\(\int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx\) [137]

Optimal result

Integrand size = 21, antiderivative size = 292 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=-\frac {\sqrt {b x+c x^2}}{4 e (d+e x)^4}+\frac {(2 c d-b e) \sqrt {b x+c x^2}}{24 d e (c d-b e) (d+e x)^3}+\frac {\left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right ) \sqrt {b x+c x^2}}{96 d^2 e (c d-b e)^2 (d+e x)^2}+\frac {(2 c d-b e) \left (8 c^2 d^2-8 b c d e+15 b^2 e^2\right ) \sqrt {b x+c x^2}}{192 d^3 e (c d-b e)^3 (d+e x)}-\frac {b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{64 d^{7/2} (c d-b e)^{7/2}} \] Output:

-1/4*(c*x^2+b*x)^(1/2)/e/(e*x+d)^4+1/24*(-b*e+2*c*d)*(c*x^2+b*x)^(1/2)/d/e 
/(-b*e+c*d)/(e*x+d)^3+1/96*(5*b^2*e^2-8*b*c*d*e+8*c^2*d^2)*(c*x^2+b*x)^(1/ 
2)/d^2/e/(-b*e+c*d)^2/(e*x+d)^2+1/192*(-b*e+2*c*d)*(15*b^2*e^2-8*b*c*d*e+8 
*c^2*d^2)*(c*x^2+b*x)^(1/2)/d^3/e/(-b*e+c*d)^3/(e*x+d)-1/64*b^2*(5*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^(7/2)/(-b*e+c*d)^(7/2)
 

Mathematica [A] (verified)

Time = 10.61 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\frac {\sqrt {x (b+c x)} \left (48 e x^{3/2} (b+c x)+\frac {40 e (2 c d-b e) x^{3/2} (b+c x) (d+e x)}{d (c d-b e)}+\frac {3 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (d+e x)^2 \left (\sqrt {d} \sqrt {c d-b e} \sqrt {x} \sqrt {b+c x} (-b d-2 c d x+b e x)+b^2 (d+e x)^2 \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )\right )}{d^{5/2} (c d-b e)^{5/2} \sqrt {b+c x}}\right )}{192 d (-c d+b e) \sqrt {x} (d+e x)^4} \] Input:

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

Output:

(Sqrt[x*(b + c*x)]*(48*e*x^(3/2)*(b + c*x) + (40*e*(2*c*d - b*e)*x^(3/2)*( 
b + c*x)*(d + e*x))/(d*(c*d - b*e)) + (3*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2* 
e^2)*(d + e*x)^2*(Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[x]*Sqrt[b + c*x]*(-(b*d) - 
2*c*d*x + b*e*x) + b^2*(d + e*x)^2*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt 
[d]*Sqrt[b + c*x])]))/(d^(5/2)*(c*d - b*e)^(5/2)*Sqrt[b + c*x])))/(192*d*( 
-(c*d) + b*e)*Sqrt[x]*(d + e*x)^4)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1167, 27, 1228, 1152, 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.

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

\(\Big \downarrow \) 1167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1228

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

\(\Big \downarrow \) 1152

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/4*(e*(b*x + c*x^2)^(3/2))/(d*(c*d - b*e)*(d + e*x)^4) + ((-5*e*(2*c*d - 
 b*e)*(b*x + c*x^2)^(3/2))/(3*d*(c*d - b*e)*(d + e*x)^3) + ((16*c^2*d^2 - 
16*b*c*d*e + 5*b^2*e^2)*(((b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(4*d* 
(c*d - b*e)*(d + e*x)^2) - (b^2*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d] 
*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(8*d^(3/2)*(c*d - b*e)^(3/2))))/(2*d 
*(c*d - b*e)))/(8*d*(c*d - b*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 1152

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/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a 
*c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)))   Int[(d + e*x)^(m + 2)*(a + b*x + 
 c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] 
 && GtQ[p, 0]
 

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 1167

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228

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

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (e x +d \right )^{4} \left (b^{2} e^{2}-\frac {16}{5} b c d e +\frac {16}{5} c^{2} d^{2}\right ) b^{2} \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )+\sqrt {x \left (c x +b \right )}\, \sqrt {d \left (b e -c d \right )}\, \left (\frac {16 c^{2} \left (2 c x +b \right ) d^{5}}{5}-\frac {16 e \left (4 c x +b \right ) c \left (-\frac {c x}{3}+b \right ) d^{4}}{5}+e^{2} \left (\frac {16}{15} c^{3} x^{3}-\frac {104}{15} b \,c^{2} x^{2}+\frac {66}{5} b^{2} c x +b^{3}\right ) d^{3}-\frac {73 \left (\frac {24}{73} c^{2} x^{2}-\frac {140}{73} c b x +b^{2}\right ) e^{3} x b \,d^{2}}{15}-\frac {11 e^{4} x^{2} \left (-\frac {38 c x}{55}+b \right ) b^{2} d}{3}-b^{3} e^{5} x^{3}\right )\right )}{64 \sqrt {d \left (b e -c d \right )}\, \left (e x +d \right )^{4} \left (b e -c d \right )^{3} d^{3}}\)	\(243\)
default	\(\text {Expression too large to display}\)	\(2184\)

