\(\int \frac {(d+e x)^{5/2}}{(b x+c x^2)^2} \, dx\) [115]

Optimal result

Integrand size = 21, antiderivative size = 159 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx=-\frac {(c d-b e) (2 c d-b e) \sqrt {d+e x}}{b^2 c (b+c x)}-\frac {d (d+e x)^{3/2}}{b x (b+c x)}+\frac {d^{3/2} (4 c d-5 b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3}-\frac {(c d-b e)^{3/2} (4 c d+b e) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 c^{3/2}} \] Output:

-(-b*e+c*d)*(-b*e+2*c*d)*(e*x+d)^(1/2)/b^2/c/(c*x+b)-d*(e*x+d)^(3/2)/b/x/( 
c*x+b)+d^(3/2)*(-5*b*e+4*c*d)*arctanh((e*x+d)^(1/2)/d^(1/2))/b^3-(-b*e+c*d 
)^(3/2)*(b*e+4*c*d)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b^3/c^ 
(3/2)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx=\frac {-\frac {b \sqrt {d+e x} \left (2 c^2 d^2 x+b^2 e^2 x+b c d (d-2 e x)\right )}{c x (b+c x)}+\frac {(-c d+b e)^{3/2} (4 c d+b e) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{c^{3/2}}+d^{3/2} (4 c d-5 b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3} \] Input:

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

Output:

(-((b*Sqrt[d + e*x]*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)))/(c*x*(b 
 + c*x))) + ((-(c*d) + b*e)^(3/2)*(4*c*d + b*e)*ArcTan[(Sqrt[c]*Sqrt[d + e 
*x])/Sqrt[-(c*d) + b*e]])/c^(3/2) + d^(3/2)*(4*c*d - 5*b*e)*ArcTanh[Sqrt[d 
 + e*x]/Sqrt[d]])/b^3
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1164, 27, 1196, 1197, 27, 1480, 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.

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

\(\Big \downarrow \) 1164

\(\displaystyle -\frac {\int \frac {\sqrt {d+e x} (d (4 c d-5 b e)-e (2 c d-b e) x)}{2 \left (c x^2+b x\right )}dx}{b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {d+e x} (d (4 c d-5 b e)-e (2 c d-b e) x)}{c x^2+b x}dx}{2 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 1196

\(\displaystyle -\frac {\frac {\int \frac {c (4 c d-5 b e) d^2+e \left (2 c^2 d^2-2 b c e d-b^2 e^2\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{c}-\frac {2 e \sqrt {d+e x} (2 c d-b e)}{c}}{2 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {\frac {2 e \left (\frac {c^2 d^2 (4 c d-5 b e) \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}-\frac {(c d-b e)^2 (b e+4 c d) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}\right )}{c}-\frac {2 e \sqrt {d+e x} (2 c d-b e)}{c}}{2 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

-(((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*(b*x + c*x^2))) - ((-2*e* 
(2*c*d - b*e)*Sqrt[d + e*x])/c + (2*e*(-((c*d^(3/2)*(4*c*d - 5*b*e)*ArcTan 
h[Sqrt[d + e*x]/Sqrt[d]])/(b*e)) + ((c*d - b*e)^(3/2)*(4*c*d + b*e)*ArcTan 
h[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*Sqrt[c]*e)))/c)/(2*b^2)
 

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 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 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(2 e^{3} \left (\frac {\left (b e -c d \right )^{2} \left (-\frac {b e \sqrt {e x +d}}{2 c \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (b e +4 c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 c \sqrt {c \left (b e -c d \right )}}\right )}{b^{3} e^{3}}-\frac {d^{2} \left (\frac {b \sqrt {e x +d}}{2 x}+\frac {\left (5 b e -4 c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}\right )}{b^{3} e^{3}}\right )\)	\(154\)
default	\(2 e^{3} \left (\frac {\left (b e -c d \right )^{2} \left (-\frac {b e \sqrt {e x +d}}{2 c \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (b e +4 c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 c \sqrt {c \left (b e -c d \right )}}\right )}{b^{3} e^{3}}-\frac {d^{2} \left (\frac {b \sqrt {e x +d}}{2 x}+\frac {\left (5 b e -4 c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}\right )}{b^{3} e^{3}}\right )\)	\(154\)
pseudoelliptic	\(-\frac {-4 \sqrt {d}\, \left (-b e +c d \right )^{2} x \left (c x +b \right ) \left (c d +\frac {b e}{4}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+\sqrt {c \left (b e -c d \right )}\, \left (5 x \left (c x +b \right ) \left (b e -\frac {4 c d}{5}\right ) c \,d^{2} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )+\sqrt {d}\, \sqrt {e x +d}\, b \left (2 c^{2} d^{2} x +b d \left (-2 e x +d \right ) c +b^{2} e^{2} x \right )\right )}{\sqrt {d}\, \sqrt {c \left (b e -c d \right )}\, b^{3} x c \left (c x +b \right )}\)	\(172\)
risch	\(-\frac {d^{2} \sqrt {e x +d}}{b^{2} x}+\frac {e \left (-\frac {d^{\frac {3}{2}} \left (5 b e -4 c d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}+\frac {-\frac {b e \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) \sqrt {e x +d}}{c \left (\left (e x +d \right ) c +b e -c d \right )}+\frac {\left (b^{3} e^{3}+2 d \,e^{2} b^{2} c -7 d^{2} e b \,c^{2}+4 d^{3} c^{3}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{c \sqrt {c \left (b e -c d \right )}}}{b e}\right )}{b^{2}}\)	\(194\)

