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

Optimal. Leaf size=68 \[ -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {2 \sqrt {x} (b B-A c)}{b c \sqrt {b x+c x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {788, 660, 207} \begin {gather*} -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {2 \sqrt {x} (b B-A c)}{b c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 788

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 69, normalized size = 1.01 \begin {gather*} -\frac {2 \sqrt {x} \left (\sqrt {b} (b B-A c)+A c \sqrt {b+c x} \tanh ^{-1}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{b^{3/2} c \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.75, size = 75, normalized size = 1.10 \begin {gather*} \frac {2 \sqrt {b x+c x^2} (A c-b B)}{b c \sqrt {x} (b+c x)}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 192, normalized size = 2.82 \begin {gather*} \left [\frac {{\left (A c^{2} x^{2} + A b c x\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (B b^{2} - A b c\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{b^{2} c^{2} x^{2} + b^{3} c x}, \frac {2 \, {\left ({\left (A c^{2} x^{2} + A b c x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (B b^{2} - A b c\right )} \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{b^{2} c^{2} x^{2} + b^{3} c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.21, size = 96, normalized size = 1.41 \begin {gather*} \frac {2 \, A \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {2 \, {\left (B b - A c\right )}}{\sqrt {c x + b} b c} - \frac {2 \, {\left (A \sqrt {b} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) - B \sqrt {-b} b + A \sqrt {-b} c\right )}}{\sqrt {-b} b^{\frac {3}{2}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

2*A*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - 2*(B*b - A*c)/(sqrt(c*x + b)*b*c) - 2*(A*sqrt(b)*c*arctan(sq
rt(b)/sqrt(-b)) - B*sqrt(-b)*b + A*sqrt(-b)*c)/(sqrt(-b)*b^(3/2)*c)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 63, normalized size = 0.93 \begin {gather*} -\frac {2 \sqrt {\left (c x +b \right ) x}\, \left (\sqrt {c x +b}\, A c \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-A \sqrt {b}\, c +B \,b^{\frac {3}{2}}\right )}{\left (c x +b \right ) b^{\frac {3}{2}} c \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x + A\right )} \sqrt {x}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*sqrt(x)/(c*x^2 + b*x)^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x)*(A + B*x)/(x*(b + c*x))**(3/2), x)

________________________________________________________________________________________