Input:

int((c*x^2+b*x)^(1/2)/(e*x+d)^5,x,method=_RETURNVERBOSE)
 

Output:

-5/64*((e*x+d)^4*(b^2*e^2-16/5*b*c*d*e+16/5*c^2*d^2)*b^2*arctan((x*(c*x+b) 
)^(1/2)/x*d/(d*(b*e-c*d))^(1/2))+(x*(c*x+b))^(1/2)*(d*(b*e-c*d))^(1/2)*(16 
/5*c^2*(2*c*x+b)*d^5-16/5*e*(4*c*x+b)*c*(-1/3*c*x+b)*d^4+e^2*(16/15*c^3*x^ 
3-104/15*b*c^2*x^2+66/5*b^2*c*x+b^3)*d^3-73/15*(24/73*c^2*x^2-140/73*c*b*x 
+b^2)*e^3*x*b*d^2-11/3*e^4*x^2*(-38/55*c*x+b)*b^2*d-b^3*e^5*x^3))/(d*(b*e- 
c*d))^(1/2)/(e*x+d)^4/(b*e-c*d)^3/d^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (264) = 528\).

Time = 0.12 (sec) , antiderivative size = 1606, normalized size of antiderivative = 5.50 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \] Input:

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^5,x, algorithm="fricas")
 

Output:

[-1/384*(3*(16*b^2*c^2*d^6 - 16*b^3*c*d^5*e + 5*b^4*d^4*e^2 + (16*b^2*c^2* 
d^2*e^4 - 16*b^3*c*d*e^5 + 5*b^4*e^6)*x^4 + 4*(16*b^2*c^2*d^3*e^3 - 16*b^3 
*c*d^2*e^4 + 5*b^4*d*e^5)*x^3 + 6*(16*b^2*c^2*d^4*e^2 - 16*b^3*c*d^3*e^3 + 
 5*b^4*d^2*e^4)*x^2 + 4*(16*b^2*c^2*d^5*e - 16*b^3*c*d^4*e^2 + 5*b^4*d^3*e 
^3)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d 
*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(48*b*c^3*d^7 - 96*b^2*c^2*d^6*e + 6 
3*b^3*c*d^5*e^2 - 15*b^4*d^4*e^3 + (16*c^4*d^5*e^2 - 40*b*c^3*d^4*e^3 + 62 
*b^2*c^2*d^3*e^4 - 53*b^3*c*d^2*e^5 + 15*b^4*d*e^6)*x^3 + (64*c^4*d^6*e - 
168*b*c^3*d^5*e^2 + 244*b^2*c^2*d^4*e^3 - 195*b^3*c*d^3*e^4 + 55*b^4*d^2*e 
^5)*x^2 + (96*c^4*d^7 - 272*b*c^3*d^6*e + 374*b^2*c^2*d^5*e^2 - 271*b^3*c* 
d^4*e^3 + 73*b^4*d^3*e^4)*x)*sqrt(c*x^2 + b*x))/(c^4*d^12 - 4*b*c^3*d^11*e 
 + 6*b^2*c^2*d^10*e^2 - 4*b^3*c*d^9*e^3 + b^4*d^8*e^4 + (c^4*d^8*e^4 - 4*b 
*c^3*d^7*e^5 + 6*b^2*c^2*d^6*e^6 - 4*b^3*c*d^5*e^7 + b^4*d^4*e^8)*x^4 + 4* 
(c^4*d^9*e^3 - 4*b*c^3*d^8*e^4 + 6*b^2*c^2*d^7*e^5 - 4*b^3*c*d^6*e^6 + b^4 
*d^5*e^7)*x^3 + 6*(c^4*d^10*e^2 - 4*b*c^3*d^9*e^3 + 6*b^2*c^2*d^8*e^4 - 4* 
b^3*c*d^7*e^5 + b^4*d^6*e^6)*x^2 + 4*(c^4*d^11*e - 4*b*c^3*d^10*e^2 + 6*b^ 
2*c^2*d^9*e^3 - 4*b^3*c*d^8*e^4 + b^4*d^7*e^5)*x), 1/192*(3*(16*b^2*c^2*d^ 
6 - 16*b^3*c*d^5*e + 5*b^4*d^4*e^2 + (16*b^2*c^2*d^2*e^4 - 16*b^3*c*d*e^5 
+ 5*b^4*e^6)*x^4 + 4*(16*b^2*c^2*d^3*e^3 - 16*b^3*c*d^2*e^4 + 5*b^4*d*e^5) 
*x^3 + 6*(16*b^2*c^2*d^4*e^2 - 16*b^3*c*d^3*e^3 + 5*b^4*d^2*e^4)*x^2 + ...
 

