3.327 \(\int \frac{1}{(d+e x)^2 (b x+c x^2)^{3/2}} \, dx\)
Optimal. Leaf size=207 \[ -\frac{e \sqrt{b x+c x^2} \left (3 b^2 e^2-4 b c d e+4 c^2 d^2\right )}{b^2 d^2 (d+e x) (c d-b e)^2}-\frac{2 (c x (2 c d-b e)+b (c d-b e))}{b^2 d \sqrt{b x+c x^2} (d+e x) (c d-b e)}+\frac{3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 d^{5/2} (c d-b e)^{5/2}} \]
[Out]
(-2*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*(d + e*x)*Sqrt[b*x + c*x^2]) - (e*(4*c^2*d^2 - 4*b
*c*d*e + 3*b^2*e^2)*Sqrt[b*x + c*x^2])/(b^2*d^2*(c*d - b*e)^2*(d + e*x)) + (3*e^2*(2*c*d - b*e)*ArcTanh[(b*d +
(2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*d^(5/2)*(c*d - b*e)^(5/2))
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Rubi [A] time = 0.235391, antiderivative size = 207, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used =
{740, 806, 724, 206} \[ -\frac{e \sqrt{b x+c x^2} \left (3 b^2 e^2-4 b c d e+4 c^2 d^2\right )}{b^2 d^2 (d+e x) (c d-b e)^2}-\frac{2 (c x (2 c d-b e)+b (c d-b e))}{b^2 d \sqrt{b x+c x^2} (d+e x) (c d-b e)}+\frac{3 e^2 (2 c d-b e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{2 d^{5/2} (c d-b e)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((d + e*x)^2*(b*x + c*x^2)^(3/2)),x]
[Out]
(-2*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*(d + e*x)*Sqrt[b*x + c*x^2]) - (e*(4*c^2*d^2 - 4*b
*c*d*e + 3*b^2*e^2)*Sqrt[b*x + c*x^2])/(b^2*d^2*(c*d - b*e)^2*(d + e*x)) + (3*e^2*(2*c*d - b*e)*ArcTanh[(b*d +
(2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*d^(5/2)*(c*d - b*e)^(5/2))
Rule 740
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 806
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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[Sim
plify[m + 2*p + 3], 0]
Rule 724
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{1}{(d+e x)^2 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) (d+e x) \sqrt{b x+c x^2}}-\frac{2 \int \frac{\frac{1}{2} b e (2 c d-3 b e)+c e (2 c d-b e) x}{(d+e x)^2 \sqrt{b x+c x^2}} \, dx}{b^2 d (c d-b e)}\\ &=-\frac{2 (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) (d+e x) \sqrt{b x+c x^2}}-\frac{e \left (4 c^2 d^2-4 b c d e+3 b^2 e^2\right ) \sqrt{b x+c x^2}}{b^2 d^2 (c d-b e)^2 (d+e x)}+\frac{\left (3 e^2 (2 c d-b e)\right ) \int \frac{1}{(d+e x) \sqrt{b x+c x^2}} \, dx}{2 d^2 (c d-b e)^2}\\ &=-\frac{2 (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) (d+e x) \sqrt{b x+c x^2}}-\frac{e \left (4 c^2 d^2-4 b c d e+3 b^2 e^2\right ) \sqrt{b x+c x^2}}{b^2 d^2 (c d-b e)^2 (d+e x)}-\frac{\left (3 e^2 (2 c d-b e)\right ) \operatorname{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 )}{d^2 (c d-b e)^2}\\ &=-\frac{2 (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) (d+e x) \sqrt{b x+c x^2}}-\frac{e \left (4 c^2 d^2-4 b c d e+3 b^2 e^2\right ) \sqrt{b x+c x^2}}{b^2 d^2 (c d-b e)^2 (d+e x)}+\frac{3 e^2 (2 c d-b e) \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 )}{2 d^{5/2} (c d-b e)^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.188565, size = 206, normalized size = 1. \[ \frac{-\sqrt{d} \sqrt{b e-c d} \left (b^2 c e \left (-4 d^2-2 d e x+3 e^2 x^2\right )+b^3 e^2 (2 d+3 e x)+2 b c^2 d \left (d^2-d e x-2 e^2 x^2\right )+4 c^3 d^2 x (d+e x)\right )-3 b^2 e^2 \sqrt{x} \sqrt{b+c x} (d+e x) (b e-2 c d) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{b^2 d^{5/2} \sqrt{x (b+c x)} (d+e x) (b e-c d)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((d + e*x)^2*(b*x + c*x^2)^(3/2)),x]
