\(\int \frac {(b x+c x^2)^{5/2}}{(d+e x)^4} \, dx\) [155]

Optimal result

Integrand size = 21, antiderivative size = 353 \[ \int \frac {\left (b x+c x^2\right )^{5/2}}{(d+e x)^4} \, dx=-\frac {5 \left (16 c^2 d^2-12 b c d e+b^2 e^2\right ) \sqrt {b x+c x^2}}{8 d e^5}+\frac {5 \left (24 c^2 d^2-16 b c d e+b^2 e^2\right ) x \sqrt {b x+c x^2}}{24 d^2 e^4}-\frac {5 \left (16 c^2 d^2-14 b c d e+b^2 e^2\right ) x^2 \sqrt {b x+c x^2}}{24 d^2 e^3 (d+e x)}-\frac {5 (2 c d-b e) x \left (b x+c x^2\right )^{3/2}}{12 d e^2 (d+e x)^2}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {5 \sqrt {c} (4 c d-3 b e) (4 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 e^6}-\frac {5 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{8 \sqrt {d} e^6 \sqrt {c d-b e}} \] Output:

-5/8*(b^2*e^2-12*b*c*d*e+16*c^2*d^2)*(c*x^2+b*x)^(1/2)/d/e^5+5/24*(b^2*e^2 
-16*b*c*d*e+24*c^2*d^2)*x*(c*x^2+b*x)^(1/2)/d^2/e^4-5/24*(b^2*e^2-14*b*c*d 
*e+16*c^2*d^2)*x^2*(c*x^2+b*x)^(1/2)/d^2/e^3/(e*x+d)-5/12*(-b*e+2*c*d)*x*( 
c*x^2+b*x)^(3/2)/d/e^2/(e*x+d)^2-1/3*(c*x^2+b*x)^(5/2)/e/(e*x+d)^3+5/4*c^( 
1/2)*(-3*b*e+4*c*d)*(-b*e+4*c*d)*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/e^6- 
5/8*(-b*e+2*c*d)*(b^2*e^2-16*b*c*d*e+16*c^2*d^2)*arctanh((-b*e+c*d)^(1/2)* 
x/d^(1/2)/(c*x^2+b*x)^(1/2))/d^(1/2)/e^6/(-b*e+c*d)^(1/2)
 

Mathematica [A] (verified)

Time = 11.58 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.96 \[ \int \frac {\left (b x+c x^2\right )^{5/2}}{(d+e x)^4} \, dx=\frac {\sqrt {x (b+c x)} \left (\frac {e (c d-b e) \sqrt {x} \left (b^2 e^2 \left (15 d^2+40 d e x+33 e^2 x^2\right )-2 b c e \left (90 d^3+230 d^2 e x+175 d e^2 x^2+27 e^3 x^3\right )+4 c^2 \left (60 d^4+150 d^3 e x+110 d^2 e^2 x^2+15 d e^3 x^3-3 e^4 x^4\right )\right )}{(d+e x)^3}+\frac {30 \sqrt {c} \left (-16 c^3 d^3+32 b c^2 d^2 e-19 b^2 c d e^2+3 b^3 e^3\right ) \text {arcsinh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {1+\frac {c x}{b}}}+\frac {15 \sqrt {c d-b e} \left (32 c^3 d^3-48 b c^2 d^2 e+18 b^2 c d e^2-b^3 e^3\right ) \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )}{\sqrt {d} \sqrt {b+c x}}\right )}{24 e^6 (-c d+b e) \sqrt {x}} \] Input:

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

Output:

(Sqrt[x*(b + c*x)]*((e*(c*d - b*e)*Sqrt[x]*(b^2*e^2*(15*d^2 + 40*d*e*x + 3 
3*e^2*x^2) - 2*b*c*e*(90*d^3 + 230*d^2*e*x + 175*d*e^2*x^2 + 27*e^3*x^3) + 
 4*c^2*(60*d^4 + 150*d^3*e*x + 110*d^2*e^2*x^2 + 15*d*e^3*x^3 - 3*e^4*x^4) 
))/(d + e*x)^3 + (30*Sqrt[c]*(-16*c^3*d^3 + 32*b*c^2*d^2*e - 19*b^2*c*d*e^ 
2 + 3*b^3*e^3)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (c*x) 
/b]) + (15*Sqrt[c*d - b*e]*(32*c^3*d^3 - 48*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 
 b^3*e^3)*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(Sqr 
t[d]*Sqrt[b + c*x])))/(24*e^6*(-(c*d) + b*e)*Sqrt[x])
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1161, 1230, 27, 1230, 1269, 1091, 219, 1154, 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.

