3.14.3 \(\int \frac {1}{(d+e x) (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=165 \[ \frac {e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]
________________________________________________________________________________________
Rubi [A] time = 0.10, antiderivative size = 165, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used =
{646, 44} \begin {gather*} \frac {e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
e/((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(e^2*(a + b*x)*Log[a + b*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^2*(a + b*x)*Log[d + e*x])/((b*
d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
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 646
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rubi steps
\begin {align*} \int \frac {1}{(d+e x) \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 {1}{\left (a b+b^2 x\right )^3 (d+e x)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{b^2 (b d-a e) (a+b x)^3}-\frac {e}{b^2 (b d-a e)^2 (a+b x)^2}+\frac {e^2}{b^2 (b d-a e)^3 (a+b x)}-\frac {e^3}{b^3 (b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {e}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2 (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*}
________________________________________________________________________________________
Mathematica [A] time = 0.05, size = 92, normalized size = 0.56 \begin {gather*} \frac {-2 e^2 (a+b x)^2 \log (d+e x)-(b d-a e) (b (d-2 e x)-3 a e)+2 e^2 (a+b x)^2 \log (a+b x)}{2 (a+b x) \sqrt {(a+b x)^2} (b d-a e)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
(-((b*d - a*e)*(-3*a*e + b*(d - 2*e*x))) + 2*e^2*(a + b*x)^2*Log[a + b*x] - 2*e^2*(a + b*x)^2*Log[d + e*x])/(2
*(b*d - a*e)^3*(a + b*x)*Sqrt[(a + b*x)^2])
________________________________________________________________________________________
IntegrateAlgebraic [B] time = 7.61, size = 4193, normalized size = 25.41 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[1/((d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]
[Out]
-((b^2*d^2*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/(a^2*(b*d - a*e)^3)) + (b^2*d^2*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^2*(b*d - a*e)^3) - (b*Sq
rt[b^2]*d^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*a^2*(b*d - a*e)^3) - (b*Sqrt[b^2]*d^2*Lo
g[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*a^2*(b*d - a*e)^3) + (b*Sqrt[b^2]*d^2*Log[2*b*d - a*e +
Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*a^2*(b*d - a*e)^3) + (b*Sqrt[b^2]*d^2*Log[2*b*d - a*e -
Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*a^2*(b*d - a*e)^3) + ((2*Sqrt[b^2]*d)/(b*d - a*e)^2 - (2*
b*Sqrt[b^2]*d*x)/(a*(b*d - a*e)^2) + (2*b*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a*(b*d - a*e)^2) - (4*b^2*d*x*ArcT
anh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(a*(b*d - a*e)^2) - (4*b^3*d*x^2*ArcTanh[(-(Sqrt[b^2]
*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(a^2*(b*d - a*e)^2) + (4*b*Sqrt[b^2]*d*x*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^2*(b*d - a*e)^2) + (4*b^2*d*x*ArcTanh[(-(Sqr
t[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) + (4*b^3*d*x^2*ArcTanh[(-(Sqr
t[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a^2*(b*d - a*e)^2) - (4*b*Sqrt[b^2]*d*x*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^2*(b*d
- a*e)^2) + (2*b*Sqrt[b^2]*d*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) + (2*
(b^2)^(3/2)*d*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a^2*(b*d - a*e)^2) - (2*b^2*d*x*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^2*(b*d - a*e)^2) + (2*b*Sq
rt[b^2]*d*x*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(b*d - a*e)^2) + (2*(b^2)^(3/2)*d*x^2*Log
[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a^2*(b*d - a*e)^2) - (2*b^2*d*x*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^2*(b*d - a*e)^2) - (2*b*Sqrt[b^2]*d*x*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) - (2*(b^2)^(3/2)*d*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^2*(b*d - a*e)^2) + (2*b^2*d*x*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^2*(b*d - a*e)^2) - (2*b*Sqrt[b
^2]*d*x*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) - (2*(b^2)^(3/2)
*d*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^2*(b*d - a*e)^2) + (2*b^2*d*x*Sq
rt[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^2*(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]))
+ ((-4*a^2*Sqrt[b^2])/(b*(-(b*d) + a*e)) - (12*(b^2)^(3/2)*x^2)/(b*(-(b*d) + a*e)) - (8*(b^2)^(3/2)*x^3)/(a*(-
(b*d) + a*e)) - (4*a*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(-(b*d) + a*e) + (4*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(-(
b*d) + a*e) + (8*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a*(-(b*d) + a*e)) - (8*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) - (16*b^3*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/a])/(a*(-(b*d) + a*e)) - (8*b^4*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a]
)/(a^2*(-(b*d) + a*e)) + (8*(b^2)^(3/2)*x^2*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*(-(b*d) + a*e)) + (8*(b^2)^(3/2)*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sq
