\(\int \frac {(d+e x)^{3/2}}{(b x+c x^2)^3} \, dx\) [125]

Optimal result

Integrand size = 21, antiderivative size = 252 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx=\frac {c (12 c d-7 b e) \sqrt {d+e x}}{4 b^3 (b+c x)^2}-\frac {d \sqrt {d+e x}}{2 b x^2 (b+c x)^2}+\frac {(8 c d-5 b e) \sqrt {d+e x}}{4 b^2 x (b+c x)^2}+\frac {3 c (2 c d-b e) \sqrt {d+e x}}{b^4 (b+c x)}-\frac {3 \left (16 c^2 d^2-12 b c d e+b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 b^5 \sqrt {d}}+\frac {3 \sqrt {c} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{4 b^5 \sqrt {c d-b e}} \] Output:

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

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx=\frac {\frac {b \sqrt {d+e x} \left (24 c^3 d x^3+b^2 c x (8 d-19 e x)-12 b c^2 x^2 (-3 d+e x)-b^3 (2 d+5 e x)\right )}{x^2 (b+c x)^2}-\frac {3 \sqrt {c} \left (16 c^2 d^2-20 b c d e+5 b^2 e^2\right ) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{\sqrt {-c d+b e}}-\frac {3 \left (16 c^2 d^2-12 b c d e+b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{4 b^5} \] Input:

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

Output:

((b*Sqrt[d + e*x]*(24*c^3*d*x^3 + b^2*c*x*(8*d - 19*e*x) - 12*b*c^2*x^2*(- 
3*d + e*x) - b^3*(2*d + 5*e*x)))/(x^2*(b + c*x)^2) - (3*Sqrt[c]*(16*c^2*d^ 
2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b 
*e]])/Sqrt[-(c*d) + b*e] - (3*(16*c^2*d^2 - 12*b*c*d*e + b^2*e^2)*ArcTanh[ 
Sqrt[d + e*x]/Sqrt[d]])/Sqrt[d])/(4*b^5)
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1164, 27, 1235, 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.

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

\(\Big \downarrow \) 1164

\(\displaystyle -\frac {\int \frac {d (12 c d-7 b e)+5 e (2 c d-b e) x}{2 \sqrt {d+e x} \left (c x^2+b x\right )^2}dx}{2 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {d (12 c d-7 b e)+5 e (2 c d-b e) x}{\sqrt {d+e x} \left (c x^2+b x\right )^2}dx}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1235

\(\displaystyle -\frac {-\frac {\int \frac {3 d (c d-b e) \left (16 c^2 d^2-12 b c e d+b^2 e^2+4 c e (2 c d-b e) x\right )}{2 \sqrt {d+e x} \left (c x^2+b x\right )}dx}{b^2 d (c d-b e)}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {3 \int \frac {16 c^2 d^2-12 b c e d+b^2 e^2+4 c e (2 c d-b e) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{2 b^2}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1197

\(\displaystyle -\frac {-\frac {3 \int \frac {e \left (8 c^2 d^2-8 b c e d+b^2 e^2+4 c (2 c d-b e) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{b^2}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {3 e \int \frac {8 c^2 d^2-8 b c e d+b^2 e^2+4 c (2 c d-b e) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{b^2}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {-\frac {3 e \left (\frac {c \left (b^2 e^2-12 b c d e+16 c^2 d^2\right ) \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}-\frac {c \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}\right )}{b^2}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {-\frac {3 e \left (\frac {\sqrt {c} \left (5 b^2 e^2-20 b c d e+16 c^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b e \sqrt {c d-b e}}-\frac {\text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (b^2 e^2-12 b c d e+16 c^2 d^2\right )}{b \sqrt {d} e}\right )}{b^2}-\frac {\sqrt {d+e x} (12 c x (2 c d-b e) (c d-b e)+b (12 c d-7 b e) (c d-b e))}{b^2 \left (b x+c x^2\right ) (c d-b e)}}{4 b^2}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{2 b^2 \left (b x+c x^2\right )^2}\)

Input:

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

Output:

