3.1706 \(\int \frac{\sqrt{d+e x}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=112 \[ \frac{2 (a+b x) \sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0532999, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {646, 50, 63, 208} \[ \frac{2 (a+b x) \sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*Sqrt[d + e*x])/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*Sqrt[b*d - a*e]*(a + b*x)*ArcTanh[(Sqrt[b]*
Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{\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{\sqrt{d+e x}}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (a+b x) \sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (\left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (a+b x) \sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right ) \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 )}{b^2 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (a+b x) \sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 \sqrt{b d-a e} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0209021, size = 81, normalized size = 0.72 \[ \frac{2 (a+b x) \left (\sqrt{b} \sqrt{d+e x}-\sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )}{b^{3/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.23, size = 104, normalized size = 0.9 \begin{align*} 2\,{\frac{bx+a}{\sqrt{ \left ( bx+a \right ) ^{2}}b\sqrt{ \left ( ae-bd \right ) b}} \left ( -\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) ae+\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) bd+\sqrt{ex+d}\sqrt{ \left ( ae-bd \right ) b} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65198, size = 306, normalized size = 2.73 \begin{align*} \left [\frac{\sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \, \sqrt{e x + d}}{b}, -\frac{2 \,{\left (\sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) - \sqrt{e x + d}\right )}}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.13563, size = 95, normalized size = 0.85 \begin{align*} \sqrt{- \frac{a e}{b^{3}} + \frac{d}{b^{2}}} \log{\left (- b \sqrt{- \frac{a e}{b^{3}} + \frac{d}{b^{2}}} + \sqrt{d + e x} \right )} - \sqrt{- \frac{a e}{b^{3}} + \frac{d}{b^{2}}} \log{\left (b \sqrt{- \frac{a e}{b^{3}} + \frac{d}{b^{2}}} + \sqrt{d + e x} \right )} + \frac{2 \sqrt{d + e x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a*e/b**3 + d/b**2)*log(-b*sqrt(-a*e/b**3 + d/b**2) + sqrt(d + e*x)) - sqrt(-a*e/b**3 + d/b**2)*log(b*sqr
t(-a*e/b**3 + d/b**2) + sqrt(d + e*x)) + 2*sqrt(d + e*x)/b

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Giac [A]  time = 1.14444, size = 99, normalized size = 0.88 \begin{align*} 2 \,{\left (\frac{{\left (b d - a e\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b} + \frac{\sqrt{x e + d}}{b}\right )} \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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