\(\int \frac {(d+e x)^{5/2}}{(b x+c x^2)^3} \, dx\) [124]

Optimal result

Integrand size = 21, antiderivative size = 280 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=\frac {\left (12 c^2 d^2-13 b c d e+2 b^2 e^2\right ) \sqrt {d+e x}}{4 b^3 (b+c x)^2}+\frac {d (8 c d-7 b e) \sqrt {d+e x}}{4 b^2 x (b+c x)^2}+\frac {3 \left (8 c^2 d^2-8 b c d e+b^2 e^2\right ) \sqrt {d+e x}}{4 b^4 (b+c x)}-\frac {d (d+e x)^{3/2}}{2 b x^2 (b+c x)^2}-\frac {3 \sqrt {d} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 b^5}+\frac {3 \sqrt {c d-b e} \left (16 c^2 d^2-12 b c d e+b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{4 b^5 \sqrt {c}} \] Output:

1/4*(2*b^2*e^2-13*b*c*d*e+12*c^2*d^2)*(e*x+d)^(1/2)/b^3/(c*x+b)^2+1/4*d*(- 
7*b*e+8*c*d)*(e*x+d)^(1/2)/b^2/x/(c*x+b)^2+3/4*(b^2*e^2-8*b*c*d*e+8*c^2*d^ 
2)*(e*x+d)^(1/2)/b^4/(c*x+b)-1/2*d*(e*x+d)^(3/2)/b/x^2/(c*x+b)^2-3/4*d^(1/ 
2)*(5*b^2*e^2-20*b*c*d*e+16*c^2*d^2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b^5+3/ 
4*(-b*e+c*d)^(1/2)*(b^2*e^2-12*b*c*d*e+16*c^2*d^2)*arctanh(c^(1/2)*(e*x+d) 
^(1/2)/(-b*e+c*d)^(1/2))/b^5/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.79 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {b \sqrt {d+e x} \left (24 c^3 d^2 x^3+12 b c^2 d x^2 (3 d-2 e x)+b^2 c x \left (8 d^2-37 d e x+3 e^2 x^2\right )+b^3 \left (-2 d^2-9 d e x+5 e^2 x^2\right )\right )}{x^2 (b+c x)^2}+\frac {3 \sqrt {-c d+b e} \left (16 c^2 d^2-12 b c d e+b^2 e^2\right ) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{\sqrt {c}}-3 \sqrt {d} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 b^5} \] Input:

Integrate[(d + e*x)^(5/2)/(b*x + c*x^2)^3,x]
 

Output:

((b*Sqrt[d + e*x]*(24*c^3*d^2*x^3 + 12*b*c^2*d*x^2*(3*d - 2*e*x) + b^2*c*x 
*(8*d^2 - 37*d*e*x + 3*e^2*x^2) + b^3*(-2*d^2 - 9*d*e*x + 5*e^2*x^2)))/(x^ 
2*(b + c*x)^2) + (3*Sqrt[-(c*d) + b*e]*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2) 
*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b*e]])/Sqrt[c] - 3*Sqrt[d]*( 
16*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(4*b^ 
5)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1164, 27, 1234, 27, 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.

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

\(\Big \downarrow \) 1164

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1234

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {d \left (16 c^2 d^2-20 b c e d+5 b^2 e^2\right )+e \left (8 c^2 d^2-8 b c e d+b^2 e^2\right ) x}{2 \sqrt {d+e x} \left (c x^2+b x\right )}dx}{b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {d \left (16 c^2 d^2-20 b c e d+5 b^2 e^2\right )+e \left (8 c^2 d^2-8 b c e d+b^2 e^2\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{2 b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1197

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {e \left (4 d (c d-b e) (2 c d-b e)+\left (8 c^2 d^2-8 b c e d+b^2 e^2\right ) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 \left (-\frac {e \int \frac {4 d (c d-b e) (2 c d-b e)+\left (8 c^2 d^2-8 b c e d+b^2 e^2\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {3 \left (-\frac {e \left (\frac {c d \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}-\frac {(c d-b e) \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}\right )}{b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {3 \left (-\frac {e \left (\frac {\sqrt {c d-b e} \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} e}-\frac {\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right )}{b e}\right )}{b^2}-\frac {\sqrt {d+e x} \left (x \left (b^2 e^2-8 b c d e+8 c^2 d^2\right )+b d (4 c d-3 b e)\right )}{b^2 \left (b x+c x^2\right )}\right )}{4 b^2}-\frac {(d+e x)^{3/2} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