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

\(\Big \downarrow \) 1161

\(\displaystyle \frac {5 \int \frac {(b+2 c x) \left (c x^2+b x\right )^{3/2}}{(d+e x)^3}dx}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 1230

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \int \frac {2 (b (4 c d-b e)+4 c (2 c d-b e) x) \sqrt {c x^2+b x}}{(d+e x)^2}dx}{8 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \int \frac {(b (4 c d-b e)+4 c (2 c d-b e) x) \sqrt {c x^2+b x}}{(d+e x)^2}dx}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 1230

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\int \frac {b \left (16 c^2 d^2-12 b c e d+b^2 e^2\right )+2 c (4 c d-3 b e) (4 c d-b e) x}{(d+e x) \sqrt {c x^2+b x}}dx}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\frac {2 c (4 c d-3 b e) (4 c d-b e) \int \frac {1}{\sqrt {c x^2+b x}}dx}{e}-\frac {(2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{e}}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\frac {4 c (4 c d-3 b e) (4 c d-b e) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{e}-\frac {(2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{e}}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (4 c d-3 b e) (4 c d-b e)}{e}-\frac {(2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{e}}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\frac {2 (2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{e}+\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (4 c d-3 b e) (4 c d-b e)}{e}}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 \left (\frac {\left (b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2}-\frac {3 \left (\frac {\sqrt {b x+c x^2} \left (b^2 e^2+4 c e x (2 c d-b e)-12 b c d e+16 c^2 d^2\right )}{e^2 (d+e x)}-\frac {\frac {4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (4 c d-3 b e) (4 c d-b e)}{e}-\frac {(2 c d-b e) \left (b^2 e^2-16 b c d e+16 c^2 d^2\right ) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{\sqrt {d} e \sqrt {c d-b e}}}{2 e^2}\right )}{4 e^2}\right )}{6 e}-\frac {\left (b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}\)

Input:

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

Output:

-1/3*(b*x + c*x^2)^(5/2)/(e*(d + e*x)^3) + (5*(((4*c*d - b*e + 2*c*e*x)*(b 
*x + c*x^2)^(3/2))/(2*e^2*(d + e*x)^2) - (3*(((16*c^2*d^2 - 12*b*c*d*e + b 
^2*e^2 + 4*c*e*(2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(e^2*(d + e*x)) - ((4*S 
qrt[c]*(4*c*d - 3*b*e)*(4*c*d - b*e)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2] 
])/e - ((2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + b^2*e^2)*ArcTanh[(b*d + ( 
2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(Sqrt[d]*e 
*Sqrt[c*d - b*e]))/(2*e^2)))/(4*e^2)))/(6*e)
 

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 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1161

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[p/(e*(m + 1))   Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^(p - 
 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && GtQ[p, 0] && (IntegerQ[p] || 
 LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, 
 c, d, e, m, p, x]
 

rule 1230

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

rule 1269

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

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (e x +d \right )^{3} \left (b^{3} e^{3} \sqrt {c}-18 b^{2} d \,e^{2} c^{\frac {3}{2}}+48 b \,d^{2} e \,c^{\frac {5}{2}}-32 c^{\frac {7}{2}} d^{3}\right ) \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )+\left (-6 \left (e x +d \right )^{3} \left (b e -4 c d \right ) c \left (b e -\frac {4 c d}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\left (\left (-\frac {4}{5} e^{4} x^{4}+16 d^{4}+\frac {88}{3} d^{2} e^{2} x^{2}+40 d^{3} e x +4 d \,e^{3} x^{3}\right ) c^{\frac {5}{2}}+e \left (\left (-\frac {70}{3} d \,e^{2} x^{2}-\frac {92}{3} d^{2} e x -\frac {18}{5} e^{3} x^{3}-12 d^{3}\right ) c^{\frac {3}{2}}+b e \sqrt {c}\, \left (\frac {8}{3} d e x +\frac {11}{5} e^{2} x^{2}+d^{2}\right )\right ) b \right ) e \sqrt {x \left (c x +b \right )}\right ) \sqrt {d \left (b e -c d \right )}\right )}{8 \sqrt {d \left (b e -c d \right )}\, \sqrt {c}\, e^{6} \left (e x +d \right )^{3}}\)	\(280\)
risch	\(\text {Expression too large to display}\)	\(1897\)
default	\(\text {Expression too large to display}\)	\(4062\)

