\(\int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx\) [293]

Optimal result

Integrand size = 21, antiderivative size = 258 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{64 d^3 (c d-b e)^3 (d+e x)^2}-\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}-\frac {b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{128 d^{7/2} (c d-b e)^{7/2}} \]

[Out]

-1/4*e*(c*x^2+b*x)^(3/2)/d/(-b*e+c*d)/(e*x+d)^4-5/24*e*(-b*e+2*c*d)*(c*x^2+b*x)^(3/2)/d^2/(-b*e+c*d)^2/(e*x+d)
^3-1/128*b^2*(5*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/(c*x^
2+b*x)^(1/2))/d^(7/2)/(-b*e+c*d)^(7/2)+1/64*(5*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*(b*d+(-b*e+2*c*d)*x)*(c*x^2+b*x)
^(1/2)/d^3/(-b*e+c*d)^3/(e*x+d)^2

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {758, 820, 734, 738, 212} \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=-\frac {b^2 \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{128 d^{7/2} (c d-b e)^{7/2}}+\frac {\sqrt {b x+c x^2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) (x (2 c d-b e)+b d)}{64 d^3 (d+e x)^2 (c d-b e)^3}-\frac {5 e \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{24 d^2 (d+e x)^3 (c d-b e)^2}-\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (d+e x)^4 (c d-b e)} \]

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x)^5,x]

[Out]

((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])/(64*d^3*(c*d - b*e)^3*(d + e
*x)^2) - (e*(b*x + c*x^2)^(3/2))/(4*d*(c*d - b*e)*(d + e*x)^4) - (5*e*(2*c*d - b*e)*(b*x + c*x^2)^(3/2))/(24*d
^2*(c*d - b*e)^2*(d + e*x)^3) - (b^2*(16*c^2*d^2 - 16*b*c*d*e + 5*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])])/(128*d^(7/2)*(c*d - b*e)^(7/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> 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] + Dist[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] /; Free
Q[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m
+ 2*p + 2, 0] && GtQ[p, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> 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] + Dist[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[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e
, 0] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(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] - Dist[
(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {\int \frac {\left (\frac {1}{2} (-8 c d+5 b e)+c e x\right ) \sqrt {b x+c 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 (c d-b e) (d+e x)^4}-\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}+\frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \int \frac {\sqrt {b x+c x^2}}{(d+e x)^3} \, dx}{16 d^2 (c d-b e)^2} \\ & = \frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{64 d^3 (c d-b e)^3 (d+e x)^2}-\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}-\frac {\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{128 d^3 (c d-b e)^3} \\ & = \frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{64 d^3 (c d-b e)^3 (d+e x)^2}-\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}+\frac {\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac {-b d-(2 c d-b e) x}{\sqrt {b x+c x^2}}\right )}{64 d^3 (c d-b e)^3} \\ & = \frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{64 d^3 (c d-b e)^3 (d+e x)^2}-\frac {e \left (b x+c x^2\right )^{3/2}}{4 d (c d-b e) (d+e x)^4}-\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 d^2 (c d-b e)^2 (d+e x)^3}-\frac {b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{128 d^{7/2} (c d-b e)^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.80 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\frac {\sqrt {x (b+c x)} \left (48 e x^{3/2} (b+c x)+\frac {40 e (2 c d-b e) x^{3/2} (b+c x) (d+e x)}{d (c d-b e)}+\frac {3 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (d+e x)^2 \left (\sqrt {d} \sqrt {c d-b e} \sqrt {x} \sqrt {b+c x} (-b d-2 c d x+b e x)+b^2 (d+e x)^2 \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )\right )}{d^{5/2} (c d-b e)^{5/2} \sqrt {b+c x}}\right )}{192 d (-c d+b e) \sqrt {x} (d+e x)^4} \]

[In]

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

[Out]

(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)

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.94

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

[In]

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

[Out]

-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))+(16
/5*c^2*(2*c*x+b)*d^5-16/5*c*e*(4*c*x+b)*(-1/3*c*x+b)*d^4+e^2*(66/5*b^2*c*x-104/15*b*c^2*x^2+b^3+16/15*c^3*x^3)
*d^3-73/15*x*e^3*b*(24/73*c^2*x^2-140/73*b*c*x+b^2)*d^2-11/3*x^2*e^4*(-38/55*c*x+b)*b^2*d-b^3*e^5*x^3)*(x*(c*x
+b))^(1/2)*(d*(b*e-c*d))^(1/2))/(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. 800 vs. \(2 (232) = 464\).

Time = 0.52 (sec) , antiderivative size = 1611, normalized size of antiderivative = 6.24 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-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
+ 63*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)*s
qrt(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 + 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)*sqr
t(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/((c*d - b*e)*x)) - (48*b*c^3*d^7 - 96*b^2*c^2
*d^6*e + 63*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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1219 vs. \(2 (232) = 464\).

Time = 0.57 (sec) , antiderivative size = 1219, normalized size of antiderivative = 4.72 \[ \int \frac {\sqrt {b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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*abs(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) - 2
4*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))*sgn(e) + 13*b^2*c*d^2*e^7*sgn(1/(e*x + d))*sgn(e) - 5*b^3*d*e^8*s
gn(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*e^8*sgn(1/(e*x + d))*sgn(e) + 3*b^2*c*d^4*e^9*sgn(1/(e*x + d))
*sgn(e) - b^3*d^3*e^10*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)*(e*x + d)*e))/((e*x + d)*e))/((e*x + d)*e)) - 3*(16*b^2*c^2*d^2*sgn(1/(e*x + d))*sgn(e) - 16*b^3*c*d*e*sg
n(1/(e*x + d))*sgn(e) + 5*b^4*e^2*sgn(1/(e*x + d))*sgn(e))*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*(sq
rt(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) + sqrt(c*d^2*e^2 - b*d*e^3)/((
e*x + d)*e))*abs(e)))/((c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e^3 - b^3*d^3*e^4)*sqrt(c*d^2 - b*d*e)*abs(e
)))*e^2

Mupad [F(-1)]

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

[In]

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

[Out]

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