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

Optimal. Leaf size=55 \[ \frac {B \sqrt {b x+c x^2}}{c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {640, 620, 206} \begin {gather*} \frac {B \sqrt {b x+c x^2}}{c}-\frac {(b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/Sqrt[b*x + c*x^2],x]

[Out]

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

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 640

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 80, normalized size = 1.45 \begin {gather*} \frac {B \sqrt {c} x (b+c x)-\sqrt {b} \sqrt {x} \sqrt {\frac {c x}{b}+1} (b B-2 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{3/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[c]*x*(b + c*x) - Sqrt[b]*(b*B - 2*A*c)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(
c^(3/2)*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.36, size = 67, normalized size = 1.22 \begin {gather*} \frac {(b B-2 A c) \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{2 c^{3/2}}+\frac {B \sqrt {b x+c x^2}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[b*x + c*x^2])/c + ((b*B - 2*A*c)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x + c*x^2]])/(2*c^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 115, normalized size = 2.09 \begin {gather*} \left [\frac {2 \, \sqrt {c x^{2} + b x} B c - {\left (B b - 2 \, A c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{2}}, \frac {\sqrt {c x^{2} + b x} B c + {\left (B b - 2 \, A c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right )}{c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*sqrt(c*x^2 + b*x)*B*c - (B*b - 2*A*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)))/c^2, (sqrt
(c*x^2 + b*x)*B*c + (B*b - 2*A*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)))/c^2]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 60, normalized size = 1.09 \begin {gather*} \frac {\sqrt {c x^{2} + b x} B}{c} + \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2 + b*x)*B/c + 1/2*(B*b - 2*A*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(3/2)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 78, normalized size = 1.42 \begin {gather*} \frac {A \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x}\, B}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(c*x^2+b*x)^(1/2)/c-1/2*B*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+A*ln((c*x+1/2*b)/c^(1/2)+(c*x^
2+b*x)^(1/2))/c^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.90, size = 75, normalized size = 1.36 \begin {gather*} -\frac {B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {A \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} + \frac {\sqrt {c x^{2} + b x} B}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*B*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + A*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))
/sqrt(c) + sqrt(c*x^2 + b*x)*B/c

________________________________________________________________________________________

mupad [B]  time = 1.35, size = 77, normalized size = 1.40 \begin {gather*} \frac {A\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{\sqrt {c}}+\frac {B\,\sqrt {c\,x^2+b\,x}}{c}-\frac {B\,b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(b*x + c*x^2)^(1/2),x)

[Out]

(A*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/c^(1/2) + (B*(b*x + c*x^2)^(1/2))/c - (B*b*log((b/2 + c*x)/
c^(1/2) + (b*x + c*x^2)^(1/2)))/(2*c^(3/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B 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)/(c*x**2+b*x)**(1/2),x)

[Out]

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

________________________________________________________________________________________