\(\int \frac {1}{(d+e x)^2 (a+b x+c x^2)^{3/2}} \, dx\) [2388]

Optimal result

Integrand size = 22, antiderivative size = 255 \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {e \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {3 e^2 (2 c d-b e) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {754, 820, 738, 212} \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {3 e^2 (2 c d-b e) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{5/2}}-\frac {e \sqrt {a+b x+c x^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right )}{\left (b^2-4 a c\right ) (d+e x) \left (a e^2-b d e+c d^2\right )^2}-\frac {2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) (d+e x) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )} \]

[In]

Int[1/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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 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 754

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(b
*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*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] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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 {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} e \left (2 b c d-3 b^2 e+8 a c e\right )+c e (2 c d-b e) x}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )} \\ & = -\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {e \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\left (3 e^2 (2 c d-b e)\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2} \\ & = -\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {e \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {\left (3 e^2 (2 c d-b e)\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^2} \\ & = -\frac {2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {e \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (-\frac {e \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {a+x (b+c x)}}{2 \left (c d^2+e (-b d+a e)\right ) (d+e x)}+\frac {b^2 e-2 c (a e+c d x)+b c (-d+e x)}{(d+e x) \sqrt {a+x (b+c x)}}+\frac {3 \left (b^2-4 a c\right ) e^2 (-2 c d+b e) \text {arctanh}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{4 \left (c d^2+e (-b d+a e)\right )^{3/2}}\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )} \]

[In]

Integrate[1/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(654\) vs. \(2(241)=482\).

Time = 0.37 (sec) , antiderivative size = 655, normalized size of antiderivative = 2.57

method	result	size
default	\(\frac {-\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {3 \left (b e -2 c d \right ) e \left (\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right )}-\frac {4 c \,e^{2} \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{e^{2}}\)	\(655\)

[In]

int(1/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/e^2*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)/((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-3/
2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(1/(a*e^2-b*d*e+c*d^2)*e^2/((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e
+c*d^2)/e^2)^(1/2)-(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2*c*(x+d/e)+(b*e-2*c*d)/e)/(4*c*(a*e^2-b*d*e+c*d^2)/e^2-
(b*e-2*c*d)^2/e^2)/((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-1/(a*e^2-b*d*e+c*d^2)*e^2
/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^
2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))-4*c/(a*e^2-b*d*e+c*d^2)*
e^2*(2*c*(x+d/e)+(b*e-2*c*d)/e)/(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/((x+d/e)^2*c+(b*e-2*c*d)/e*(x+
d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1481 vs. \(2 (241) = 482\).

