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

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

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Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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*}

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Mathematica [A]  time = 0.06, size = 76, normalized size = 1.00 \begin {gather*} \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)} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.21, size = 98, normalized size = 1.29 \begin {gather*} \frac {e \sqrt {d+e x}}{(b d-a e) (-a e-b (d+e x)+b d)}-\frac {e \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{\sqrt {b} (a e-b d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 280, normalized size = 3.68 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 97, normalized size = 1.28 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.07, size = 77, normalized size = 1.01 \begin {gather*} \frac {e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}+\frac {\sqrt {e x +d}\, e}{\left (a e -b d \right ) \left (b e x +a e \right )} \end {gather*}

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

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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(1/(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^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 positive or negative?

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mupad [B]  time = 0.57, size = 74, normalized size = 0.97 \begin {gather*} \frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}}+\frac {e\,\sqrt {d+e\,x}}{\left (a\,e-b\,d\right )\,\left (a\,e-b\,d+b\,\left (d+e\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{2} \sqrt {d + e x}}\, dx \end {gather*}

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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