Sympy [F]

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

integrate((c*x**2+b*x)**(1/2)/(e*x+d)**5,x)
 

Output:

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

Maxima [F(-2)]

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

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^5,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. 1219 vs. \(2 (264) = 528\).

Time = 0.34 (sec) , antiderivative size = 1219, normalized size of antiderivative = 4.17 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \] Input:

integrate((c*x^2+b*x)^(1/2)/(e*x+d)^5,x, algorithm="giac")
 

Output:

-1/384*((48*b^2*c^2*d^2*e^3*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e 
)*sqrt(c)*abs(e))) - 48*b^3*c*d*e^4*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 
 - b*d*e)*sqrt(c)*abs(e))) + 15*b^4*e^5*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c 
*d^2 - b*d*e)*sqrt(c)*abs(e))) + 32*sqrt(c*d^2 - b*d*e)*c^(7/2)*d^3*abs(e) 
 - 48*sqrt(c*d^2 - b*d*e)*b*c^(5/2)*d^2*e*abs(e) + 76*sqrt(c*d^2 - b*d*e)* 
b^2*c^(3/2)*d*e^2*abs(e) - 30*sqrt(c*d^2 - b*d*e)*b^3*sqrt(c)*e^3*abs(e))* 
sgn(1/(e*x + d))*sgn(e)/(sqrt(c*d^2 - b*d*e)*c^3*d^6*e^4*abs(e) - 3*sqrt(c 
*d^2 - b*d*e)*b*c^2*d^5*e^5*abs(e) + 3*sqrt(c*d^2 - b*d*e)*b^2*c*d^4*e^6*a 
bs(e) - sqrt(c*d^2 - b*d*e)*b^3*d^3*e^7*abs(e)) - 2*sqrt(c - 2*c*d/(e*x + 
d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2)*((16*c^3*d^3*e 
^4*sgn(1/(e*x + d))*sgn(e) - 24*b*c^2*d^2*e^5*sgn(1/(e*x + d))*sgn(e) + 38 
*b^2*c*d*e^6*sgn(1/(e*x + d))*sgn(e) - 15*b^3*e^7*sgn(1/(e*x + d))*sgn(e)) 
/(c^3*d^6*e^8 - 3*b*c^2*d^5*e^9 + 3*b^2*c*d^4*e^10 - b^3*d^3*e^11) + 2*((8 
*c^3*d^4*e^5*sgn(1/(e*x + d))*sgn(e) - 16*b*c^2*d^3*e^6*sgn(1/(e*x + d))*s 
gn(e) + 13*b^2*c*d^2*e^7*sgn(1/(e*x + d))*sgn(e) - 5*b^3*d*e^8*sgn(1/(e*x 
+ d))*sgn(e))/(c^3*d^6*e^8 - 3*b*c^2*d^5*e^9 + 3*b^2*c*d^4*e^10 - b^3*d^3* 
e^11) + 4*((2*c^3*d^5*e^6*sgn(1/(e*x + d))*sgn(e) - 5*b*c^2*d^4*e^7*sgn(1/ 
(e*x + d))*sgn(e) + 4*b^2*c*d^3*e^8*sgn(1/(e*x + d))*sgn(e) - b^3*d^2*e^9* 
sgn(1/(e*x + d))*sgn(e))/(c^3*d^6*e^8 - 3*b*c^2*d^5*e^9 + 3*b^2*c*d^4*e^10 
 - b^3*d^3*e^11) - 6*(c^3*d^6*e^7*sgn(1/(e*x + d))*sgn(e) - 3*b*c^2*d^5...
 

Mupad [F(-1)]

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

int((b*x + c*x^2)^(1/2)/(d + e*x)^5,x)
 

Output:

int((b*x + c*x^2)^(1/2)/(d + e*x)^5, x)
 

Reduce [F]

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

int((c*x^2+b*x)^(1/2)/(e*x+d)^5,x)
 

Output:

int((c*x^2+b*x)^(1/2)/(e*x+d)^5,x)