Time = 1.17 (sec) , antiderivative size = 3004, normalized size of antiderivative = 11.78 \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(3*(2*(a*b^2*c - 4*a^2*c^2)*d^2*e^2 - (a*b^3 - 4*a^2*b*c)*d*e^3 + (2*(b^2*c^2 - 4*a*c^3)*d*e^3 - (b^3*c
- 4*a*b*c^2)*e^4)*x^3 + (2*(b^2*c^2 - 4*a*c^3)*d^2*e^2 + (b^3*c - 4*a*b*c^2)*d*e^3 - (b^4 - 4*a*b^2*c)*e^4)*x^
2 + (2*(b^3*c - 4*a*b*c^2)*d^2*e^2 - (b^4 - 6*a*b^2*c + 8*a^2*c^2)*d*e^3 - (a*b^3 - 4*a^2*b*c)*e^4)*x)*sqrt(c*
d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e
^2)*x^2 + 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 +
4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) + 4*(2*b*c^3*d^5 - 2*(3*b^2*c^2 - 4*a*c^3)*d^4
*e + 6*(b^3*c - 2*a*b*c^2)*d^3*e^2 - (2*b^4 - 3*a*b^2*c - 4*a^2*c^2)*d^2*e^3 + (a*b^3 - 2*a^2*b*c)*d*e^4 + (a^
2*b^2 - 4*a^3*c)*e^5 + (4*c^4*d^4*e - 8*b*c^3*d^3*e^2 + (7*b^2*c^2 - 4*a*c^3)*d^2*e^3 - (3*b^3*c - 4*a*b*c^2)*
d*e^4 + (3*a*b^2*c - 8*a^2*c^2)*e^5)*x^2 + (4*c^4*d^5 - 6*b*c^3*d^4*e + 8*a*c^3*d^3*e^2 + (5*b^3*c - 16*a*b*c^
2)*d^2*e^3 - (3*b^4 - 8*a*b^2*c - 4*a^2*c^2)*d*e^4 + (3*a*b^3 - 10*a^2*b*c)*e^5)*x)*sqrt(c*x^2 + b*x + a))/((a
*b^2*c^3 - 4*a^2*c^4)*d^7 - 3*(a*b^3*c^2 - 4*a^2*b*c^3)*d^6*e + 3*(a*b^4*c - 3*a^2*b^2*c^2 - 4*a^3*c^3)*d^5*e^
2 - (a*b^5 + 2*a^2*b^3*c - 24*a^3*b*c^2)*d^4*e^3 + 3*(a^2*b^4 - 3*a^3*b^2*c - 4*a^4*c^2)*d^3*e^4 - 3*(a^3*b^3
- 4*a^4*b*c)*d^2*e^5 + (a^4*b^2 - 4*a^5*c)*d*e^6 + ((b^2*c^4 - 4*a*c^5)*d^6*e - 3*(b^3*c^3 - 4*a*b*c^4)*d^5*e^
2 + 3*(b^4*c^2 - 3*a*b^2*c^3 - 4*a^2*c^4)*d^4*e^3 - (b^5*c + 2*a*b^3*c^2 - 24*a^2*b*c^3)*d^3*e^4 + 3*(a*b^4*c
- 3*a^2*b^2*c^2 - 4*a^3*c^3)*d^2*e^5 - 3*(a^2*b^3*c - 4*a^3*b*c^2)*d*e^6 + (a^3*b^2*c - 4*a^4*c^2)*e^7)*x^3 +
((b^2*c^4 - 4*a*c^5)*d^7 - 2*(b^3*c^3 - 4*a*b*c^4)*d^6*e + 3*(a*b^2*c^3 - 4*a^2*c^4)*d^5*e^2 + (2*b^5*c - 11*a
*b^3*c^2 + 12*a^2*b*c^3)*d^4*e^3 - (b^6 - a*b^4*c - 15*a^2*b^2*c^2 + 12*a^3*c^3)*d^3*e^4 + 3*(a*b^5 - 4*a^2*b^
3*c)*d^2*e^5 - (3*a^2*b^4 - 13*a^3*b^2*c + 4*a^4*c^2)*d*e^6 + (a^3*b^3 - 4*a^4*b*c)*e^7)*x^2 + ((b^3*c^3 - 4*a
*b*c^4)*d^7 - (3*b^4*c^2 - 13*a*b^2*c^3 + 4*a^2*c^4)*d^6*e + 3*(b^5*c - 4*a*b^3*c^2)*d^5*e^2 - (b^6 - a*b^4*c
- 15*a^2*b^2*c^2 + 12*a^3*c^3)*d^4*e^3 + (2*a*b^5 - 11*a^2*b^3*c + 12*a^3*b*c^2)*d^3*e^4 + 3*(a^3*b^2*c - 4*a^
4*c^2)*d^2*e^5 - 2*(a^3*b^3 - 4*a^4*b*c)*d*e^6 + (a^4*b^2 - 4*a^5*c)*e^7)*x), 1/2*(3*(2*(a*b^2*c - 4*a^2*c^2)*
d^2*e^2 - (a*b^3 - 4*a^2*b*c)*d*e^3 + (2*(b^2*c^2 - 4*a*c^3)*d*e^3 - (b^3*c - 4*a*b*c^2)*e^4)*x^3 + (2*(b^2*c^