Input:

Int[(d + e*x)^(5/2)/(b*x + c*x^2)^3,x]
 

Output:

-1/2*((d + e*x)^(3/2)*(b*d + (2*c*d - b*e)*x))/(b^2*(b*x + c*x^2)^2) - (3* 
(-((Sqrt[d + e*x]*(b*d*(4*c*d - 3*b*e) + (8*c^2*d^2 - 8*b*c*d*e + b^2*e^2) 
*x))/(b^2*(b*x + c*x^2))) - (e*(-((Sqrt[d]*(16*c^2*d^2 - 20*b*c*d*e + 5*b^ 
2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*e)) + (Sqrt[c*d - b*e]*(16*c^2*d 
^2 - 12*b*c*d*e + b^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e] 
])/(b*Sqrt[c]*e)))/b^2))/(4*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 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 1234

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*( 
(f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 
 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g 
*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* 
(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 
] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 = 0.74 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(-\frac {12 \left (\left (-\frac {b^{3} e^{3} \sqrt {d}}{16}+c \,d^{\frac {3}{2}} \left (c^{2} d^{2}-\frac {7}{4} b c d e +\frac {13}{16} b^{2} e^{2}\right )\right ) x^{2} \left (c x +b \right )^{2} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+\frac {\left (\frac {15 x^{2} \left (c x +b \right )^{2} d \left (b^{2} e^{2}-4 b c d e +\frac {16}{5} c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2}+\left (-\frac {5 e^{2} \left (\frac {3 c x}{5}+b \right ) x^{2} b^{2} \sqrt {d}}{2}+\left (12 \left (b \,c^{2} e -d \,c^{3}\right ) x^{3}+\left (-18 b \,c^{2} d +\frac {37}{2} c e \,b^{2}\right ) x^{2}+\left (-4 c d \,b^{2}+\frac {9}{2} e \,b^{3}\right ) x +b^{3} d \right ) d^{\frac {3}{2}}\right ) \sqrt {e x +d}\, b \right ) \sqrt {c \left (b e -c d \right )}}{24}\right )}{\sqrt {d}\, \sqrt {c \left (b e -c d \right )}\, b^{5} x^{2} \left (c x +b \right )^{2}}\)	\(259\)
derivativedivides	\(2 e^{5} \left (-\frac {d \left (\frac {\left (\frac {9}{8} b^{2} e^{2}-\frac {3}{2} b c d e \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} d \,e^{2} b^{2}+\frac {3}{2} d^{2} e b c \right ) \sqrt {e x +d}}{e^{2} x^{2}}+\frac {3 \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{8 \sqrt {d}}\right )}{b^{5} e^{5}}+\frac {\left (b e -c d \right ) \left (\frac {\left (\frac {3}{8} e^{2} b^{2} c -\frac {3}{2} c^{2} d e b \right ) \left (e x +d \right )^{\frac {3}{2}}+\frac {b e \left (5 b^{2} e^{2}-17 b c d e +12 c^{2} d^{2}\right ) \sqrt {e x +d}}{8}}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}+\frac {3 \left (b^{2} e^{2}-12 b c d e +16 c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{8 \sqrt {c \left (b e -c d \right )}}\right )}{b^{5} e^{5}}\right )\)	\(267\)
default	\(2 e^{5} \left (-\frac {d \left (\frac {\left (\frac {9}{8} b^{2} e^{2}-\frac {3}{2} b c d e \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} d \,e^{2} b^{2}+\frac {3}{2} d^{2} e b c \right ) \sqrt {e x +d}}{e^{2} x^{2}}+\frac {3 \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{8 \sqrt {d}}\right )}{b^{5} e^{5}}+\frac {\left (b e -c d \right ) \left (\frac {\left (\frac {3}{8} e^{2} b^{2} c -\frac {3}{2} c^{2} d e b \right ) \left (e x +d \right )^{\frac {3}{2}}+\frac {b e \left (5 b^{2} e^{2}-17 b c d e +12 c^{2} d^{2}\right ) \sqrt {e x +d}}{8}}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}+\frac {3 \left (b^{2} e^{2}-12 b c d e +16 c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{8 \sqrt {c \left (b e -c d \right )}}\right )}{b^{5} e^{5}}\right )\)	\(267\)
risch	\(-\frac {d \sqrt {e x +d}\, \left (9 b e x -12 c d x +2 b d \right )}{4 b^{4} x^{2}}-\frac {e \left (\frac {3 \sqrt {d}\, \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{e b}+\frac {\frac {8 \left (\left (-\frac {3}{8} b^{3} c \,e^{3}+\frac {15}{8} d \,e^{2} b^{2} c^{2}-\frac {3}{2} b \,c^{3} d^{2} e \right ) \left (e x +d \right )^{\frac {3}{2}}-\frac {b e \left (5 b^{3} e^{3}-22 d \,e^{2} b^{2} c +29 d^{2} e b \,c^{2}-12 d^{3} c^{3}\right ) \sqrt {e x +d}}{8}\right )}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}-\frac {3 \left (b^{3} e^{3}-13 d \,e^{2} b^{2} c +28 d^{2} e b \,c^{2}-16 d^{3} c^{3}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\sqrt {c \left (b e -c d \right )}}}{b e}\right )}{4 b^{4}}\)	\(274\)