Input:

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

Output:

2*e^3*((b*e-c*d)^2/b^3/e^3*(-1/2*b*e/c*(e*x+d)^(1/2)/((e*x+d)*c+b*e-c*d)+1 
/2*(b*e+4*c*d)/c/(c*(b*e-c*d))^(1/2)*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^ 
(1/2)))-d^2/b^3/e^3*(1/2*b*(e*x+d)^(1/2)/x+1/2*(5*b*e-4*c*d)/d^(1/2)*arcta 
nh((e*x+d)^(1/2)/d^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 996, normalized size of antiderivative = 6.26 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*(((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c 
*d*e - b^3*e^2)*x)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e 
*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 
 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt( 
d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqr 
t(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d^2 - 3*b*c^2*d*e - b 
^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e) 
/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^ 
2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(d)*log((e*x - 2 
*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*(b^2*c*d^2 + (2*b*c^2*d^2 - 2*b^2*c*d 
*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^4*c*x), -1/2*(2*((4*c^3*d 
^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 - 5*b^2*c*d*e)*x)*sqrt(-d)*arctan(sqr 
t(-d)/sqrt(e*x + d)) + ((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c 
^2*d^2 - 3*b^2*c*d*e - b^3*e^2)*x)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d 
- b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(b^2*c*d^2 + 
 (2*b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2)*x)*sqrt(e*x + d))/(b^3*c^2*x^2 + b^ 
4*c*x), -(((4*c^3*d^2 - 3*b*c^2*d*e - b^2*c*e^2)*x^2 + (4*b*c^2*d^2 - 3*b^ 
2*c*d*e - b^3*e^2)*x)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-( 
c*d - b*e)/c)/(c*d - b*e)) + ((4*c^3*d^2 - 5*b*c^2*d*e)*x^2 + (4*b*c^2*d^2 
 - 5*b^2*c*d*e)*x)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x + d)) + (b^2*c*d^2...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.64 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx=-\frac {{\left (4 \, c d^{3} - 5 \, b d^{2} e\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d}} + \frac {{\left (4 \, c^{3} d^{3} - 7 \, b c^{2} d^{2} e + 2 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b^{3} c} - \frac {2 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{2} e - 2 \, \sqrt {e x + d} c^{2} d^{3} e - 2 \, {\left (e x + d\right )}^{\frac {3}{2}} b c d e^{2} + 3 \, \sqrt {e x + d} b c d^{2} e^{2} + {\left (e x + d\right )}^{\frac {3}{2}} b^{2} e^{3} - \sqrt {e x + d} b^{2} d e^{3}}{{\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )} b^{2} c} \] Input:

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

Output:

-(4*c*d^3 - 5*b*d^2*e)*arctan(sqrt(e*x + d)/sqrt(-d))/(b^3*sqrt(-d)) + (4* 
c^3*d^3 - 7*b*c^2*d^2*e + 2*b^2*c*d*e^2 + b^3*e^3)*arctan(sqrt(e*x + d)*c/ 
sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b^3*c) - (2*(e*x + d)^(3/2)*c^ 
2*d^2*e - 2*sqrt(e*x + d)*c^2*d^3*e - 2*(e*x + d)^(3/2)*b*c*d*e^2 + 3*sqrt 
(e*x + d)*b*c*d^2*e^2 + (e*x + d)^(3/2)*b^2*e^3 - sqrt(e*x + d)*b^2*d*e^3) 
/(((e*x + d)^2*c - 2*(e*x + d)*c*d + c*d^2 + (e*x + d)*b*e - b*d*e)*b^2*c)
 