Input:

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

Output:

-5/8/(d*(b*e-c*d))^(1/2)/c^(1/2)*((e*x+d)^3*(b^3*e^3*c^(1/2)-18*b^2*d*e^2* 
c^(3/2)+48*b*d^2*e*c^(5/2)-32*c^(7/2)*d^3)*arctan((x*(c*x+b))^(1/2)/x*d/(d 
*(b*e-c*d))^(1/2))+(-6*(e*x+d)^3*(b*e-4*c*d)*c*(b*e-4/3*c*d)*arctanh((x*(c 
*x+b))^(1/2)/x/c^(1/2))+((-4/5*e^4*x^4+16*d^4+88/3*d^2*e^2*x^2+40*d^3*e*x+ 
4*d*e^3*x^3)*c^(5/2)+e*((-70/3*d*e^2*x^2-92/3*d^2*e*x-18/5*e^3*x^3-12*d^3) 
*c^(3/2)+b*e*c^(1/2)*(8/3*d*e*x+11/5*e^2*x^2+d^2))*b)*e*(x*(c*x+b))^(1/2)) 
*(d*(b*e-c*d))^(1/2))/e^6/(e*x+d)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 769 vs. \(2 (313) = 626\).

Time = 0.28 (sec) , antiderivative size = 3097, normalized size of antiderivative = 8.77 \[ \int \frac {\left (b x+c x^2\right )^{5/2}}{(d+e x)^4} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((c*x^2+b*x)^(5/2)/(e*x+d)^4,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 977 vs. \(2 (313) = 626\).

Time = 0.25 (sec) , antiderivative size = 977, normalized size of antiderivative = 2.77 \[ \int \frac {\left (b x+c x^2\right )^{5/2}}{(d+e x)^4} \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*sqrt(c*x^2 + b*x)*(2*c^2*x/e^4 - (16*c^3*d*e^10 - 9*b*c^2*e^11)/(c*e^1 
5)) - 5/8*(32*c^3*d^3 - 48*b*c^2*d^2*e + 18*b^2*c*d*e^2 - b^3*e^3)*arctan( 
-((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/(sq 
rt(-c*d^2 + b*d*e)*e^6) - 5/8*(16*c^3*d^2 - 16*b*c^2*d*e + 3*b^2*c*e^2)*lo 
g(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/(sqrt(c)*e^6) - 1/24 
*(480*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*c^3*d^3*e^2 - 720*(sqrt(c)*x - sqr 
t(c*x^2 + b*x))^5*b*c^2*d^2*e^3 + 306*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^ 
2*c*d*e^4 - 33*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b^3*e^5 + 1680*(sqrt(c)*x 
 - sqrt(c*x^2 + b*x))^4*c^(7/2)*d^4*e - 2160*(sqrt(c)*x - sqrt(c*x^2 + b*x 
))^4*b*c^(5/2)*d^3*e^2 + 666*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^2*c^(3/2) 
*d^2*e^3 - 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^3*sqrt(c)*d*e^4 + 1504*( 
sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^4*d^5 - 400*(sqrt(c)*x - sqrt(c*x^2 + b 
*x))^3*b*c^3*d^4*e - 1308*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^2*c^2*d^3*e^ 
2 + 574*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^3*c*d^2*e^3 - 40*(sqrt(c)*x - 
sqrt(c*x^2 + b*x))^3*b^4*d*e^4 + 2256*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b* 
c^(7/2)*d^5 - 2412*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^2*c^(5/2)*d^4*e + 4 
62*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^3*c^(3/2)*d^3*e^2 + 24*(sqrt(c)*x - 
 sqrt(c*x^2 + b*x))^2*b^4*sqrt(c)*d^2*e^3 + 1128*(sqrt(c)*x - sqrt(c*x^2 + 
 b*x))*b^2*c^3*d^5 - 1272*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*c^2*d^4*e + 
324*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^4*c*d^3*e^2 - 15*(sqrt(c)*x - sqr...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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