\(\int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx\) [77]

Optimal result

Integrand size = 19, antiderivative size = 53 \[ \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, 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 = {678, 674, 213} \[ \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \sqrt {b x+c x^2}}{\sqrt {x}}-2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right ) \]

[In]

Int[Sqrt[b*x + c*x^2]/x^(3/2),x]

[Out]

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

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 678

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \sqrt {x} \sqrt {b+c x} \left (\sqrt {b+c x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{\sqrt {x (b+c x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (\sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-\sqrt {c x +b}\right )}{\sqrt {x}\, \sqrt {c x +b}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.09 \[ \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx=\left [\frac {\sqrt {b} x \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, \sqrt {c x^{2} + b x} \sqrt {x}}{x}, \frac {2 \, {\left (\sqrt {-b} x \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + \sqrt {c x^{2} + b x} \sqrt {x}\right )}}{x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {b x+c x^2}}{x^{3/2}} \, dx=\frac {2 \, b \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {c x + b} - \frac {2 \, {\left (b \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b} \sqrt {b}\right )}}{\sqrt {-b}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b*x + c*x^2)^(1/2)/x^(3/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/x^(3/2), x)