\(\int \frac {x^2}{\sqrt {b x+c x^2}} \, dx\) [44]

Optimal result

Integrand size = 17, antiderivative size = 76 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=-\frac {3 b \sqrt {b x+c x^2}}{4 c^2}+\frac {x \sqrt {b x+c x^2}}{2 c}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {684, 654, 634, 212} \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}}-\frac {3 b \sqrt {b x+c x^2}}{4 c^2}+\frac {x \sqrt {b x+c x^2}}{2 c} \]

[In]

Int[x^2/Sqrt[b*x + c*x^2],x]

[Out]

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

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 654

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

Rule 684

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

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.28 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x \left (-3 b^2-b c x+2 c^2 x^2\right )+6 b^2 \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{4 c^{5/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x^2/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*x*(-3*b^2 - b*c*x + 2*c^2*x^2) + 6*b^2*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sq
rt[b + c*x])])/(4*c^(5/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {2 c^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}\, x +3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{2}-3 b \sqrt {c}\, \sqrt {x \left (c x +b \right )}}{4 c^{\frac {5}{2}}}\)	\(59\)
risch	\(-\frac {\left (-2 c x +3 b \right ) x \left (c x +b \right )}{4 c^{2} \sqrt {x \left (c x +b \right )}}+\frac {3 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}\)	\(62\)
default	\(\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\)	\(71\)

[In]

int(x^2/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\left [\frac {3 \, b^{2} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{2} x - 3 \, b c\right )} \sqrt {c x^{2} + b x}}{8 \, c^{3}}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{2} x - 3 \, b c\right )} \sqrt {c x^{2} + b x}}{4 \, c^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.53 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \frac {3 b^{2} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \left (- \frac {3 b}{4 c^{2}} + \frac {x}{2 c}\right ) \sqrt {b x + c x^{2}} & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {5}{2}}}{5 b^{3}} & \text {for}\: b \neq 0 \\\tilde {\infty } x^{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*b**2*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c)
+ x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(8*c**2) + (-3*b/(4*c**2) + x/(2*c))*sqrt(b*x + c*x**2)
, Ne(c, 0)), (2*(b*x)**(5/2)/(5*b**3), Ne(b, 0)), (zoo*x**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} x}{2 \, c} + \frac {3 \, b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} b}{4 \, c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, x}{c} - \frac {3 \, b}{c^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {b x+c x^2}} \, dx=\int \frac {x^2}{\sqrt {c\,x^2+b\,x}} \,d x \]

[In]

int(x^2/(b*x + c*x^2)^(1/2),x)

[Out]

int(x^2/(b*x + c*x^2)^(1/2), x)