-1/2*(Sqrt[d + e*x]*(b*d + (2*c*d - b*e)*x))/(b^2*(b*x + c*x^2)^2) - (-((S 
qrt[d + e*x]*(b*(12*c*d - 7*b*e)*(c*d - b*e) + 12*c*(c*d - b*e)*(2*c*d - b 
*e)*x))/(b^2*(c*d - b*e)*(b*x + c*x^2))) - (3*e*(-(((16*c^2*d^2 - 12*b*c*d 
*e + b^2*e^2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d]*e)) + (Sqrt[c]*(1 
6*c^2*d^2 - 20*b*c*d*e + 5*b^2*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c 
*d - b*e]])/(b*e*Sqrt[c*d - b*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 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a 
+ b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] 
 + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^m 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -1] && (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 = 0.63 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {12 \left (x^{2} \left (c x +b \right )^{2} c \left (\frac {5 b^{2} e^{2} \sqrt {d}}{16}+c \left (c d -\frac {5 b e}{4}\right ) d^{\frac {3}{2}}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+\frac {\left (\frac {3 x^{2} \left (c x +b \right )^{2} \left (b^{2} e^{2}-12 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2}+\left (\frac {5 \left (3 c x +b \right ) e x b \left (\frac {4 c x}{5}+b \right ) \sqrt {d}}{2}+\left (-6 c^{2} x^{2}-6 c b x +b^{2}\right ) d^{\frac {3}{2}} \left (2 c x +b \right )\right ) \sqrt {e x +d}\, b \right ) \sqrt {c \left (b e -c d \right )}}{24}\right )}{\sqrt {d}\, \sqrt {c \left (b e -c d \right )}\, b^{5} x^{2} \left (c x +b \right )^{2}}\)	\(209\)
risch	\(-\frac {\sqrt {e x +d}\, \left (5 b e x -12 c d x +2 b d \right )}{4 b^{4} x^{2}}-\frac {e \left (-\frac {\left (-3 b^{2} e^{2}+36 b c d e -48 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}+\frac {8 c \left (\frac {\left (\frac {7}{8} e^{2} b^{2} c -\frac {3}{2} c^{2} d e b \right ) \left (e x +d \right )^{\frac {3}{2}}+\frac {3 b e \left (3 b^{2} e^{2}-7 b c d e +4 c^{2} d^{2}\right ) \sqrt {e x +d}}{8}}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}+\frac {3 \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{8 \sqrt {c \left (b e -c d \right )}}\right )}{b e}\right )}{4 b^{4}}\)	\(233\)
derivativedivides	\(2 e^{5} \left (-\frac {\frac {\frac {b e \left (5 b e -12 c d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {3}{2} d^{2} e b c -\frac {3}{8} d \,e^{2} b^{2}\right ) \sqrt {e x +d}}{e^{2} x^{2}}+\frac {3 \left (b^{2} e^{2}-12 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{8 \sqrt {d}}}{b^{5} e^{5}}-\frac {c \left (\frac {\left (\frac {7}{8} e^{2} b^{2} c -\frac {3}{2} c^{2} d e b \right ) \left (e x +d \right )^{\frac {3}{2}}+\frac {3 b e \left (3 b^{2} e^{2}-7 b c d e +4 c^{2} d^{2}\right ) \sqrt {e x +d}}{8}}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}+\frac {3 \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{8 \sqrt {c \left (b e -c d \right )}}\right )}{b^{5} e^{5}}\right )\)	\(257\)
default	\(2 e^{5} \left (-\frac {\frac {\frac {b e \left (5 b e -12 c d \right ) \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {3}{2} d^{2} e b c -\frac {3}{8} d \,e^{2} b^{2}\right ) \sqrt {e x +d}}{e^{2} x^{2}}+\frac {3 \left (b^{2} e^{2}-12 b c d e +16 c^{2} d^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{8 \sqrt {d}}}{b^{5} e^{5}}-\frac {c \left (\frac {\left (\frac {7}{8} e^{2} b^{2} c -\frac {3}{2} c^{2} d e b \right ) \left (e x +d \right )^{\frac {3}{2}}+\frac {3 b e \left (3 b^{2} e^{2}-7 b c d e +4 c^{2} d^{2}\right ) \sqrt {e x +d}}{8}}{\left (\left (e x +d \right ) c +b e -c d \right )^{2}}+\frac {3 \left (5 b^{2} e^{2}-20 b c d e +16 c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{8 \sqrt {c \left (b e -c d \right )}}\right )}{b^{5} e^{5}}\right )\)	\(257\)

Input:

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

Output:

