\(\int \frac {(d+e x)^4}{(b x+c x^2)^{5/2}} \, dx\) [171]

Optimal result

Integrand size = 21, antiderivative size = 234 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^4}{3 b \left (b x+c x^2\right )^{3/2}}+\frac {2 \left (2 c^4 d^4-4 b c^3 d^3 e+6 b^2 c^2 d^2 e^2-4 b^3 c d e^3+b^4 e^4\right ) x^2}{3 b^3 c^2 \left (b x+c x^2\right )^{3/2}}+\frac {4 d^3 (c d-2 b e)}{b^3 \sqrt {b x+c x^2}}+\frac {8 \left (4 c^4 d^4-8 b c^3 d^3 e+3 b^2 c^2 d^2 e^2+b^3 c d e^3-b^4 e^4\right ) x}{3 b^4 c^2 \sqrt {b x+c x^2}}+\frac {2 e^4 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}} \] Output:

-2/3*d^4/b/(c*x^2+b*x)^(3/2)+2/3*(b^4*e^4-4*b^3*c*d*e^3+6*b^2*c^2*d^2*e^2- 
4*b*c^3*d^3*e+2*c^4*d^4)*x^2/b^3/c^2/(c*x^2+b*x)^(3/2)+4*d^3*(-2*b*e+c*d)/ 
b^3/(c*x^2+b*x)^(1/2)+8/3*(-b^4*e^4+b^3*c*d*e^3+3*b^2*c^2*d^2*e^2-8*b*c^3* 
d^3*e+4*c^4*d^4)*x/b^4/c^2/(c*x^2+b*x)^(1/2)+2*e^4*arctanh(c^(1/2)*x/(c*x^ 
2+b*x)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.83 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \left (\sqrt {c} \left (3 b^5 e^4 x^2-16 c^5 d^4 x^3+4 b^4 c e^4 x^3+8 b c^4 d^3 x^2 (-3 d+4 e x)-6 b^2 c^3 d^2 x \left (d^2-8 d e x+2 e^2 x^2\right )+b^3 c^2 d \left (d^3+12 d^2 e x-18 d e^2 x^2-4 e^3 x^3\right )\right )+3 b^4 e^4 x^{3/2} (b+c x)^{3/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{3 b^4 c^{5/2} (x (b+c x))^{3/2}} \] Input:

Integrate[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]
 

Output:

(-2*(Sqrt[c]*(3*b^5*e^4*x^2 - 16*c^5*d^4*x^3 + 4*b^4*c*e^4*x^3 + 8*b*c^4*d 
^3*x^2*(-3*d + 4*e*x) - 6*b^2*c^3*d^2*x*(d^2 - 8*d*e*x + 2*e^2*x^2) + b^3* 
c^2*d*(d^3 + 12*d^2*e*x - 18*d*e^2*x^2 - 4*e^3*x^3)) + 3*b^4*e^4*x^(3/2)*( 
b + c*x)^(3/2)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(3*b^4*c^(5/2)*(x 
*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1164, 1233, 27, 1160, 1091, 219}

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.

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

\(\Big \downarrow \) 1164

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

\(\Big \downarrow \) 1233

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

Int[(d + e*x)^4/(b*x + c*x^2)^(5/2),x]
 

Output:

(-2*(d + e*x)^3*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) - (2* 
((-2*(d + e*x)*(b*c*d^2*(4*c*d - 5*b*e) + (2*c*d - b*e)*(4*c^2*d^2 - 4*b*c 
*d*e - b^2*e^2)*x))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(((2*c*d - b*e)*(8*c^2* 
d^2 - 8*b*c*d*e - 3*b^2*e^2)*Sqrt[b*x + c*x^2])/c - (3*b^4*e^3*ArcTanh[(Sq 
rt[c]*x)/Sqrt[b*x + c*x^2]])/c^(3/2)))/(b^2*c)))/(3*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1160

