3.2133 \(\int \frac{(a+b x) \sqrt{d+e x}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac{\sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

[Out]

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

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Rubi [A]  time = 0.0697587, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {768, 646, 63, 208} \[ -\frac{\sqrt{d+e x}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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

Mathematica [A]  time = 0.0896922, size = 81, normalized size = 0.75 \[ \frac{\frac{e (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{\sqrt{a e-b d}}-\sqrt{b} \sqrt{d+e x}}{b^{3/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 108, normalized size = 1. \begin{align*} -{\frac{ \left ( bx+a \right ) ^{2}}{b} \left ( -\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ) xbe-\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ) ae+\sqrt{ex+d}\sqrt{ \left ( ae-bd \right ) b} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )} \sqrt{e x + d}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*sqrt(e*x + d)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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Fricas [A]  time = 1.02333, size = 498, normalized size = 4.61 \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 - a^{2} b^{2} e +{\left (b^{4} d - a b^{3} e\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 - a^{2} b^{2} e +{\left (b^{4} d - a b^{3} e\right )} x}\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)^(3/2),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 - a^2*b^2*e + (b^4*d - a*b^3*e)*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
- a^2*b^2*e + (b^4*d - a*b^3*e)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \sqrt{d + e x}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral((a + b*x)*sqrt(d + e*x)/((a + b*x)**2)**(3/2), x)

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Giac [A]  time = 1.17479, size = 173, normalized size = 1.6 \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} b \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{\sqrt{x e + d} e}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\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)^(3/2),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*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - s
qrt(x*e + d)*e/(((x*e + d)*b - b*d + a*e)*b*sgn((x*e + d)*b*e - b*d*e + a*e^2))