[Out]
(-(Sqrt[d]*Sqrt[-(c*d) + b*e]*(4*c^3*d^2*x*(d + e*x) + b^3*e^2*(2*d + 3*e*x) + 2*b*c^2*d*(d^2 - d*e*x - 2*e^2*
x^2) + b^2*c*e*(-4*d^2 - 2*d*e*x + 3*e^2*x^2))) - 3*b^2*e^2*(-2*c*d + b*e)*Sqrt[x]*Sqrt[b + c*x]*(d + e*x)*Arc
Tan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(b^2*d^(5/2)*(-(c*d) + b*e)^(5/2)*Sqrt[x*(b + c*x)]
*(d + e*x))
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Maple [B] time = 0.228, size = 893, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x+d)^2/(c*x^2+b*x)^(3/2),x)
[Out]
1/d/(b*e-c*d)/(d/e+x)/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)-3*e^2/d^2/(b*e-c*d)^2/(c*(d/e+
x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*b+9*e/d/(b*e-c*d)^2/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*
e-c*d)/e^2)^(1/2)*c-3*e^2/d^2/(b*e-c*d)^2/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*c+12*e/d
/(b*e-c*d)^2/b/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*c^2-12/(b*e-c*d)^2/b^2/(c*(d/e+x)^2
+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x*c^3-6/(b*e-c*d)^2/b/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-
c*d)/e^2)^(1/2)*c^2+3/2*e^2/d^2/(b*e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e
+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*b-3*e/d/(b*
e-c*d)^2/(-d*(b*e-c*d)/e^2)^(1/2)*ln((-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(
d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*c-8*c^2/d/(b*e-c*d)/b^2/(c*(d/e+x)^2+(b*e-2*c*
d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)*x-4*c/d/(b*e-c*d)/b/(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1
/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.13507, size = 1791, normalized size = 8.65 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
[Out]
[-1/2*(3*((2*b^2*c^2*d*e^3 - b^3*c*e^4)*x^3 + (2*b^2*c^2*d^2*e^2 + b^3*c*d*e^3 - b^4*e^4)*x^2 + (2*b^3*c*d^2*e
^2 - b^4*d*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*(2*b*c^3*d^5 - 6*b^2*c^2*d^4*e + 6*b^3*c*d^3*e^2 - 2*b^4*d^2*e^3 + (4*c^4*d^4*e - 8*b*c^3*d^3*e^
2 + 7*b^2*c^2*d^2*e^3 - 3*b^3*c*d*e^4)*x^2 + (4*c^4*d^5 - 6*b*c^3*d^4*e + 5*b^3*c*d^2*e^3 - 3*b^4*d*e^4)*x)*sq
rt(c*x^2 + b*x))/((b^2*c^4*d^6*e - 3*b^3*c^3*d^5*e^2 + 3*b^4*c^2*d^4*e^3 - b^5*c*d^3*e^4)*x^3 + (b^2*c^4*d^7 -
2*b^3*c^3*d^6*e + 2*b^5*c*d^4*e^3 - b^6*d^3*e^4)*x^2 + (b^3*c^3*d^7 - 3*b^4*c^2*d^6*e + 3*b^5*c*d^5*e^2 - b^6
*d^4*e^3)*x), (3*((2*b^2*c^2*d*e^3 - b^3*c*e^4)*x^3 + (2*b^2*c^2*d^2*e^2 + b^3*c*d*e^3 - b^4*e^4)*x^2 + (2*b^3
*c*d^2*e^2 - b^4*d*e^3)*x)*sqrt(-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)
) - (2*b*c^3*d^5 - 6*b^2*c^2*d^4*e + 6*b^3*c*d^3*e^2 - 2*b^4*d^2*e^3 + (4*c^4*d^4*e - 8*b*c^3*d^3*e^2 + 7*b^2*
c^2*d^2*e^3 - 3*b^3*c*d*e^4)*x^2 + (4*c^4*d^5 - 6*b*c^3*d^4*e + 5*b^3*c*d^2*e^3 - 3*b^4*d*e^4)*x)*sqrt(c*x^2 +
b*x))/((b^2*c^4*d^6*e - 3*b^3*c^3*d^5*e^2 + 3*b^4*c^2*d^4*e^3 - b^5*c*d^3*e^4)*x^3 + (b^2*c^4*d^7 - 2*b^3*c^3
*d^6*e + 2*b^5*c*d^4*e^3 - b^6*d^3*e^4)*x^2 + (b^3*c^3*d^7 - 3*b^4*c^2*d^6*e + 3*b^5*c*d^5*e^2 - b^6*d^4*e^3)*
x)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)**2/(c*x**2+b*x)**(3/2),x)
[Out]
Timed out
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^2/(c*x^2+b*x)^(3/2),x, algorithm="giac")
[Out]
Timed out