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

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 1233

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

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {-\frac {2 b^{3} d \left (-4 e^{3} x^{3}-18 d \,e^{2} x^{2}+12 d^{2} e x +d^{3}\right ) c^{\frac {5}{2}}}{3}+2 x \left (2 b^{2} d^{2} \left (2 e^{2} x^{2}-8 d e x +d^{2}\right ) c^{\frac {7}{2}}+8 x \left (-\frac {4 e x}{3}+d \right ) b \,d^{3} c^{\frac {9}{2}}+\frac {16 c^{\frac {11}{2}} d^{4} x^{2}}{3}+e^{4} \left (-\frac {4 c^{\frac {3}{2}} x^{2}}{3}-b x \sqrt {c}+\sqrt {x \left (c x +b \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) \left (c x +b \right )\right ) b^{4}\right )}{c^{\frac {5}{2}} x \left (c x +b \right ) \sqrt {x \left (c x +b \right )}\, b^{4}}\)	\(179\)
risch	\(-\frac {2 d^{3} \left (c x +b \right ) \left (12 b e x -8 c d x +b d \right )}{3 b^{4} x \sqrt {x \left (c x +b \right )}}+\frac {\frac {e^{4} b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {5}{2}}}-\frac {4 \left (b^{4} e^{4}-2 d \,e^{3} b^{3} c +2 d^{3} e b \,c^{3}-d^{4} c^{4}\right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c^{3} b \left (\frac {b}{c}+x \right )}+\frac {b \left (b^{4} e^{4}-4 d \,e^{3} b^{3} c +6 d^{2} e^{2} b^{2} c^{2}-4 d^{3} e b \,c^{3}+d^{4} c^{4}\right ) \left (\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b \left (\frac {b}{c}+x \right )^{2}}+\frac {4 c \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b^{2} \left (\frac {b}{c}+x \right )}\right )}{c^{4}}}{b^{3}}\)	\(289\)
default	\(d^{4} \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )+e^{4} \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )}{2 c}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}}{c}\right )+4 d \,e^{3} \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )+6 d^{2} e^{2} \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )+4 d^{3} e \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )\)	\(599\)

Input:

int((e*x+d)^4/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/c^(5/2)/(x*(c*x+b))^(1/2)*(-1/3*b^3*d*(-4*e^3*x^3-18*d*e^2*x^2+12*d^2*e* 
x+d^3)*c^(5/2)+x*(2*b^2*d^2*(2*e^2*x^2-8*d*e*x+d^2)*c^(7/2)+8*x*(-4/3*e*x+ 
d)*b*d^3*c^(9/2)+16/3*c^(11/2)*d^4*x^2+e^4*(-4/3*c^(3/2)*x^2-b*x*c^(1/2)+( 
x*(c*x+b))^(1/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*(c*x+b))*b^4))/x/(c* 
x+b)/b^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 526, normalized size of antiderivative = 2.25 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (b^{3} c^{3} d^{4} - 4 \, {\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \, {\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{4} c^{2} e^{4} x^{4} + 2 \, b^{5} c e^{4} x^{3} + b^{6} e^{4} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (b^{3} c^{3} d^{4} - 4 \, {\left (4 \, c^{6} d^{4} - 8 \, b c^{5} d^{3} e + 3 \, b^{2} c^{4} d^{2} e^{2} + b^{3} c^{3} d e^{3} - b^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (8 \, b c^{5} d^{4} - 16 \, b^{2} c^{4} d^{3} e + 6 \, b^{3} c^{3} d^{2} e^{2} - b^{5} c e^{4}\right )} x^{2} - 6 \, {\left (b^{2} c^{4} d^{4} - 2 \, b^{3} c^{3} d^{3} e\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{3 \, {\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}\right ] \] Input:

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
 

Output:

[1/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x^2)*sqrt(c)*log(2*c* 
x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(b^3*c^3*d^4 - 4*(4*c^6*d^4 - 8*b 
*c^5*d^3*e + 3*b^2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^2*e^4)*x^3 - 3*(8*b 
*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5*c*e^4)*x^2 - 6*(b^2* 
c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4* 
x^3 + b^6*c^3*x^2), -2/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x 
^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x + b)) + (b^3*c^3*d^4 - 
 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*b^2*c^4*d^2*e^2 + b^3*c^3*d*e^3 - b^4*c^ 
2*e^4)*x^3 - 3*(8*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*b^3*c^3*d^2*e^2 - b^5*c 
*e^4)*x^2 - 6*(b^2*c^4*d^4 - 2*b^3*c^3*d^3*e)*x)*sqrt(c*x^2 + b*x))/(b^4*c 
^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x^2)]
 

Sympy [F]

\[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x+d)**4/(c*x**2+b*x)**(5/2),x)
 

Output:

Integral((d + e*x)**4/(x*(b + c*x))**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (214) = 428\).