Input:

int((e*x+d)^(5/2)/(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)
 

Output:

-12/d^(1/2)*((-1/16*b^3*e^3*d^(1/2)+c*d^(3/2)*(c^2*d^2-7/4*b*c*d*e+13/16*b 
^2*e^2))*x^2*(c*x+b)^2*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2))+1/24*(1 
5/2*x^2*(c*x+b)^2*d*(b^2*e^2-4*b*c*d*e+16/5*c^2*d^2)*arctanh((e*x+d)^(1/2) 
/d^(1/2))+(-5/2*e^2*(3/5*c*x+b)*x^2*b^2*d^(1/2)+(12*(b*c^2*e-c^3*d)*x^3+(- 
18*b*c^2*d+37/2*c*e*b^2)*x^2+(-4*c*d*b^2+9/2*e*b^3)*x+b^3*d)*d^(3/2))*(e*x 
+d)^(1/2)*b)*(c*(b*e-c*d))^(1/2))/(c*(b*e-c*d))^(1/2)/b^5/x^2/(c*x+b)^2
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1655, normalized size of antiderivative = 5.91 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((e*x+d)^(5/2)/(c*x^2+b*x)^3,x, algorithm="fricas")
 

Output:

[1/8*(3*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 
 12*b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^ 
2)*x^2)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*s 
qrt((c*d - b*e)/c))/(c*x + b)) + 3*((16*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2 
*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b^2*c^2*d*e + 5*b^3*c*e^2)*x^3 + (16*b^2* 
c^2*d^2 - 20*b^3*c*d*e + 5*b^4*e^2)*x^2)*sqrt(d)*log((e*x - 2*sqrt(e*x + d 
)*sqrt(d) + 2*d)/x) - 2*(2*b^4*d^2 - 3*(8*b*c^3*d^2 - 8*b^2*c^2*d*e + b^3* 
c*e^2)*x^3 - (36*b^2*c^2*d^2 - 37*b^3*c*d*e + 5*b^4*e^2)*x^2 - (8*b^3*c*d^ 
2 - 9*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2), 1/ 
8*(6*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 12 
*b^2*c^2*d*e + b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 12*b^3*c*d*e + b^4*e^2)* 
x^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c* 
d - b*e)) + 3*((16*c^4*d^2 - 20*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c 
^3*d^2 - 20*b^2*c^2*d*e + 5*b^3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 20*b^3*c*d* 
e + 5*b^4*e^2)*x^2)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 
 2*(2*b^4*d^2 - 3*(8*b*c^3*d^2 - 8*b^2*c^2*d*e + b^3*c*e^2)*x^3 - (36*b^2* 
c^2*d^2 - 37*b^3*c*d*e + 5*b^4*e^2)*x^2 - (8*b^3*c*d^2 - 9*b^4*d*e)*x)*sqr 
t(e*x + d))/(b^5*c^2*x^4 + 2*b^6*c*x^3 + b^7*x^2), 1/8*(6*((16*c^4*d^2 - 2 
0*b*c^3*d*e + 5*b^2*c^2*e^2)*x^4 + 2*(16*b*c^3*d^2 - 20*b^2*c^2*d*e + 5*b^ 
3*c*e^2)*x^3 + (16*b^2*c^2*d^2 - 20*b^3*c*d*e + 5*b^4*e^2)*x^2)*sqrt(-d...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(5/2)/(c*x**2+b*x)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(5/2)/(c*x^2+b*x)^3,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 = 0.14 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.54 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx=-\frac {3 \, {\left (16 \, c^{3} d^{3} - 28 \, b c^{2} d^{2} e + 13 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{4 \, \sqrt {-c^{2} d + b c e} b^{5}} + \frac {3 \, {\left (16 \, c^{2} d^{3} - 20 \, b c d^{2} e + 5 \, b^{2} d e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{4 \, b^{5} \sqrt {-d}} + \frac {24 \, {\left (e x + d\right )}^{\frac {7}{2}} c^{3} d^{2} e - 72 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{3} d^{3} e + 72 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{3} d^{4} e - 24 \, \sqrt {e x + d} c^{3} d^{5} e - 24 \, {\left (e x + d\right )}^{\frac {7}{2}} b c^{2} d e^{2} + 108 \, {\left (e x + d\right )}^{\frac {5}{2}} b c^{2} d^{2} e^{2} - 144 \, {\left (e x + d\right )}^{\frac {3}{2}} b c^{2} d^{3} e^{2} + 60 \, \sqrt {e x + d} b c^{2} d^{4} e^{2} + 3 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{2} c e^{3} - 46 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} c d e^{3} + 91 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} c d^{2} e^{3} - 48 \, \sqrt {e x + d} b^{2} c d^{3} e^{3} + 5 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} e^{4} - 19 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d e^{4} + 12 \, \sqrt {e x + d} b^{3} d^{2} e^{4}}{4 \, {\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )}^{2} b^{4}} \] Input:

