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

Optimal. Leaf size=52 \[ \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 A \sqrt {b x+c x^2}}{b x} \]

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 = {792, 620, 206} \begin {gather*} \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 A \sqrt {b x+c x^2}}{b x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 69, normalized size = 1.33 \begin {gather*} \frac {2 \sqrt {x (b+c x)} \left (\frac {\sqrt {b} B \sqrt {x} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c} \sqrt {\frac {c x}{b}+1}}-A\right )}{b x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(-A + (Sqrt[b]*B*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[c]*Sqrt[1 + (c*x)/b]))
)/(b*x)

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IntegrateAlgebraic [A]  time = 0.30, size = 58, normalized size = 1.12 \begin {gather*} -\frac {2 A \sqrt {b x+c x^2}}{b x}-\frac {B \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 116, normalized size = 2.23 \begin {gather*} \left [\frac {B b \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} A c}{b c x}, -\frac {2 \, {\left (B b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} A c\right )}}{b c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 59, normalized size = 1.13 \begin {gather*} -\frac {B \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{\sqrt {c}} + \frac {2 \, A}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/sqrt(c) + 2*A/(sqrt(c)*x - sqrt(c*x^2 + b*x))

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maple [A]  time = 0.05, size = 51, normalized size = 0.98 \begin {gather*} \frac {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {2 \sqrt {c \,x^{2}+b x}\, A}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.73, size = 49, normalized size = 0.94 \begin {gather*} \frac {B \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) - 2*sqrt(c*x^2 + b*x)*A/(b*x)

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mupad [B]  time = 1.27, size = 50, normalized size = 0.96 \begin {gather*} \frac {B\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{\sqrt {c}}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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