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

Optimal. Leaf size=83 \[ \frac {\sqrt {b x+c x^2} (2 A c+b B)}{b}+\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {792, 664, 620, 206} \[ \frac {\sqrt {b x+c x^2} (2 A c+b B)}{b}+\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + 2*A*c)*Sqrt[b*x + c*x^2])/b - (2*A*(b*x + c*x^2)^(3/2))/(b*x^2) + ((b*B + 2*A*c)*ArcTanh[(Sqrt[c]*x)/S
qrt[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 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), 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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 75, normalized size = 0.90 \[ \frac {\sqrt {x (b+c x)} \left (\frac {\sqrt {x} (2 A c+b B) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c} \sqrt {\frac {c x}{b}+1}}-2 A+B x\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 138, normalized size = 1.66 \[ \left [\frac {{\left (B b + 2 \, A c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (B c x - 2 \, A c\right )} \sqrt {c x^{2} + b x}}{2 \, c x}, -\frac {{\left (B b + 2 \, A c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (B c x - 2 \, A c\right )} \sqrt {c x^{2} + b x}}{c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.21, size = 82, normalized size = 0.99 \[ \sqrt {c x^{2} + b x} B - \frac {{\left (B b + 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} + \frac {2 \, A b}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 113, normalized size = 1.36 \[ A \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}+\frac {2 \sqrt {c \,x^{2}+b x}\, A c}{b}+\sqrt {c \,x^{2}+b x}\, B -\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A}{b \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 89, normalized size = 1.07 \[ \frac {B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, \sqrt {c}} + A \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \sqrt {c x^{2} + b x} B - \frac {2 \, \sqrt {c x^{2} + b x} A}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**2, x)

________________________________________________________________________________________