Time = 0.04 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.96 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, e^{4} x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )} - \frac {4 \, d e^{3} x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, c d^{4} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, c^{2} d^{4} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {8 \, d^{3} e x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {64 \, c d^{3} e x}{3 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {8 \, d^{2} e^{2} x}{\sqrt {c x^{2} + b x} b^{2}} - \frac {4 \, d^{2} e^{2} x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {4 \, b d e^{3} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {8 \, d e^{3} x}{3 \, \sqrt {c x^{2} + b x} b c} - \frac {4 \, e^{4} x}{3 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {e^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {2 \, d^{4}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, c d^{4}}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {32 \, d^{3} e}{3 \, \sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, d^{2} e^{2}}{\sqrt {c x^{2} + b x} b c} + \frac {4 \, d e^{3}}{3 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} e^{4}}{3 \, b c^{2}} \] Input:

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
 

Output:

-1/3*e^4*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) + b*x/((c*x^2 + b*x)^(3/2)*c^2) 
- 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/(sqrt(c*x^2 + b*x)*c^2)) - 4*d*e^3*x^2/( 
(c*x^2 + b*x)^(3/2)*c) - 4/3*c*d^4*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^2* 
d^4*x/(sqrt(c*x^2 + b*x)*b^4) + 8/3*d^3*e*x/((c*x^2 + b*x)^(3/2)*b) - 64/3 
*c*d^3*e*x/(sqrt(c*x^2 + b*x)*b^3) + 8*d^2*e^2*x/(sqrt(c*x^2 + b*x)*b^2) - 
 4*d^2*e^2*x/((c*x^2 + b*x)^(3/2)*c) - 4/3*b*d*e^3*x/((c*x^2 + b*x)^(3/2)* 
c^2) + 8/3*d*e^3*x/(sqrt(c*x^2 + b*x)*b*c) - 4/3*e^4*x/(sqrt(c*x^2 + b*x)* 
c^2) + e^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 2/3*d^4/ 
((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d^4/(sqrt(c*x^2 + b*x)*b^3) - 32/3*d^3*e/ 
(sqrt(c*x^2 + b*x)*b^2) + 4*d^2*e^2/(sqrt(c*x^2 + b*x)*b*c) + 4/3*d*e^3/(s 
qrt(c*x^2 + b*x)*c^2) - 2/3*sqrt(c*x^2 + b*x)*e^4/(b*c^2)
 

Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {e^{4} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} - \frac {2 \, {\left (\frac {d^{4}}{b} - {\left (x {\left (\frac {4 \, {\left (4 \, c^{5} d^{4} - 8 \, b c^{4} d^{3} e + 3 \, b^{2} c^{3} d^{2} e^{2} + b^{3} c^{2} d e^{3} - b^{4} c e^{4}\right )} x}{b^{4} c^{2}} + \frac {3 \, {\left (8 \, b c^{4} d^{4} - 16 \, b^{2} c^{3} d^{3} e + 6 \, b^{3} c^{2} d^{2} e^{2} - b^{5} e^{4}\right )}}{b^{4} c^{2}}\right )} + \frac {6 \, {\left (b^{2} c^{3} d^{4} - 2 \, b^{3} c^{2} d^{3} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \] Input:

integrate((e*x+d)^4/(c*x^2+b*x)^(5/2),x, algorithm="giac")
 

Output:

-e^4*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2) - 2/3 
*(d^4/b - (x*(4*(4*c^5*d^4 - 8*b*c^4*d^3*e + 3*b^2*c^3*d^2*e^2 + b^3*c^2*d 
*e^3 - b^4*c*e^4)*x/(b^4*c^2) + 3*(8*b*c^4*d^4 - 16*b^2*c^3*d^3*e + 6*b^3* 
c^2*d^2*e^2 - b^5*e^4)/(b^4*c^2)) + 6*(b^2*c^3*d^4 - 2*b^3*c^2*d^3*e)/(b^4 
*c^2))*x)/(c*x^2 + b*x)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \] Input:

int((d + e*x)^4/(b*x + c*x^2)^(5/2),x)
 

