\(\int \frac {\sqrt {x}}{(b x+c x^2)^{3/2}} \, dx\) [108]

Optimal result

Integrand size = 19, antiderivative size = 56 \[ \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {680, 674, 213} \[ \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}} \]

[In]

Int[Sqrt[x]/(b*x + c*x^2)^(3/2),x]

[Out]

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

Rule 213

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

Rule 674

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 680

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (\sqrt {b}-\sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{b^{3/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[Sqrt[x]/(b*x + c*x^2)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}-\sqrt {b}\right )}{b^{\frac {3}{2}} \sqrt {x}\, \left (c x +b \right )}\)	\(51\)

[In]

int(x^(1/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.77 \[ \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (c x^{2} + b x\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} b \sqrt {x}}{b^{2} c x^{2} + b^{3} x}, \frac {2 \, {\left ({\left (c x^{2} + b x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} b \sqrt {x}\right )}}{b^{2} c x^{2} + b^{3} x}\right ] \]

[In]

integrate(x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate(x**(1/2)/(c*x**2+b*x)**(3/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate(x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {2 \, {\left (\sqrt {b} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b}\right )}}{\sqrt {-b} b^{\frac {3}{2}}} + \frac {2}{\sqrt {c x + b} b} \]

[In]

integrate(x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^(1/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(1/2)/(b*x + c*x^2)^(3/2), x)