rt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(a^2*(-(b*d) + a*e)) + (8*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) + (16*b^3*x^3*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a*(-(b*d) + a*e)) + (8*b^4*x^4*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a^2*(-(b*d) + a*e)) - (8*(b^2)^(3/2)*x^2*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*(-(b*d) + a*e)) - (8*(
b^2)^(3/2)*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)])/(a^2*(-(b*d) + a*e)) + (4*(b^2)^(3/2)*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(
b*(-(b*d) + a*e)) + (8*(b^2)^(3/2)*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*e
)) + (4*b^3*Sqrt[b^2]*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a^2*(-(b*d) + a*e)) - (4*b^2
*x^2*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)) -
(4*b^3*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]])/(a^2*(-(b*d)
+ a*e)) + (4*(b^2)^(3/2)*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*(-(b*d) + a*e)) + (8*(b^
2)^(3/2)*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*e)) + (4*b^3*Sqrt[b^2]*x^4*L
og[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a^2*(-(b*d) + a*e)) - (4*b^2*x^2*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)) - (4*b^3*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]])/(a^2*(-(b*d) + a*e)) - (4*(b^2)^(3/2)*x^2*L
og[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*(-(b*d) + a*e)) - (8*(b^2)^(3/2)*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]])/(a*(-(b*d) + a*e)) - (4*b^3*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]])/(a^2*(-(b*d) + a*e)) + (4*b^2*x^2*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)) + (4
*b^3*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]])/(a^
2*(-(b*d) + a*e)) - (4*(b^2)^(3/2)*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*
(-(b*d) + a*e)) - (8*(b^2)^(3/2)*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]])/(a*(-
(b*d) + a*e)) - (4*b^3*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]])/(a^2*
(-(b*d) + a*e)) + (4*b^2*x^2*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)) + (4*b^3*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]])/(a^2*(-(b*d) + a*e)))/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])
^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)
________________________________________________________________________________________
fricas [B] time = 0.41, size = 242, normalized size = 1.47 \begin {gather*} -\frac {b^{2} d^{2} - 4 \, a b d e + 3 \, a^{2} e^{2} - 2 \, {\left (b^{2} d e - a b e^{2}\right )} x - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (b x + a\right ) + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} b^{3} d^{3} - 3 \, a^{3} b^{2} d^{2} e + 3 \, a^{4} b d e^{2} - a^{5} e^{3} + {\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} x^{2} + 2 \, {\left (a b^{4} d^{3} - 3 \, a^{2} b^{3} d^{2} e + 3 \, a^{3} b^{2} d e^{2} - a^{4} b e^{3}\right )} x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
[Out]
-1/2*(b^2*d^2 - 4*a*b*d*e + 3*a^2*e^2 - 2*(b^2*d*e - a*b*e^2)*x - 2*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*log(
b*x + a) + 2*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*log(e*x + d))/(a^2*b^3*d^3 - 3*a^3*b^2*d^2*e + 3*a^4*b*d*e^
2 - a^5*e^3 + (b^5*d^3 - 3*a*b^4*d^2*e + 3*a^2*b^3*d*e^2 - a^3*b^2*e^3)*x^2 + 2*(a*b^4*d^3 - 3*a^2*b^3*d^2*e +
3*a^3*b^2*d*e^2 - a^4*b*e^3)*x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
[Out]
sage0*x
________________________________________________________________________________________
maple [A] time = 0.07, size = 156, normalized size = 0.95 \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 )+4 a b \,e^{2} x \ln \left (b x +a \right )-4 a b \,e^{2} x \ln \left (e x +d \right )+2 a^{2} e^{2} \ln \left (b x +a \right )-2 a^{2} e^{2} \ln \left (e x +d \right )-2 a b \,e^{2} x +2 b^{2} d e x -3 a^{2} e^{2}+4 a b d e -b^{2} d^{2}\right ) \left (b x +a \right )}{2 \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(1/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
[Out]
-1/2*(2*b^2*e^2*x^2*ln(b*x+a)-2*b^2*e^2*x^2*ln(e*x+d)+4*a*b*e^2*x*ln(b*x+a)-4*ln(e*x+d)*x*a*b*e^2+2*a^2*e^2*ln
(b*x+a)-2*ln(e*x+d)*a^2*e^2-2*a*b*e^2*x+2*b^2*d*e*x-3*a^2*e^2+4*a*b*d*e-b^2*d^2)*(b*x+a)/(a*e-b*d)^3/((b*x+a)^
2)^(3/2)
________________________________________________________________________________________
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(1/(e*x+d)/(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?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (d+e\,x\right )\,{\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(1/((d + e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)
[Out]
int(1/((d + e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right ) \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(1/(e*x+d)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
[Out]
Integral(1/((d + e*x)*((a + b*x)**2)**(3/2)), x)
________________________________________________________________________________________