3.229 \(\int \frac{A+B x}{\sqrt{e x} \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 B \sqrt{b x+c x^2}}{c \sqrt{e x}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{b x+c x^2}}{\sqrt{b} \sqrt{e x}}\right )}{\sqrt{b} \sqrt{e}} \]

[Out]

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

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Rubi [A]  time = 0.0611555, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {794, 660, 208} \[ \frac{2 B \sqrt{b x+c x^2}}{c \sqrt{e x}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{b x+c x^2}}{\sqrt{b} \sqrt{e x}}\right )}{\sqrt{b} \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

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

Mathematica [A]  time = 0.0354381, size = 71, normalized size = 0.99 \[ \frac{2 x \left (\sqrt{b} B (b+c x)-A c \sqrt{b+c x} \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{\sqrt{b} c \sqrt{e x} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(Sqrt[b]*B*(b + c*x) - A*c*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(Sqrt[b]*c*Sqrt[e*x]*Sqrt[x*(b
+ c*x)])

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Maple [A]  time = 0.017, size = 72, normalized size = 1. \begin{align*} -2\,{\frac{\sqrt{x \left ( cx+b \right ) }}{\sqrt{ex}\sqrt{e \left ( cx+b \right ) }c\sqrt{be}} \left ( Ace{\it Artanh} \left ({\frac{\sqrt{e \left ( cx+b \right ) }}{\sqrt{be}}} \right ) -B\sqrt{e \left ( cx+b \right ) }\sqrt{be} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x)^(1/2),x)

[Out]

-2/(e*x)^(1/2)*(x*(c*x+b))^(1/2)*(A*c*e*arctanh((e*(c*x+b))^(1/2)/(b*e)^(1/2))-B*(e*(c*x+b))^(1/2)*(b*e)^(1/2)
)/(e*(c*x+b))^(1/2)/c/(b*e)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{\sqrt{c x^{2} + b x} \sqrt{e x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65939, size = 367, normalized size = 5.1 \begin{align*} \left [\frac{\sqrt{b e} A c x \log \left (-\frac{c e x^{2} + 2 \, b e x - 2 \, \sqrt{c x^{2} + b x} \sqrt{b e} \sqrt{e x}}{x^{2}}\right ) + 2 \, \sqrt{c x^{2} + b x} \sqrt{e x} B b}{b c e x}, \frac{2 \,{\left (\sqrt{-b e} A c x \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-b e} \sqrt{e x}}{c e x^{2} + b e x}\right ) + \sqrt{c x^{2} + b x} \sqrt{e x} B b\right )}}{b c e x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(1/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[(sqrt(b*e)*A*c*x*log(-(c*e*x^2 + 2*b*e*x - 2*sqrt(c*x^2 + b*x)*sqrt(b*e)*sqrt(e*x))/x^2) + 2*sqrt(c*x^2 + b*x
)*sqrt(e*x)*B*b)/(b*c*e*x), 2*(sqrt(-b*e)*A*c*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-b*e)*sqrt(e*x)/(c*e*x^2 + b*e*x
)) + sqrt(c*x^2 + b*x)*sqrt(e*x)*B*b)/(b*c*e*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{\sqrt{e x} \sqrt{x \left (b + c x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/(sqrt(e*x)*sqrt(x*(b + c*x))), x)

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Giac [A]  time = 1.15076, size = 139, normalized size = 1.93 \begin{align*} 2 \,{\left (\frac{A \arctan \left (\frac{\sqrt{c x e + b e}}{\sqrt{-b e}}\right ) e}{\sqrt{-b e}} + \frac{\sqrt{c x e + b e} B}{c}\right )} e^{\left (-1\right )} - \frac{2 \,{\left (A c \arctan \left (\frac{\sqrt{b} e^{\frac{1}{2}}}{\sqrt{-b e}}\right ) e + \sqrt{-b e} B \sqrt{b} e^{\frac{1}{2}}\right )} e^{\left (-1\right )}}{\sqrt{-b e} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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