Mupad [B] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 1127, normalized size of antiderivative = 7.09 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(((d + e*x)^(1/2)*(b^2*d*e^3 + 2*c^2*d^3*e - 3*b*c*d^2*e^2))/(b^2*c) - (e* 
(d + e*x)^(3/2)*(b^2*e^2 + 2*c^2*d^2 - 2*b*c*d*e))/(b^2*c))/((b*e - 2*c*d) 
*(d + e*x) + c*(d + e*x)^2 + c*d^2 - b*d*e) - (atanh((10*e^9*(d^3)^(1/2)*( 
d + e*x)^(1/2))/(10*d^2*e^9 + (32*c*d^3*e^8)/b - (132*c^2*d^4*e^7)/b^2 + ( 
130*c^3*d^5*e^6)/b^3 - (40*c^4*d^6*e^5)/b^4) + (32*d*e^8*(d^3)^(1/2)*(d + 
e*x)^(1/2))/(32*d^3*e^8 + (10*b*d^2*e^9)/c - (132*c*d^4*e^7)/b + (130*c^2* 
d^5*e^6)/b^2 - (40*c^3*d^6*e^5)/b^3) - (132*c*d^2*e^7*(d^3)^(1/2)*(d + e*x 
)^(1/2))/(32*b*d^3*e^8 - 132*c*d^4*e^7 + (130*c^2*d^5*e^6)/b + (10*b^2*d^2 
*e^9)/c - (40*c^3*d^6*e^5)/b^2) + (130*c^2*d^3*e^6*(d^3)^(1/2)*(d + e*x)^( 
1/2))/(32*b^2*d^3*e^8 + 130*c^2*d^5*e^6 - (40*c^3*d^6*e^5)/b + (10*b^3*d^2 
*e^9)/c - 132*b*c*d^4*e^7) - (40*c^3*d^4*e^5*(d^3)^(1/2)*(d + e*x)^(1/2))/ 
(32*b^3*d^3*e^8 - 40*c^3*d^6*e^5 + 130*b*c^2*d^5*e^6 - 132*b^2*c*d^4*e^7 + 
 (10*b^4*d^2*e^9)/c))*(5*b*e - 4*c*d)*(d^3)^(1/2))/b^3 - (atanh((30*d^3*e^ 
6*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e 
)^(1/2))/(14*b^3*d^2*e^9 + 110*c^3*d^5*e^6 - 82*b*c^2*d^4*e^7 - 4*b^2*c*d^ 
3*e^8 + (2*b^4*d*e^10)/c - (40*c^4*d^6*e^5)/b) - (2*d*e^8*(d + e*x)^(1/2)* 
(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(4*c^3*d^ 
3*e^8 - 14*b*c^2*d^2*e^9 + (82*c^4*d^4*e^7)/b - (110*c^5*d^5*e^6)/b^2 + (4 
0*c^6*d^6*e^5)/b^3 - 2*b^2*c*d*e^10) + (18*d^2*e^7*(d + e*x)^(1/2)*(c^6*d^ 
3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(2*b^3*d*e^10...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 547, normalized size of antiderivative = 3.44 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^2} \, dx=\frac {2 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{3} e^{2} x +6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{2} c d e x +2 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{2} c \,e^{2} x^{2}-8 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b \,c^{2} d^{2} x +6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b \,c^{2} d e \,x^{2}-8 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) c^{3} d^{2} x^{2}-2 \sqrt {e x +d}\, b^{3} c \,e^{2} x -2 \sqrt {e x +d}\, b^{2} c^{2} d^{2}+4 \sqrt {e x +d}\, b^{2} c^{2} d e x -4 \sqrt {e x +d}\, b \,c^{3} d^{2} x +5 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) b^{2} c^{2} d e x -4 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) b \,c^{3} d^{2} x +5 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) b \,c^{3} d e \,x^{2}-4 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) c^{4} d^{2} x^{2}-5 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) b^{2} c^{2} d e x +4 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) b \,c^{3} d^{2} x -5 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) b \,c^{3} d e \,x^{2}+4 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) c^{4} d^{2} x^{2}}{2 b^{3} c^{2} x \left (c x +b \right )} \] Input:

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

Output:

(2*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d) 
))*b**3*e**2*x + 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c) 
*sqrt(b*e - c*d)))*b**2*c*d*e*x + 2*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + 
 e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c*e**2*x**2 - 8*sqrt(c)*sqrt(b*e 
- c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b*c**2*d**2*x + 6 
*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d))) 
*b*c**2*d*e*x**2 - 8*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt( 
c)*sqrt(b*e - c*d)))*c**3*d**2*x**2 - 2*sqrt(d + e*x)*b**3*c*e**2*x - 2*sq 
rt(d + e*x)*b**2*c**2*d**2 + 4*sqrt(d + e*x)*b**2*c**2*d*e*x - 4*sqrt(d + 
e*x)*b*c**3*d**2*x + 5*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**2*c**2*d*e* 
x - 4*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b*c**3*d**2*x + 5*sqrt(d)*log(s 
qrt(d + e*x) - sqrt(d))*b*c**3*d*e*x**2 - 4*sqrt(d)*log(sqrt(d + e*x) - sq 
rt(d))*c**4*d**2*x**2 - 5*sqrt(d)*log(sqrt(d + e*x) + sqrt(d))*b**2*c**2*d 
*e*x + 4*sqrt(d)*log(sqrt(d + e*x) + sqrt(d))*b*c**3*d**2*x - 5*sqrt(d)*lo 
g(sqrt(d + e*x) + sqrt(d))*b*c**3*d*e*x**2 + 4*sqrt(d)*log(sqrt(d + e*x) + 
 sqrt(d))*c**4*d**2*x**2)/(2*b**3*c**2*x*(b + c*x))