2 - 4*a*c^3)*d^2*e^2 + (b^3*c - 4*a*b*c^2)*d*e^3 - (b^4 - 4*a*b^2*c)*e^4)*x^2 + (2*(b^3*c - 4*a*b*c^2)*d^2*e^2
- (b^4 - 6*a*b^2*c + 8*a^2*c^2)*d*e^3 - (a*b^3 - 4*a^2*b*c)*e^4)*x)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*
sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^
2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 2*(2*b*c^3*d^5 - 2*(3*b^2*c^2 - 4*
a*c^3)*d^4*e + 6*(b^3*c - 2*a*b*c^2)*d^3*e^2 - (2*b^4 - 3*a*b^2*c - 4*a^2*c^2)*d^2*e^3 + (a*b^3 - 2*a^2*b*c)*d
*e^4 + (a^2*b^2 - 4*a^3*c)*e^5 + (4*c^4*d^4*e - 8*b*c^3*d^3*e^2 + (7*b^2*c^2 - 4*a*c^3)*d^2*e^3 - (3*b^3*c - 4
*a*b*c^2)*d*e^4 + (3*a*b^2*c - 8*a^2*c^2)*e^5)*x^2 + (4*c^4*d^5 - 6*b*c^3*d^4*e + 8*a*c^3*d^3*e^2 + (5*b^3*c -
16*a*b*c^2)*d^2*e^3 - (3*b^4 - 8*a*b^2*c - 4*a^2*c^2)*d*e^4 + (3*a*b^3 - 10*a^2*b*c)*e^5)*x)*sqrt(c*x^2 + b*x
+ a))/((a*b^2*c^3 - 4*a^2*c^4)*d^7 - 3*(a*b^3*c^2 - 4*a^2*b*c^3)*d^6*e + 3*(a*b^4*c - 3*a^2*b^2*c^2 - 4*a^3*c
^3)*d^5*e^2 - (a*b^5 + 2*a^2*b^3*c - 24*a^3*b*c^2)*d^4*e^3 + 3*(a^2*b^4 - 3*a^3*b^2*c - 4*a^4*c^2)*d^3*e^4 - 3
*(a^3*b^3 - 4*a^4*b*c)*d^2*e^5 + (a^4*b^2 - 4*a^5*c)*d*e^6 + ((b^2*c^4 - 4*a*c^5)*d^6*e - 3*(b^3*c^3 - 4*a*b*c
^4)*d^5*e^2 + 3*(b^4*c^2 - 3*a*b^2*c^3 - 4*a^2*c^4)*d^4*e^3 - (b^5*c + 2*a*b^3*c^2 - 24*a^2*b*c^3)*d^3*e^4 + 3
*(a*b^4*c - 3*a^2*b^2*c^2 - 4*a^3*c^3)*d^2*e^5 - 3*(a^2*b^3*c - 4*a^3*b*c^2)*d*e^6 + (a^3*b^2*c - 4*a^4*c^2)*e
^7)*x^3 + ((b^2*c^4 - 4*a*c^5)*d^7 - 2*(b^3*c^3 - 4*a*b*c^4)*d^6*e + 3*(a*b^2*c^3 - 4*a^2*c^4)*d^5*e^2 + (2*b^
5*c - 11*a*b^3*c^2 + 12*a^2*b*c^3)*d^4*e^3 - (b^6 - a*b^4*c - 15*a^2*b^2*c^2 + 12*a^3*c^3)*d^3*e^4 + 3*(a*b^5
- 4*a^2*b^3*c)*d^2*e^5 - (3*a^2*b^4 - 13*a^3*b^2*c + 4*a^4*c^2)*d*e^6 + (a^3*b^3 - 4*a^4*b*c)*e^7)*x^2 + ((b^3
*c^3 - 4*a*b*c^4)*d^7 - (3*b^4*c^2 - 13*a*b^2*c^3 + 4*a^2*c^4)*d^6*e + 3*(b^5*c - 4*a*b^3*c^2)*d^5*e^2 - (b^6
- a*b^4*c - 15*a^2*b^2*c^2 + 12*a^3*c^3)*d^4*e^3 + (2*a*b^5 - 11*a^2*b^3*c + 12*a^3*b*c^2)*d^3*e^4 + 3*(a^3*b^
2*c - 4*a^4*c^2)*d^2*e^5 - 2*(a^3*b^3 - 4*a^4*b*c)*d*e^6 + (a^4*b^2 - 4*a^5*c)*e^7)*x)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x\right )^{2} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/((d + e*x)**2*(a + b*x + c*x**2)**(3/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),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(a*e^2-b*d*e>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{2}} \,d x } \]

[In]

integrate(1/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

[In]

int(1/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x)

[Out]

int(1/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)), x)