### 3.1650 $$\int \frac{1}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)} \, dx$$

Optimal. Leaf size=76 $\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{b} (b d-a e)^{3/2}}-\frac{\sqrt{d+e x}}{(a+b x) (b d-a e)}$

[Out]

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

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Rubi [A]  time = 0.0444152, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 51, 63, 208} $\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{b} (b d-a e)^{3/2}}-\frac{\sqrt{d+e x}}{(a+b x) (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

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] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) (a+b x)}-\frac{e \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 (b d-a e)}\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) (a+b x)}-\frac{\operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b d-a e}\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) (a+b x)}+\frac{e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{b} (b d-a e)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0556035, size = 76, normalized size = 1. $\frac{e \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{\sqrt{b} (a e-b d)^{3/2}}-\frac{\sqrt{d+e x}}{(a+b x) (b d-a e)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.196, size = 77, normalized size = 1. \begin{align*}{\frac{e}{ \left ( ae-bd \right ) \left ( bxe+ae \right ) }\sqrt{ex+d}}+{\frac{e}{ae-bd}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99259, size = 603, normalized size = 7.93 \begin{align*} \left [-\frac{\sqrt{b^{2} d - a b e}{\left (b e x + a e\right )} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) + 2 \,{\left (b^{2} d - a b e\right )} \sqrt{e x + d}}{2 \,{\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} +{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x\right )}}, -\frac{\sqrt{-b^{2} d + a b e}{\left (b e x + a e\right )} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) +{\left (b^{2} d - a b e\right )} \sqrt{e x + d}}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} +{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(b^2*d - a*b*e)*(b*e*x + a*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x
+ a)) + 2*(b^2*d - a*b*e)*sqrt(e*x + d))/(a*b^3*d^2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2
*b^2*e^2)*x), -(sqrt(-b^2*d + a*b*e)*(b*e*x + a*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) +
(b^2*d - a*b*e)*sqrt(e*x + d))/(a*b^3*d^2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*
x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{2} \sqrt{d + e x}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1706, size = 131, normalized size = 1.72 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e}{\sqrt{-b^{2} d + a b e}{\left (b d - a e\right )}} - \frac{\sqrt{x e + d} e}{{\left ({\left (x e + d\right )} b - b d + a e\right )}{\left (b d - a e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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