integrate((e*x+d)^(5/2)/(c*x^2+b*x)^3,x, algorithm="giac")
 

Output:

-3/4*(16*c^3*d^3 - 28*b*c^2*d^2*e + 13*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^5) + 3/4*(16*c^2* 
d^3 - 20*b*c*d^2*e + 5*b^2*d*e^2)*arctan(sqrt(e*x + d)/sqrt(-d))/(b^5*sqrt 
(-d)) + 1/4*(24*(e*x + d)^(7/2)*c^3*d^2*e - 72*(e*x + d)^(5/2)*c^3*d^3*e + 
 72*(e*x + d)^(3/2)*c^3*d^4*e - 24*sqrt(e*x + d)*c^3*d^5*e - 24*(e*x + d)^ 
(7/2)*b*c^2*d*e^2 + 108*(e*x + d)^(5/2)*b*c^2*d^2*e^2 - 144*(e*x + d)^(3/2 
)*b*c^2*d^3*e^2 + 60*sqrt(e*x + d)*b*c^2*d^4*e^2 + 3*(e*x + d)^(7/2)*b^2*c 
*e^3 - 46*(e*x + d)^(5/2)*b^2*c*d*e^3 + 91*(e*x + d)^(3/2)*b^2*c*d^2*e^3 - 
 48*sqrt(e*x + d)*b^2*c*d^3*e^3 + 5*(e*x + d)^(5/2)*b^3*e^4 - 19*(e*x + d) 
^(3/2)*b^3*d*e^4 + 12*sqrt(e*x + d)*b^3*d^2*e^4)/(((e*x + d)^2*c - 2*(e*x 
+ d)*c*d + c*d^2 + (e*x + d)*b*e - b*d*e)^2*b^4)
 

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 910, normalized size of antiderivative = 3.25 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((d + e*x)^(5/2)/(b*x + c*x^2)^3,x)
 

Output:

