3.1746 \(\int \frac{A+B x}{(a+b x) \sqrt{d+e x}} \, dx\)

Optimal. Leaf size=74 \[ \frac{2 B \sqrt{d+e x}}{b e}-\frac{2 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{b d-a e}} \]

[Out]

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

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Rubi [A]  time = 0.0414876, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {80, 63, 208} \[ \frac{2 B \sqrt{d+e x}}{b e}-\frac{2 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{b d-a e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.137004, size = 74, normalized size = 1. \[ \frac{2 (a B-A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} \sqrt{b d-a e}}+\frac{2 B \sqrt{d+e x}}{b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 96, normalized size = 1.3 \begin{align*} 2\,{\frac{B\sqrt{ex+d}}{be}}+2\,{\frac{A}{\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }-2\,{\frac{Ba}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{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)/(b*x+a)/(e*x+d)^(1/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38184, size = 450, normalized size = 6.08 \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)/(b*x+a)/(e*x+d)^(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 [A]  time = 10.6422, size = 66, normalized size = 0.89 \begin{align*} \frac{2 B \sqrt{d + e x}}{b e} + \frac{2 \left (- A b + B a\right ) \operatorname{atan}{\left (\frac{1}{\sqrt{\frac{b}{a e - b d}} \sqrt{d + e x}} \right )}}{b \sqrt{\frac{b}{a e - b d}} \left (a e - b d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29385, size = 93, normalized size = 1.26 \begin{align*} \frac{2 \, \sqrt{x e + d} B e^{\left (-1\right )}}{b} - \frac{2 \,{\left (B a - A b\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)/(b*x+a)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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