3.54 \(\int \frac {x^2}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 x}{c \sqrt {b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {652, 620, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 x}{c \sqrt {b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*x)/(c*Sqrt[b*x + c*x^2]) + (2*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 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p,
-1]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 67, normalized size = 1.40 \[ \frac {2 \sqrt {b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )-2 \sqrt {c} x}{c^{3/2} \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x + c*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.81, size = 126, normalized size = 2.62 \[ \left [\frac {{\left (c x + b\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} c}{c^{3} x + b c^{2}}, -\frac {2 \, {\left ({\left (c x + b\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c\right )}}{c^{3} x + b c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2,2]%%%}+%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[
1,3]%%%}+%%%{1,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1]
%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 47, normalized size = 0.98 \[ -\frac {2 x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {\ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x/c/(c*x^2+b*x)^(1/2)+1/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 = 1.42, size = 45, normalized size = 0.94 \[ -\frac {2 \, x}{\sqrt {c x^{2} + b x} c} + \frac {\log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.20, size = 46, normalized size = 0.96 \[ \frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}-\frac {2\,x}{c\,\sqrt {c\,x^2+b\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x + c*x^2)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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