(3*atanh((81*c^2*d^2*e^8*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1/2))/(8*((189*c 
^3*d^3*e^8)/8 - (351*b*c^2*d^2*e^9)/32 - (27*c^4*d^4*e^7)/(2*b) + (27*b^2* 
c*d*e^10)/32)) + (27*c^3*d^3*e^7*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1/2))/(2 
*((27*c^4*d^4*e^7)/2 - (189*b*c^3*d^3*e^8)/8 + (351*b^2*c^2*d^2*e^9)/32 - 
(27*b^3*c*d*e^10)/32)) + (27*c*d*e^9*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1/2) 
)/(32*((351*c^2*d^2*e^9)/32 - (27*b*c*d*e^10)/32 - (189*c^3*d^3*e^8)/(8*b) 
 + (27*c^4*d^4*e^7)/(2*b^2))))*(-c*(b*e - c*d))^(1/2)*(b^2*e^2 + 16*c^2*d^ 
2 - 12*b*c*d*e))/(4*b^5*c) - (3*d^(1/2)*atanh((135*c*d^(1/2)*e^10*(d + e*x 
)^(1/2))/(32*((135*c*d*e^10)/32 - (675*c^2*d^2*e^9)/(32*b) + (243*c^3*d^3* 
e^8)/(8*b^2) - (27*c^4*d^4*e^7)/(2*b^3))) + (675*c^2*d^(3/2)*e^9*(d + e*x) 
^(1/2))/(32*((675*c^2*d^2*e^9)/32 - (135*b*c*d*e^10)/32 - (243*c^3*d^3*e^8 
)/(8*b) + (27*c^4*d^4*e^7)/(2*b^2))) + (243*c^3*d^(5/2)*e^8*(d + e*x)^(1/2 
))/(8*((243*c^3*d^3*e^8)/8 - (675*b*c^2*d^2*e^9)/32 - (27*c^4*d^4*e^7)/(2* 
b) + (135*b^2*c*d*e^10)/32)) + (27*c^4*d^(7/2)*e^7*(d + e*x)^(1/2))/(2*((2 
7*c^4*d^4*e^7)/2 - (243*b*c^3*d^3*e^8)/8 + (675*b^2*c^2*d^2*e^9)/32 - (135 
*b^3*c*d*e^10)/32)))*(5*b^2*e^2 + 16*c^2*d^2 - 20*b*c*d*e))/(4*b^5) - (((d 
 + e*x)^(3/2)*(19*b^3*d*e^4 - 72*c^3*d^4*e + 144*b*c^2*d^3*e^2 - 91*b^2*c* 
d^2*e^3))/(4*b^4) + (3*(d + e*x)^(1/2)*(2*c^3*d^5*e - b^3*d^2*e^4 - 5*b*c^ 
2*d^4*e^2 + 4*b^2*c*d^3*e^3))/b^4 - ((b*e - 2*c*d)*(d + e*x)^(5/2)*(5*b^2* 
e^3 + 36*c^2*d^2*e - 36*b*c*d*e^2))/(4*b^4) - (3*c*e*(d + e*x)^(7/2)*(b...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1078, normalized size of antiderivative = 3.85 \[ \int \frac {(d+e x)^{5/2}}{\left (b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((e*x+d)^(5/2)/(c*x^2+b*x)^3,x)
 

Output:

(6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d) 
))*b**4*e**2*x**2 - 72*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqr 
t(c)*sqrt(b*e - c*d)))*b**3*c*d*e*x**2 + 12*sqrt(c)*sqrt(b*e - c*d)*atan(( 
sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c*e**2*x**3 + 96*sqrt(c)* 
sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c** 
2*d**2*x**2 - 144*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)* 
sqrt(b*e - c*d)))*b**2*c**2*d*e*x**3 + 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqr 
t(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c**2*e**2*x**4 + 192*sqrt(c) 
*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b*c**3* 
d**2*x**3 - 72*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqr 
t(b*e - c*d)))*b*c**3*d*e*x**4 + 96*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + 
 e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*c**4*d**2*x**4 - 4*sqrt(d + e*x)*b**4* 
c*d**2 - 18*sqrt(d + e*x)*b**4*c*d*e*x + 10*sqrt(d + e*x)*b**4*c*e**2*x**2 
 + 16*sqrt(d + e*x)*b**3*c**2*d**2*x - 74*sqrt(d + e*x)*b**3*c**2*d*e*x**2 
 + 6*sqrt(d + e*x)*b**3*c**2*e**2*x**3 + 72*sqrt(d + e*x)*b**2*c**3*d**2*x 
**2 - 48*sqrt(d + e*x)*b**2*c**3*d*e*x**3 + 48*sqrt(d + e*x)*b*c**4*d**2*x 
**3 + 15*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**4*c*e**2*x**2 - 60*sqrt(d 
)*log(sqrt(d + e*x) - sqrt(d))*b**3*c**2*d*e*x**2 + 30*sqrt(d)*log(sqrt(d 
+ e*x) - sqrt(d))*b**3*c**2*e**2*x**3 + 48*sqrt(d)*log(sqrt(d + e*x) - sqr 
t(d))*b**2*c**3*d**2*x**2 - 120*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b*...