3.19.6 \(\int \frac {a+b x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=169 \[ -\frac {b}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac {2 b e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {2 b e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]
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Rubi [A] time = 0.11, antiderivative size = 169, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used =
{770, 21, 44} \begin {gather*} -\frac {b}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac {2 b e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {2 b e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*x)/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
-(b/((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (e*(a + b*x))/((b*d - a*e)^2*(d + e*x)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (2*b*e*(a + b*x)*Log[a + b*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*b*e*(a + b*x)
*Log[d + e*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
Rule 44
Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])
Rule 770
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]
Rubi steps
\begin {align*} \int \frac {a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {a+b x}{\left (a b+b^2 x\right )^3 (d+e x)^2} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{(a+b x)^2 (d+e x)^2} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b^2}{(b d-a e)^2 (a+b x)^2}-\frac {2 b^2 e}{(b d-a e)^3 (a+b x)}+\frac {e^2}{(b d-a e)^2 (d+e x)^2}+\frac {2 b e^2}{(b d-a e)^3 (d+e x)}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {b}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b e (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b e (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 92, normalized size = 0.54 \begin {gather*} \frac {-((b d-a e) (a e+b (d+2 e x)))-2 b e (a+b x) (d+e x) \log (a+b x)+2 b e (a+b x) (d+e x) \log (d+e x)}{\sqrt {(a+b x)^2} (d+e x) (b d-a e)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
(-((b*d - a*e)*(a*e + b*(d + 2*e*x))) - 2*b*e*(a + b*x)*(d + e*x)*Log[a + b*x] + 2*b*e*(a + b*x)*(d + e*x)*Log
[d + e*x])/((b*d - a*e)^3*Sqrt[(a + b*x)^2]*(d + e*x))
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IntegrateAlgebraic [B] time = 18.08, size = 4119, normalized size = 24.37 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(a + b*x)/((d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
(-8*b^2*x^2*(-(b^2*d) + a*b*e + b^2*e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 8*(b^2)^(3/2)*x^2*(a*b*d - a^2*e + b^
2*d*x - 2*a*b*e*x - b^2*e*x^2))/(Sqrt[b^2]*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a*b^3*d^2*x^2 + 8*a^2
*b^2*d*e*x^2 - 8*b^4*d^2*x^3 + 8*a*b^3*d*e*x^3 + 8*a^2*b^2*e^2*x^3 + 16*a*b^3*e^2*x^4 + 8*b^4*e^2*x^5) + (b*d
- a*e)*(8*a^2*b^4*d^2*x^2 - 8*a^3*b^3*d*e*x^2 + 16*a*b^5*d^2*x^3 - 16*a^2*b^4*d*e*x^3 - 8*a^3*b^3*e^2*x^3 + 8*
b^6*d^2*x^4 - 8*a*b^5*d*e*x^4 - 24*a^2*b^4*e^2*x^4 - 24*a*b^5*e^2*x^5 - 8*b^6*e^2*x^6)) + (2*b^2*d*ArcTanh[(Sq
rt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/(a*(b*d - a*e)^3) - (2*b^2*d*ArcTanh[(Sqrt[b^2]*e*x)/(2*b*d -
a*e) - (e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a*(b*d - a*e)^3) + (b*Sqrt[b^2]*d*Log[-a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^3) + (b*Sqrt[b^2]*d*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*
b*x + b^2*x^2]])/(a*(b*d - a*e)^3) - (b*Sqrt[b^2]*d*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]])/(a*(b*d - a*e)^3) - (b*Sqrt[b^2]*d*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
])/(a*(b*d - a*e)^3) + ((-16*a^2*b^3*Sqrt[b^2]*x^3)/(b*d - a*e)^2 - (32*a*b^4*Sqrt[b^2]*x^4)/(b*d - a*e)^2 - (
16*b^5*Sqrt[b^2]*x^5)/(b*d - a*e)^2 + (16*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b*d - a*e)^2 + (16*b^5*x^4
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(b*d - a*e)^2 - (16*a^2*b^4*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/a])/(b*d - a*e)^2 - (48*a*b^5*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(b*d
- a*e)^2 - (48*b^6*x^5*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(b*d - a*e)^2 - (16*b^7*x
^6*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(a*(b*d - a*e)^2) + (16*a*b^3*Sqrt[b^2]*x^3*Sq
rt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(b*d - a*e)^2 + (32*b
^4*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(b
*d - a*e)^2 + (16*b^5*Sqrt[b^2]*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/a])/(a*(b*d - a*e)^2) + (16*a^2*b^4*x^3*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(2*b*d - a*e)])/(b*d - a*e)^2 + (48*a*b^5*x^4*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(2*b*d - a*e)])/(b*d - a*e)^2 + (48*b^6*x^5*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b
*d - a*e)])/(b*d - a*e)^2 + (16*b^7*x^6*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d -
a*e)])/(a*(b*d - a*e)^2) - (16*a*b^3*Sqrt[b^2]*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*e*x) + e
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(b*d - a*e)^2 - (32*b^4*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(b*d - a*e)^2 - (16*b^5*Sqr
t[b^2]*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d -
a*e)])/(a*(b*d - a*e)^2) + (8*a^2*b^3*Sqrt[b^2]*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b
*d - a*e)^2 + (24*a*b^4*Sqrt[b^2]*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (
24*b^5*Sqrt[b^2]*x^5*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (8*b^6*Sqrt[b^2]*x
^6*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) - (8*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (16*b^5*x^4*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (8*b^6*x^5*Sqrt[a^2 + 2*a
*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) + (8*a^2*b^3*Sqrt[b^2
]*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (24*a*b^4*Sqrt[b^2]*x^4*Log[a - Sq
rt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (24*b^5*Sqrt[b^2]*x^5*Log[a - Sqrt[b^2]*x + Sqrt[a
^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (8*b^6*Sqrt[b^2]*x^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*
