\(\int \frac {x^2}{(b x+c x^2)^{3/2}} \, dx\) [54]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 x}{c \sqrt {b x+c x^2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {666, 634, 212} \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}-\frac {2 x}{c \sqrt {b x+c x^2}} \]

[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 212

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

Rule 634

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 666

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*} \text {integral}& = -\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 \text {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*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {c} x+4 \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{c^{3/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x^2/(b*x + c*x^2)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\frac {2 x}{c \sqrt {x \left (c x +b \right )}}+\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{c^{\frac {3}{2}}}\)	\(39\)
default	\(-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.62 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (c x + b\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, \sqrt {c x^{2} + b x} c}{c^{3} x + b c^{2}}, -\frac {2 \, {\left ({\left (c x + b\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c\right )}}{c^{3} x + b c^{2}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{2}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, x}{\sqrt {c x^{2} + b x} c} + \frac {\log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.40 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}} - \frac {2 \, b}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {x^2}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{c^{3/2}}-\frac {2\,x}{c\,\sqrt {c\,x^2+b\,x}} \]

[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))