\(\int \sqrt {b x+c x^2} \, dx\) [288]

Optimal result

Integrand size = 13, antiderivative size = 60 \[ \int \sqrt {b x+c x^2} \, dx=\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {626, 634, 212} \[ \int \sqrt {b x+c x^2} \, dx=\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}} \]

[In]

Int[Sqrt[b*x + c*x^2],x]

[Out]

((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*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 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \sqrt {b x+c x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} (b+2 c x)+\frac {2 b^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{4 c^{3/2}} \]

[In]

Integrate[Sqrt[b*x + c*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\)	\(56\)
pseudoelliptic	\(\frac {2 c^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}\, x +b \sqrt {c}\, \sqrt {x \left (c x +b \right )}-\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b^{2}}{4 c^{\frac {3}{2}}}\)	\(58\)
risch	\(\frac {\left (2 c x +b \right ) x \left (c x +b \right )}{4 c \sqrt {x \left (c x +b \right )}}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.02 \[ \int \sqrt {b x+c x^2} \, dx=\left [\frac {b^{2} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{2} x + b c\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2}}, \frac {b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, c^{2} x + b c\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.70 \[ \int \sqrt {b x+c x^2} \, dx=\begin {cases} - \frac {b^{2} \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c} + \left (\frac {b}{4 c} + \frac {x}{2}\right ) \sqrt {b x + c x^{2}} & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05 \[ \int \sqrt {b x+c x^2} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b x} x - \frac {b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x} b}{4 \, c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98 \[ \int \sqrt {b x+c x^2} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (2 \, x + \frac {b}{c}\right )} + \frac {b^{2} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \sqrt {b x+c x^2} \, dx=\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}} \]

[In]

int((b*x + c*x^2)^(1/2),x)

[Out]

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