3.1602 \(\int \frac {1}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac {e^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {3 b e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {3 b e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {2 b e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 217, 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} \[ \frac {e^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}+\frac {3 b e^2 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {3 b e^2 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {2 b e}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*e)/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*(b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) + (e^2*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*e^2*(a + b*x)*Log[a + b
*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*e^2*(a + b*x)*Log[d + e*x])/((b*d - a*e)^4*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)^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 {1}{\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 (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{b (b d-a e)^2 (a+b x)^3}-\frac {2 e}{b (b d-a e)^3 (a+b x)^2}+\frac {3 e^2}{b (b d-a e)^4 (a+b x)}-\frac {e^3}{b^3 (b d-a e)^3 (d+e x)^2}-\frac {3 e^3}{b^2 (b d-a e)^4 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 b e}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{2 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e^2 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b e^2 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 141, normalized size = 0.65 \[ \frac {-(b d-a e) \left (-2 a^2 e^2-a b e (5 d+9 e x)+b^2 \left (d^2-3 d e x-6 e^2 x^2\right )\right )+6 b e^2 (a+b x)^2 (d+e x) \log (a+b x)-6 b e^2 (a+b x)^2 (d+e x) \log (d+e x)}{2 (a+b x) \sqrt {(a+b x)^2} (d+e x) (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*d - a*e)*(-2*a^2*e^2 - a*b*e*(5*d + 9*e*x) + b^2*(d^2 - 3*d*e*x - 6*e^2*x^2))) + 6*b*e^2*(a + b*x)^2*(d
+ e*x)*Log[a + b*x] - 6*b*e^2*(a + b*x)^2*(d + e*x)*Log[d + e*x])/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[(a + b*x)^2]
*(d + e*x))

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fricas [B]  time = 1.07, size = 494, normalized size = 2.28 \[ -\frac {b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} + 2 \, a^{3} e^{3} - 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} - 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} e^{3} x^{3} + a^{2} b d e^{2} + {\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + {\left (2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} e^{3} x^{3} + a^{2} b d e^{2} + {\left (b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + {\left (2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} b^{4} d^{5} - 4 \, a^{3} b^{3} d^{4} e + 6 \, a^{4} b^{2} d^{3} e^{2} - 4 \, a^{5} b d^{2} e^{3} + a^{6} d e^{4} + {\left (b^{6} d^{4} e - 4 \, a b^{5} d^{3} e^{2} + 6 \, a^{2} b^{4} d^{2} e^{3} - 4 \, a^{3} b^{3} d e^{4} + a^{4} b^{2} e^{5}\right )} x^{3} + {\left (b^{6} d^{5} - 2 \, a b^{5} d^{4} e - 2 \, a^{2} b^{4} d^{3} e^{2} + 8 \, a^{3} b^{3} d^{2} e^{3} - 7 \, a^{4} b^{2} d e^{4} + 2 \, a^{5} b e^{5}\right )} x^{2} + {\left (2 \, a b^{5} d^{5} - 7 \, a^{2} b^{4} d^{4} e + 8 \, a^{3} b^{3} d^{3} e^{2} - 2 \, a^{4} b^{2} d^{2} e^{3} - 2 \, a^{5} b d e^{4} + a^{6} e^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*d^3 - 6*a*b^2*d^2*e + 3*a^2*b*d*e^2 + 2*a^3*e^3 - 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 - 3*(b^3*d^2*e + 2*a
*b^2*d*e^2 - 3*a^2*b*e^3)*x - 6*(b^3*e^3*x^3 + a^2*b*d*e^2 + (b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + (2*a*b^2*d*e^2 +
a^2*b*e^3)*x)*log(b*x + a) + 6*(b^3*e^3*x^3 + a^2*b*d*e^2 + (b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + (2*a*b^2*d*e^2 + a
^2*b*e^3)*x)*log(e*x + d))/(a^2*b^4*d^5 - 4*a^3*b^3*d^4*e + 6*a^4*b^2*d^3*e^2 - 4*a^5*b*d^2*e^3 + a^6*d*e^4 +