Output:

int((d + e*x)^4/(b*x + c*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.15 \[ \int \frac {(d+e x)^4}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{5} e^{4} x^{2}+2 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{4} c \,e^{4} x^{3}-\frac {8 \sqrt {c}\, \sqrt {c x +b}\, b^{5} e^{4} x^{2}}{3}+\frac {40 \sqrt {c}\, \sqrt {c x +b}\, b^{4} c d \,e^{3} x^{2}}{3}-\frac {8 \sqrt {c}\, \sqrt {c x +b}\, b^{4} c \,e^{4} x^{3}}{3}-24 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{2} d^{2} e^{2} x^{2}+\frac {40 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{2} d \,e^{3} x^{3}}{3}+\frac {64 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} d^{3} e \,x^{2}}{3}-24 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} d^{2} e^{2} x^{3}-\frac {32 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{4} d^{4} x^{2}}{3}+\frac {64 \sqrt {c}\, \sqrt {c x +b}\, b \,c^{4} d^{3} e \,x^{3}}{3}-\frac {32 \sqrt {c}\, \sqrt {c x +b}\, c^{5} d^{4} x^{3}}{3}-2 \sqrt {x}\, b^{5} c \,e^{4} x^{2}-\frac {8 \sqrt {x}\, b^{4} c^{2} e^{4} x^{3}}{3}-\frac {2 \sqrt {x}\, b^{3} c^{3} d^{4}}{3}-8 \sqrt {x}\, b^{3} c^{3} d^{3} e x +12 \sqrt {x}\, b^{3} c^{3} d^{2} e^{2} x^{2}+\frac {8 \sqrt {x}\, b^{3} c^{3} d \,e^{3} x^{3}}{3}+4 \sqrt {x}\, b^{2} c^{4} d^{4} x -32 \sqrt {x}\, b^{2} c^{4} d^{3} e \,x^{2}+8 \sqrt {x}\, b^{2} c^{4} d^{2} e^{2} x^{3}+16 \sqrt {x}\, b \,c^{5} d^{4} x^{2}-\frac {64 \sqrt {x}\, b \,c^{5} d^{3} e \,x^{3}}{3}+\frac {32 \sqrt {x}\, c^{6} d^{4} x^{3}}{3}}{\sqrt {c x +b}\, b^{4} c^{3} x^{2} \left (c x +b \right )} \] Input:

int((e*x+d)^4/(c*x^2+b*x)^(5/2),x)
 

Output:

(2*(3*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*b**5*e**4*x**2 + 3*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqr 
t(c))/sqrt(b))*b**4*c*e**4*x**3 - 4*sqrt(c)*sqrt(b + c*x)*b**5*e**4*x**2 + 
 20*sqrt(c)*sqrt(b + c*x)*b**4*c*d*e**3*x**2 - 4*sqrt(c)*sqrt(b + c*x)*b** 
4*c*e**4*x**3 - 36*sqrt(c)*sqrt(b + c*x)*b**3*c**2*d**2*e**2*x**2 + 20*sqr 
t(c)*sqrt(b + c*x)*b**3*c**2*d*e**3*x**3 + 32*sqrt(c)*sqrt(b + c*x)*b**2*c 
**3*d**3*e*x**2 - 36*sqrt(c)*sqrt(b + c*x)*b**2*c**3*d**2*e**2*x**3 - 16*s 
qrt(c)*sqrt(b + c*x)*b*c**4*d**4*x**2 + 32*sqrt(c)*sqrt(b + c*x)*b*c**4*d* 
*3*e*x**3 - 16*sqrt(c)*sqrt(b + c*x)*c**5*d**4*x**3 - 3*sqrt(x)*b**5*c*e** 
4*x**2 - 4*sqrt(x)*b**4*c**2*e**4*x**3 - sqrt(x)*b**3*c**3*d**4 - 12*sqrt( 
x)*b**3*c**3*d**3*e*x + 18*sqrt(x)*b**3*c**3*d**2*e**2*x**2 + 4*sqrt(x)*b* 
*3*c**3*d*e**3*x**3 + 6*sqrt(x)*b**2*c**4*d**4*x - 48*sqrt(x)*b**2*c**4*d* 
*3*e*x**2 + 12*sqrt(x)*b**2*c**4*d**2*e**2*x**3 + 24*sqrt(x)*b*c**5*d**4*x 
**2 - 32*sqrt(x)*b*c**5*d**3*e*x**3 + 16*sqrt(x)*c**6*d**4*x**3))/(3*sqrt( 
b + c*x)*b**4*c**3*x**2*(b + c*x))