3.1.73 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^3} \, dx\)

Optimal. Leaf size=73 \[ -\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}-\frac {2 B \sqrt {b x+c x^2}}{x}+2 B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 73, 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, 662, 620, 206} \begin {gather*} -\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}-\frac {2 B \sqrt {b x+c x^2}}{x}+2 B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*Sqrt[b*x + c*x^2])/x - (2*A*(b*x + c*x^2)^(3/2))/(3*b*x^3) + 2*B*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + 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^3} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}+B \int \frac {\sqrt {b x+c x^2}}{x^2} \, dx\\ &=-\frac {2 B \sqrt {b x+c x^2}}{x}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}+(B c) \int \frac {1}{\sqrt {b x+c x^2}} \, dx\\ &=-\frac {2 B \sqrt {b x+c x^2}}{x}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}+(2 B c) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )\\ &=-\frac {2 B \sqrt {b x+c x^2}}{x}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}+2 B \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 86, normalized size = 1.18 \begin {gather*} \frac {2 \sqrt {x (b+c x)} \left ((b+c x) \sqrt {\frac {c x}{b}+1} (b B-A c)-b^2 B \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x}{b}\right )\right )}{3 b c x^2 \sqrt {\frac {c x}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*((b*B - A*c)*(b + c*x)*Sqrt[1 + (c*x)/b] - b^2*B*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c
*x)/b)]))/(3*b*c*x^2*Sqrt[1 + (c*x)/b])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.32, size = 72, normalized size = 0.99 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} (A b+A c x+3 b B x)}{3 b x^2}-B \sqrt {c} \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 141, normalized size = 1.93 \begin {gather*} \left [\frac {3 \, B b \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} {\left (A b + {\left (3 \, B b + A c\right )} x\right )}}{3 \, b x^{2}}, -\frac {2 \, {\left (3 \, B b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} {\left (A b + {\left (3 \, B b + A c\right )} x\right )}\right )}}{3 \, b x^{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^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.26, size = 151, normalized size = 2.07 \begin {gather*} -B \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b \sqrt {c} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A c^{\frac {3}{2}} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b c + A b^{2} \sqrt {c}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} \sqrt {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^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 89, normalized size = 1.22 \begin {gather*} B \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {2 \sqrt {c \,x^{2}+b x}\, B c}{b}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} B}{b \,x^{2}}-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} A}{3 b \,x^{3}} \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^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 85, normalized size = 1.16 \begin {gather*} {\left (\sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {2 \, \sqrt {c x^{2} + b x}}{x}\right )} B - \frac {2}{3} \, A {\left (\frac {\sqrt {c x^{2} + b x} c}{b x} + \frac {\sqrt {c x^{2} + b x}}{x^{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^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^3,x)

[Out]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{3}}\, 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**3,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**3, x)

________________________________________________________________________________________