3.2069 \(\int \frac{(a+b x) \sqrt{d+e x}}{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 \sqrt{d+e x}}{b}-\frac{2 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0292775, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 50, 63, 208} \[ \frac{2 \sqrt{d+e x}}{b}-\frac{2 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0249448, size = 62, normalized size = 1. \[ \frac{2 \sqrt{d+e x}}{b}-\frac{2 \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 92, normalized size = 1.5 \begin{align*} 2\,{\frac{\sqrt{ex+d}}{b}}-2\,{\frac{ae}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+2\,{\frac{d}{\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(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 [A]  time = 1.0368, size = 306, normalized size = 4.94 \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((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^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 = 22.0554, size = 61, normalized size = 0.98 \begin{align*} \frac{2 \left (\frac{e \sqrt{d + e x}}{b} - \frac{e \left (a e - b d\right ) \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e - b d}{b}}} \right )}}{b^{2} \sqrt{\frac{a e - b d}{b}}}\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15411, size = 90, normalized size = 1.45 \begin{align*} \frac{2 \,{\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{2 \, \sqrt{x e + d}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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