-12/d^(1/2)*(x^2*(c*x+b)^2*c*(5/16*b^2*e^2*d^(1/2)+c*(c*d-5/4*b*e)*d^(3/2) 
)*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2))+1/24*(3/2*x^2*(c*x+b)^2*(b^2 
*e^2-12*b*c*d*e+16*c^2*d^2)*arctanh((e*x+d)^(1/2)/d^(1/2))+(5/2*(3*c*x+b)* 
e*x*b*(4/5*c*x+b)*d^(1/2)+(-6*c^2*x^2-6*b*c*x+b^2)*d^(3/2)*(2*c*x+b))*(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 = 0.18 (sec) , antiderivative size = 1628, normalized size of antiderivative = 6.46 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/8*(3*((16*c^4*d^3 - 20*b*c^3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^3 
*d^3 - 20*b^2*c^2*d^2*e + 5*b^3*c*d*e^2)*x^3 + (16*b^2*c^2*d^3 - 20*b^3*c* 
d^2*e + 5*b^4*d*e^2)*x^2)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2 
*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + 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(d)*log 
((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 2*(2*b^4*d^2 - 12*(2*b*c^3*d^2 
 - b^2*c^2*d*e)*x^3 - (36*b^2*c^2*d^2 - 19*b^3*c*d*e)*x^2 - (8*b^3*c*d^2 - 
 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x^4 + 2*b^6*c*d*x^3 + b^7*d*x^2), 
 -1/8*(6*((16*c^4*d^3 - 20*b*c^3*d^2*e + 5*b^2*c^2*d*e^2)*x^4 + 2*(16*b*c^ 
3*d^3 - 20*b^2*c^2*d^2*e + 5*b^3*c*d*e^2)*x^3 + (16*b^2*c^2*d^3 - 20*b^3*c 
*d^2*e + 5*b^4*d*e^2)*x^2)*sqrt(-c/(c*d - b*e))*arctan(sqrt(e*x + d)*sqrt( 
-c/(c*d - b*e))) - 3*((16*c^4*d^2 - 12*b*c^3*d*e + b^2*c^2*e^2)*x^4 + 2*(1 
6*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(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) 
+ 2*(2*b^4*d^2 - 12*(2*b*c^3*d^2 - b^2*c^2*d*e)*x^3 - (36*b^2*c^2*d^2 - 19 
*b^3*c*d*e)*x^2 - (8*b^3*c*d^2 - 5*b^4*d*e)*x)*sqrt(e*x + d))/(b^5*c^2*d*x 
^4 + 2*b^6*c*d*x^3 + b^7*d*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(-d)*arctan(sqrt(-d)/sqrt(e...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((e*x+d)^(3/2)/(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 = 0.27 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.49 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx=-\frac {3 \, {\left (16 \, c^{3} d^{2} - 20 \, b c^{2} d e + 5 \, b^{2} c e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{4 \, \sqrt {-c^{2} d + b c e} b^{5}} + \frac {3 \, {\left (16 \, c^{2} d^{2} - 12 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{4 \, b^{5} \sqrt {-d}} + \frac {24 \, {\left (e x + d\right )}^{\frac {7}{2}} c^{3} d e - 72 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{3} d^{2} e + 72 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{3} d^{3} e - 24 \, \sqrt {e x + d} c^{3} d^{4} e - 12 \, {\left (e x + d\right )}^{\frac {7}{2}} b c^{2} e^{2} + 72 \, {\left (e x + d\right )}^{\frac {5}{2}} b c^{2} d e^{2} - 108 \, {\left (e x + d\right )}^{\frac {3}{2}} b c^{2} d^{2} e^{2} + 48 \, \sqrt {e x + d} b c^{2} d^{3} e^{2} - 19 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} c e^{3} + 46 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} c d e^{3} - 27 \, \sqrt {e x + d} b^{2} c d^{2} e^{3} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} e^{4} + 3 \, \sqrt {e x + d} b^{3} d e^{4}}{4 \, {\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )}^{2} b^{4}} \] Input:

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

Output:

