3.1862 \(\int \frac{A+B x}{\sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=124 \[ \frac{2 B (a+b x) \sqrt{d+e x}}{b e \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

[Out]

(2*B*(a + b*x)*Sqrt[d + e*x])/(b*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*(A*b - a*B)*(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 = 0.0758117, antiderivative size = 124, 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 = {770, 80, 63, 208} \[ \frac{2 B (a+b x) \sqrt{d+e x}}{b e \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*(a + b*x)*Sqrt[d + e*x])/(b*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*(A*b - a*B)*(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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{A+B x}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B (a+b x) \sqrt{d+e x}}{b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (\frac{1}{2} A b^2 e-\frac{1}{2} a b 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 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B (a+b x) \sqrt{d+e x}}{b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (4 \left (\frac{1}{2} A b^2 e-\frac{1}{2} a b 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^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B (a+b x) \sqrt{d+e x}}{b e \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (A b-a B) (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 = 0.0698704, size = 111, normalized size = 0.9 \[ \frac{2 (a+b x) \left (e (a B-A b) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )+\sqrt{b} B \sqrt{d+e x} (b d-a e)\right )}{b^{3/2} e \sqrt{(a+b x)^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 110, normalized size = 0.9 \begin{align*} 2\,{\frac{bx+a}{\sqrt{ \left ( bx+a \right ) ^{2}}eb\sqrt{ \left ( ae-bd \right ) b}} \left ( A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) be-B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) ae+B\sqrt{ex+d}\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*x+a)^2)^(1/2),x)

[Out]

2*(b*x+a)*(A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*b*e-B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*e
+B*(e*x+d)^(1/2)*((a*e-b*d)*b)^(1/2))/((b*x+a)^2)^(1/2)/e/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{B x + A}{\sqrt{{\left (b x + a\right )}^{2}} \sqrt{e x + d}}\,{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*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64944, size = 450, normalized size = 3.63 \begin{align*} \left [-\frac{\sqrt{b^{2} d - a b e}{\left (B a - A b\right )} e \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 b^{2} d - B a b e\right )} \sqrt{e x + d}}{b^{3} d e - a b^{2} e^{2}}, -\frac{2 \,{\left (\sqrt{-b^{2} d + a b e}{\left (B a - A b\right )} e \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) -{\left (B b^{2} d - B a b e\right )} \sqrt{e x + d}\right )}}{b^{3} d e - a b^{2} 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*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[-(sqrt(b^2*d - a*b*e)*(B*a - A*b)*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*b^2*d - B*a*b*e)*sqrt(e*x + d))/(b^3*d*e - a*b^2*e^2), -2*(sqrt(-b^2*d + a*b*e)*(B*a - A*b)*e*arctan(
sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (B*b^2*d - B*a*b*e)*sqrt(e*x + d))/(b^3*d*e - a*b^2*e^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{\sqrt{d + e x} \sqrt{\left (a + b x\right )^{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*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.1382, size = 117, normalized size = 0.94 \begin{align*} \frac{2 \, \sqrt{x e + d} B e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right )}{b} - \frac{2 \,{\left (B a \mathrm{sgn}\left (b x + a\right ) - A b \mathrm{sgn}\left (b x + a\right )\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*B*e^(-1)*sgn(b*x + a)/b - 2*(B*a*sgn(b*x + a) - A*b*sgn(b*x + a))*arctan(sqrt(x*e + d)*b/sqrt(
-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b)