(b^6*d^4*e - 4*a*b^5*d^3*e^2 + 6*a^2*b^4*d^2*e^3 - 4*a^3*b^3*d*e^4 + a^4*b^2*e^5)*x^3 + (b^6*d^5 - 2*a*b^5*d^4
*e - 2*a^2*b^4*d^3*e^2 + 8*a^3*b^3*d^2*e^3 - 7*a^4*b^2*d*e^4 + 2*a^5*b*e^5)*x^2 + (2*a*b^5*d^5 - 7*a^2*b^4*d^4
*e + 8*a^3*b^3*d^3*e^2 - 2*a^4*b^2*d^2*e^3 - 2*a^5*b*d*e^4 + a^6*e^5)*x)

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giac [B]  time = 0.43, size = 627, normalized size = 2.89 \[ -\frac {3 \, b e^{3} \log \left ({\left | -b + \frac {b d}{x e + d} - \frac {a e}{x e + d} \right |}\right )}{b^{4} d^{4} 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 ) - 4 \, a b^{3} d^{3} 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 ) + 6 \, a^{2} b^{2} d^{2} 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 ) - 4 \, a^{3} b d 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 ) + a^{4} e^{5} \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^{5}}{{\left (b^{3} d^{3} 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 ) - 3 \, a b^{2} d^{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 ) + 3 \, a^{2} b d e^{5} \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^{6} \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 {5 \, b^{3} e^{2} - \frac {6 \, {\left (b^{3} d e^{3} - a b^{2} e^{4}\right )} e^{\left (-1\right )}}{x e + d}}{2 \, {\left (b d - a e\right )}^{4} {\left (b - \frac {b d}{x e + d} + \frac {a e}{x e + d}\right )}^{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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*e^3*log(abs(-b + b*d/(x*e + d) - a*e/(x*e + d)))/(b^4*d^4*e*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^
2/(x*e + d)^2) - 4*a*b^3*d^3*e^2*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) + 6*a^2*b^2*d^2*e
^3*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - 4*a^3*b*d*e^4*sgn(-b*e/(x*e + d) + b*d*e/(x*e
+ d)^2 - a*e^2/(x*e + d)^2) + a^4*e^5*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2)) - e^5/((b^
3*d^3*e^3*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^4*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^5*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)
^2) - a^3*e^6*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2))*(x*e + d)) - 1/2*(5*b^3*e^2 - 6*(b^
3*d*e^3 - a*b^2*e^4)*e^(-1)/(x*e + d))/((b*d - a*e)^4*(b - b*d/(x*e + d) + a*e/(x*e + d))^2*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 [B]  time = 0.07, size = 330, normalized size = 1.52 \[ \frac {\left (6 b^{3} e^{3} x^{3} \ln \left (b x +a \right )-6 b^{3} e^{3} x^{3} \ln \left (e x +d \right )+12 a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-12 a \,b^{2} e^{3} x^{2} \ln \left (e x +d \right )+6 b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )-6 b^{3} d \,e^{2} x^{2} \ln \left (e x +d \right )+6 a^{2} b \,e^{3} x \ln \left (b x +a \right )-6 a^{2} b \,e^{3} x \ln \left (e x +d \right )+12 a \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-12 a \,b^{2} d \,e^{2} x \ln \left (e x +d \right )-6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+6 a^{2} b d \,e^{2} \ln \left (b x +a \right )-6 a^{2} b d \,e^{2} \ln \left (e x +d \right )-9 a^{2} b \,e^{3} x +6 a \,b^{2} d \,e^{2} x +3 b^{3} d^{2} e x -2 a^{3} e^{3}-3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (b x +a \right )}{2 \left (e x +d \right ) \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*b^3*e^3*x^3*ln(b*x+a)-6*b^3*e^3*x^3*ln(e*x+d)+12*a*b^2*e^3*x^2*ln(b*x+a)+6*b^3*d*e^2*x^2*ln(b*x+a)-12*l
n(e*x+d)*x^2*a*b^2*e^3-6*b^3*d*e^2*x^2*ln(e*x+d)+6*a^2*b*e^3*x*ln(b*x+a)+12*a*b^2*d*e^2*x*ln(b*x+a)-6*ln(e*x+d
)*x*a^2*b*e^3-12*ln(e*x+d)*x*a*b^2*d*e^2-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*a^2*b*d*e^2*ln(b*x+a)-6*a^2*b*d*e^2
*ln(e*x+d)-9*a^2*b*e^3*x+6*a*b^2*d*e^2*x+3*b^3*d^2*e*x-2*a^3*e^3-3*a^2*b*d*e^2+6*a*b^2*d^2*e-b^3*d^3)*(b*x+a)/
(e*x+d)/(a*e-b*d)^4/((b*x+a)^2)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = -0.00 \[ \int \frac {1}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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