-3/4*(16*c^3*d^2 - 20*b*c^2*d*e + 5*b^2*c*e^2)*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^2 - 12*b*c*d* 
e + b^2*e^2)*arctan(sqrt(e*x + d)/sqrt(-d))/(b^5*sqrt(-d)) + 1/4*(24*(e*x 
+ d)^(7/2)*c^3*d*e - 72*(e*x + d)^(5/2)*c^3*d^2*e + 72*(e*x + d)^(3/2)*c^3 
*d^3*e - 24*sqrt(e*x + d)*c^3*d^4*e - 12*(e*x + d)^(7/2)*b*c^2*e^2 + 72*(e 
*x + d)^(5/2)*b*c^2*d*e^2 - 108*(e*x + d)^(3/2)*b*c^2*d^2*e^2 + 48*sqrt(e* 
x + d)*b*c^2*d^3*e^2 - 19*(e*x + d)^(5/2)*b^2*c*e^3 + 46*(e*x + d)^(3/2)*b 
^2*c*d*e^3 - 27*sqrt(e*x + d)*b^2*c*d^2*e^3 - 5*(e*x + d)^(3/2)*b^3*e^4 + 
3*sqrt(e*x + d)*b^3*d*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 = 5.73 (sec) , antiderivative size = 1880, normalized size of antiderivative = 7.46 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((3*(d + e*x)^(1/2)*(b^3*d*e^4 - 8*c^3*d^4*e + 16*b*c^2*d^3*e^2 - 9*b^2*c* 
d^2*e^3))/(4*b^4) - ((d + e*x)^(3/2)*(5*b^3*e^4 - 72*c^3*d^3*e + 108*b*c^2 
*d^2*e^2 - 46*b^2*c*d*e^3))/(4*b^4) - (e*(d + e*x)^(5/2)*(72*c^3*d^2 + 19* 
b^2*c*e^2 - 72*b*c^2*d*e))/(4*b^4) + (3*c*e*(2*c^2*d - b*c*e)*(d + e*x)^(7 
/2))/b^4)/(c^2*(d + e*x)^4 - (d + e*x)*(4*c^2*d^3 + 2*b^2*d*e^2 - 6*b*c*d^ 
2*e) - (4*c^2*d - 2*b*c*e)*(d + e*x)^3 + (d + e*x)^2*(b^2*e^2 + 6*c^2*d^2 
- 6*b*c*d*e) + c^2*d^4 + b^2*d^2*e^2 - 2*b*c*d^3*e) - (3*atanh((27*c^2*e^9 
*(d + e*x)^(1/2))/(32*d^(3/2)*((27*c^2*e^9)/(32*d) - (81*c^3*e^8)/(8*b) + 
(27*c^4*d*e^7)/(2*b^2))) - (81*c^3*e^8*(d + e*x)^(1/2))/(8*d^(1/2)*((27*b* 
c^2*e^9)/(32*d) - (81*c^3*e^8)/8 + (27*c^4*d*e^7)/(2*b))) + (27*c^4*d^(1/2 
)*e^7*(d + e*x)^(1/2))/(2*((27*c^4*d*e^7)/2 - (81*b*c^3*e^8)/8 + (27*b^2*c 
^2*e^9)/(32*d))))*(b^2*e^2 + 16*c^2*d^2 - 12*b*c*d*e))/(4*b^5*d^(1/2)) + ( 
atan((((((d + e*x)^(1/2)*(117*b^4*c^3*e^6 + 2304*c^7*d^4*e^2 - 4608*b*c^6* 
d^3*e^3 - 1008*b^3*c^4*d*e^5 + 3312*b^2*c^5*d^2*e^4))/(4*b^8) - (3*(-c*(b* 
e - c*d))^(1/2)*((3*b^12*c^2*e^5 - 24*b^11*c^3*d*e^4 + 24*b^10*c^4*d^2*e^3 
)/b^12 - (3*(32*b^11*c^2*e^3 - 64*b^10*c^3*d*e^2)*(-c*(b*e - c*d))^(1/2)*( 
d + e*x)^(1/2)*(5*b^2*e^2 + 16*c^2*d^2 - 20*b*c*d*e))/(32*b^8*(b^6*e - b^5 
*c*d)))*(5*b^2*e^2 + 16*c^2*d^2 - 20*b*c*d*e))/(8*(b^6*e - b^5*c*d)))*(-c* 
(b*e - c*d))^(1/2)*(5*b^2*e^2 + 16*c^2*d^2 - 20*b*c*d*e)*3i)/(8*(b^6*e - b 
^5*c*d)) + ((((d + e*x)^(1/2)*(117*b^4*c^3*e^6 + 2304*c^7*d^4*e^2 - 460...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1327, normalized size of antiderivative = 5.27 \[ \int \frac {(d+e x)^{3/2}}{\left (b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 30*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - 
c*d)))*b**4*d*e**2*x**2 + 120*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)* 
c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c*d**2*e*x**2 - 60*sqrt(c)*sqrt(b*e - c 
*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c*d*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**3*x**2 + 240*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*e*x**3 - 30*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**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**3*x**3 + 120*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*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**3*x**4 - 
 4*sqrt(d + e*x)*b**5*d**2*e - 10*sqrt(d + e*x)*b**5*d*e**2*x + 4*sqrt(d + 
 e*x)*b**4*c*d**3 + 26*sqrt(d + e*x)*b**4*c*d**2*e*x - 38*sqrt(d + e*x)*b* 
*4*c*d*e**2*x**2 - 16*sqrt(d + e*x)*b**3*c**2*d**3*x + 110*sqrt(d + e*x)*b 
**3*c**2*d**2*e*x**2 - 24*sqrt(d + e*x)*b**3*c**2*d*e**2*x**3 - 72*sqrt(d 
+ e*x)*b**2*c**3*d**3*x**2 + 72*sqrt(d + e*x)*b**2*c**3*d**2*e*x**3 - 48*s 
qrt(d + e*x)*b*c**4*d**3*x**3 + 3*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b** 
5*e**3*x**2 - 39*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**4*c*d*e**2*x**2 + 
 6*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b**4*c*e**3*x**3 + 84*sqrt(d)*l...