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

Optimal. Leaf size=75 \[ \frac {b (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}}-\frac {\sqrt {b x+c x^2} (-4 A c+3 b B-2 B c x)}{4 c^2} \]

[Out]

1/4*b*(-4*A*c+3*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-1/4*(-2*B*c*x-4*A*c+3*B*b)*(c*x^2+b*x)^(1/2)
/c^2

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Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {779, 620, 206} \[ \frac {b (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}}-\frac {\sqrt {b x+c x^2} (-4 A c+3 b B-2 B c x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*B - 4*A*c - 2*B*c*x)*Sqrt[b*x + c*x^2])/(4*c^2) + (b*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x
^2]])/(4*c^(5/2))

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x (A+B x)}{\sqrt {b x+c x^2}} \, dx &=-\frac {(3 b B-4 A c-2 B c x) \sqrt {b x+c x^2}}{4 c^2}+\frac {(b (3 b B-4 A c)) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c^2}\\ &=-\frac {(3 b B-4 A c-2 B c x) \sqrt {b x+c x^2}}{4 c^2}+\frac {(b (3 b B-4 A c)) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c^2}\\ &=-\frac {(3 b B-4 A c-2 B c x) \sqrt {b x+c x^2}}{4 c^2}+\frac {b (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 96, normalized size = 1.28 \[ \frac {b^{3/2} \sqrt {x} \sqrt {\frac {c x}{b}+1} (3 b B-4 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )+\sqrt {c} x (b+c x) (4 A c-3 b B+2 B c x)}{4 c^{5/2} \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(b + c*x)*(-3*b*B + 4*A*c + 2*B*c*x) + b^(3/2)*(3*b*B - 4*A*c)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(S
qrt[c]*Sqrt[x])/Sqrt[b]])/(4*c^(5/2)*Sqrt[x*(b + c*x)])

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fricas [A]  time = 0.69, size = 158, normalized size = 2.11 \[ \left [-\frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} x - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{8 \, c^{3}}, -\frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B c^{2} x - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{4 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((3*B*b^2 - 4*A*b*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*B*c^2*x - 3*B*b*c + 4*A
*c^2)*sqrt(c*x^2 + b*x))/c^3, -1/4*((3*B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2
*B*c^2*x - 3*B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x))/c^3]

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giac [A]  time = 0.24, size = 83, normalized size = 1.11 \[ \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B x}{c} - \frac {3 \, B b - 4 \, A c}{c^{2}}\right )} - \frac {{\left (3 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*B*x/c - (3*B*b - 4*A*c)/c^2) - 1/8*(3*B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(c)*x - sqrt(c
*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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maple [A]  time = 0.05, size = 118, normalized size = 1.57 \[ -\frac {A b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}+\frac {3 B \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+b x}\, B x}{2 c}+\frac {\sqrt {c \,x^{2}+b x}\, A}{c}-\frac {3 \sqrt {c \,x^{2}+b x}\, B b}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x/c*(c*x^2+b*x)^(1/2)-3/4*B*b/c^2*(c*x^2+b*x)^(1/2)+3/8*B*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)
^(1/2))+A/c*(c*x^2+b*x)^(1/2)-1/2*A*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 0.79, size = 115, normalized size = 1.53 \[ \frac {\sqrt {c x^{2} + b x} B x}{2 \, c} + \frac {3 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} B b}{4 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + b*x)*B*x/c + 3/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 1/2*A*b*log(2*c
*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) - 3/4*sqrt(c*x^2 + b*x)*B*b/c^2 + sqrt(c*x^2 + b*x)*A/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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