3.1707 \(\int \frac {1}{\sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=72 \[ -\frac {2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {646, 63, 208} \[ -\frac {2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.02, size = 63, normalized size = 0.88 \[ -\frac {2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {(a+b x)^2} \sqrt {b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.99, size = 119, normalized size = 1.65 \[ \left [\frac {\log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right )}{\sqrt {b^{2} d - a b e}}, \frac {2 \, \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right )}{b^{2} d - a b e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 47, normalized size = 0.65 \[ \frac {2 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) \mathrm {sgn}\left (b x + a\right )}{\sqrt {-b^{2} d + a b e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 51, normalized size = 0.71 \[ \frac {2 \left (b x +a \right ) \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (b x +a \right )^{2}}\, \sqrt {\left (a e -b d \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d+e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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