### 3.1550 $$\int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx$$

Optimal. Leaf size=46 $\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 (d+e x)^2 (b d-a e)}$

[Out]

((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(b*d - a*e)*(d + e*x)^2)

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Rubi [A]  time = 0.0198825, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.071, Rules used = {646, 37} $\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 (d+e x)^2 (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(b*d - a*e)*(d + e*x)^2)

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]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{a b+b^2 x}{(d+e x)^3} \, dx}{a b+b^2 x}\\ &=\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 (b d-a e) (d+e x)^2}\\ \end{align*}

Mathematica [A]  time = 0.0163794, size = 44, normalized size = 0.96 $-\frac{\sqrt{(a+b x)^2} (a e+b (d+2 e x))}{2 e^2 (a+b x) (d+e x)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[(a + b*x)^2]*(a*e + b*(d + 2*e*x)))/(2*e^2*(a + b*x)*(d + e*x)^2)

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Maple [A]  time = 0.041, size = 41, normalized size = 0.9 \begin{align*} -{\frac{2\,bxe+ae+bd}{2\, \left ( ex+d \right ) ^{2}{e}^{2} \left ( bx+a \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55283, size = 81, normalized size = 1.76 \begin{align*} -\frac{2 \, b e x + b d + a e}{2 \,{\left (e^{4} x^{2} + 2 \, d e^{3} x + d^{2} e^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.451762, size = 39, normalized size = 0.85 \begin{align*} - \frac{a e + b d + 2 b e x}{2 d^{2} e^{2} + 4 d e^{3} x + 2 e^{4} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13743, size = 59, normalized size = 1.28 \begin{align*} -\frac{{\left (2 \, b x e \mathrm{sgn}\left (b x + a\right ) + b d \mathrm{sgn}\left (b x + a\right ) + a e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{2 \,{\left (x e + d\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*x*e*sgn(b*x + a) + b*d*sgn(b*x + a) + a*e*sgn(b*x + a))*e^(-2)/(x*e + d)^2