x^2]])/(a*(b*d - a*e)^2) - (8*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/(b*d - a*e)^2 - (16*b^5*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2]])/(b*d - a*e)^2 - (8*b^6*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*
b*x + b^2*x^2]])/(a*(b*d - a*e)^2) - (8*a^2*b^3*Sqrt[b^2]*x^3*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (24*a*b^4*Sqrt[b^2]*x^4*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a
*b*x + b^2*x^2]])/(b*d - a*e)^2 - (24*b^5*Sqrt[b^2]*x^5*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/(b*d - a*e)^2 - (8*b^6*Sqrt[b^2]*x^6*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^
2*x^2]])/(a*(b*d - a*e)^2) + (8*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (16*b^5*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e + Sqrt
[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (8*b^6*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b
*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) - (8*a^2*b^3*Sqrt[b^2]*x^3*Log[
2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (24*a*b^4*Sqrt[b^2]*x^4*Log[2*
b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (24*b^5*Sqrt[b^2]*x^5*Log[2*b*d
- a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 - (8*b^6*Sqrt[b^2]*x^6*Log[2*b*d - a*e
- Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) + (8*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (16*b^5*x^4*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*d - a*e)^2 + (8*b
^6*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b
*d - a*e)^2))/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*
x^2])*(a^2 + a*b*x - a*Sqrt[b^2]*x + b^2*x^2 + a*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - Sqrt[b^2]*x*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])*(-a^2 - a*b*x - a*Sqrt[b^2]*x - b^2*x^2 + a*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + Sqrt[b^2]*x*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]))
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fricas [A] time = 0.44, size = 241, normalized size = 1.43 \begin {gather*} -\frac {b^{2} d^{2} - a^{2} e^{2} + 2 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{4} - 3 \, a^{2} b^{2} d^{3} e + 3 \, a^{3} b d^{2} e^{2} - a^{4} d e^{3} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x^{2} + {\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + 2 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
[Out]
-(b^2*d^2 - a^2*e^2 + 2*(b^2*d*e - a*b*e^2)*x + 2*(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)*log(b*x + a)
- 2*(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)*log(e*x + d))/(a*b^3*d^4 - 3*a^2*b^2*d^3*e + 3*a^3*b*d^2*
e^2 - a^4*d*e^3 + (b^4*d^3*e - 3*a*b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x^2 + (b^4*d^4 - 2*a*b^3*d^3*e +
2*a^3*b*d*e^3 - a^4*e^4)*x)
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giac [B] time = 0.46, size = 486, normalized size = 2.88 \begin {gather*} \frac {2 \, b e^{2} \log \left ({\left | -b + \frac {b d}{x e + d} - \frac {a e}{x e + d} \right |}\right )}{b^{3} d^{3} e \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 3 \, a^{2} b d e^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - a^{3} e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} + \frac {e^{3}}{{\left (b^{2} d^{2} e^{2} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 2 \, a b d e^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + a^{2} e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )\right )} {\left (x e + d\right )}} + \frac {b^{2} e}{{\left (b d - a e\right )}^{3} {\left (b - \frac {b d}{x e + d} + \frac {a e}{x e + d}\right )} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
[Out]
2*b*e^2*log(abs(-b + b*d/(x*e + d) - a*e/(x*e + d)))/(b^3*d^3*e*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2
/(x*e + d)^2) - 3*a*b^2*d^2*e^2*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) + 3*a^2*b*d*e^3*sg
n(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - a^3*e^4*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a
*e^2/(x*e + d)^2)) + e^3/((b^2*d^2*e^2*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - 2*a*b*d*e
^3*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) + a^2*e^4*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^
2 - a*e^2/(x*e + d)^2))*(x*e + d)) + b^2*e/((b*d - a*e)^3*(b - b*d/(x*e + d) + a*e/(x*e + d))*sgn(-b*e/(x*e +
d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2))
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maple [A] time = 0.09, size = 181, normalized size = 1.07 \begin {gather*} \frac {\left (2 b^{2} e^{2} x^{2} \ln \left (b x +a \right )-2 b^{2} e^{2} x^{2} \ln \left (e x +d \right )+2 a b \,e^{2} x \ln \left (b x +a \right )-2 a b \,e^{2} x \ln \left (e x +d \right )+2 b^{2} d e x \ln \left (b x +a \right )-2 b^{2} d e x \ln \left (e x +d \right )+2 a b d e \ln \left (b x +a \right )-2 a b d e \ln \left (e x +d \right )-2 a b \,e^{2} x +2 b^{2} d e x -a^{2} e^{2}+b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{\left (e x +d \right ) \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
[Out]
(2*ln(b*x+a)*x^2*b^2*e^2-2*ln(e*x+d)*x^2*b^2*e^2+2*a*b*e^2*x*ln(b*x+a)+2*b^2*d*e*x*ln(b*x+a)-2*ln(e*x+d)*x*a*b
*e^2-2*ln(e*x+d)*x*b^2*d*e+2*a*b*d*e*ln(b*x+a)-2*ln(e*x+d)*a*b*d*e-2*a*b*e^2*x+2*b^2*d*e*x-a^2*e^2+b^2*d^2)*(b
*x+a)^2/(e*x+d)/(a*e-b*d)^3/((b*x+a)^2)^(3/2)
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)/(e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(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-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,x}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x)/((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)
[Out]
int((a + b*x)/((d + e*x)^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x}{\left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)/(e*x+d)**2/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
[Out]
Integral((a + b*x)/((d + e*x)**2*((a + b*x)**2)**(3/2)), x)
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