*d)*x)) - (3*a^2*c^2 + (23*b^2*c^2 - 35*a*b*c*d + 1 
5*a^2*d^2)*x^2 + (11*a*b*c^2 - 5*a^2*c*d)*x)*sqrt(b*x^2 + a*x))/(c^3*x^3)]
 

Sympy [F]

\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{6} \left (c + d x\right )}\, dx \] Input:

integrate((b*x**2+a*x)**(5/2)/x**6/(d*x+c),x)
 

Output:

Integral((x*(a + b*x))**(5/2)/(x**6*(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\int { \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{{\left (d x + c\right )} x^{6}} \,d x } \] Input:

integrate((b*x^2+a*x)^(5/2)/x^6/(d*x+c),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a*x)^(5/2)/((d*x + c)*x^6), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (118) = 236\).

Time = 0.14 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.70 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (-\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} d + \sqrt {b} c}{\sqrt {-b c^{2} + a c d}}\right )}{\sqrt {-b c^{2} + a c d} c^{3}} + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a b^{2} c^{2} - 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a^{2} b c d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{4} a^{3} d^{2} + 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a^{2} b^{\frac {3}{2}} c^{2} - 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} a^{3} \sqrt {b} c d + 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{3} b c^{2} - 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{4} c d + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{4} \sqrt {b} c^{2} + 3 \, a^{5} c^{2}\right )}}{15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{5} c^{3}} \] Input:

integrate((b*x^2+a*x)^(5/2)/x^6/(d*x+c),x, algorithm="giac")
 

Output:

2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(-((sqrt(b)*x 
- sqrt(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*(45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^4*a*b^2*c^2 - 45*(sq 
rt(b)*x - sqrt(b*x^2 + a*x))^4*a^2*b*c*d + 15*(sqrt(b)*x - sqrt(b*x^2 + a* 
x))^4*a^3*d^2 + 45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^2*b^(3/2)*c^2 - 15* 
(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^3*sqrt(b)*c*d + 35*(sqrt(b)*x - sqrt(b 
*x^2 + a*x))^2*a^3*b*c^2 - 5*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^4*c*d + 1 
5*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^4*sqrt(b)*c^2 + 3*a^5*c^2)/((sqrt(b)*x 
 - sqrt(b*x^2 + a*x))^5*c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^6\,\left (c+d\,x\right )} \,d x \] Input:

int((a*x + b*x^2)^(5/2)/(x^6*(c + d*x)),x)
 

Output:

int((a*x + b*x^2)^(5/2)/(x^6*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.67 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^6 (c+d x)} \, dx=\frac {2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x^{3}-4 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d \,x^{3}+2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{3}+2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a^{2} d^{2} x^{3}-4 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) a b c d \,x^{3}+2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{3}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c^{3}}{5}+\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c^{2} d x}{3}-2 \sqrt {x}\, \sqrt {b x +a}\, a^{2} c \,d^{2} x^{2}-\frac {22 \sqrt {x}\, \sqrt {b x +a}\, a b \,c^{3} x}{15}+\frac {14 \sqrt {x}\, \sqrt {b x +a}\, a b \,c^{2} d \,x^{2}}{3}-\frac {46 \sqrt {x}\, \sqrt {b x +a}\, b^{2} c^{3} x^{2}}{15}+\frac {6 \sqrt {b}\, a^{2} c \,d^{2} x^{3}}{5}-\frac {34 \sqrt {b}\, a b \,c^{2} d \,x^{3}}{15}+\frac {2 \sqrt {b}\, b^{2} c^{3} x^{3}}{3}}{c^{4} x^{3}} \] Input:

int((b*x^2+a*x)^(5/2)/x^6/(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 - 30*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)))*b**2*c**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**2*d**2*x**3 - 30*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)*s 
qrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b)))*b**2*c**2*x**3 
- 3*sqrt(x)*sqrt(a + b*x)*a**2*c**3 + 5*sqrt(x)*sqrt(a + b*x)*a**2*c**2*d* 
x - 15*sqrt(x)*sqrt(a + b*x)*a**2*c*d**2*x**2 - 11*sqrt(x)*sqrt(a + b*x)*a 
*b*c**3*x + 35*sqrt(x)*sqrt(a + b*x)*a*b*c**2*d*x**2 - 23*sqrt(x)*sqrt(a + 
 b*x)*b**2*c**3*x**2 + 9*sqrt(b)*a**2*c*d**2*x**3 - 17*sqrt(b)*a*b*c**2*d* 
x**3 + 5*sqrt(b)*b**2*c**3*x**3))/(15*c**4*x**3)