\(\int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx\) [78]

Optimal result

Integrand size = 19, antiderivative size = 54 \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=-\frac {\sqrt {b x+c x^2}}{x^{3/2}}-\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, 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 = {676, 674, 213} \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=-\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}}-\frac {\sqrt {b x+c x^2}}{x^{3/2}} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^(5/2),x]

[Out]

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

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 676

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 + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=-\frac {\sqrt {x (b+c x)}}{x^{3/2}}-\frac {c \sqrt {x (b+c x)} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {x} \sqrt {b+c x}} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\left (-\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c x -\sqrt {c x +b}\, \sqrt {b}\right ) \sqrt {x \left (c x +b \right )}}{x^{\frac {3}{2}} \sqrt {c x +b}\, \sqrt {b}}\)	\(53\)
risch	\(-\frac {c x +b}{\sqrt {x}\, \sqrt {x \left (c x +b \right )}}-\frac {c \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{\sqrt {b}\, \sqrt {x \left (c x +b \right )}}\)	\(58\)

[In]

int((c*x^2+b*x)^(1/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.33 \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=\left [\frac {\sqrt {b} c x^{2} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, \sqrt {c x^{2} + b x} b \sqrt {x}}{2 \, b x^{2}}, \frac {\sqrt {-b} c x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - \sqrt {c x^{2} + b x} b \sqrt {x}}{b x^{2}}\right ] \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

[1/2*(sqrt(b)*c*x^2*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*sq
rt(x))/(b*x^2), (sqrt(-b)*c*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - sqrt(c*x^2 + b*x)*b*sqrt(x))/(b*x
^2)]

Sympy [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{x^{\frac {5}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**(1/2)/x**(5/2),x)

[Out]

Integral(sqrt(x*(b + c*x))/x**(5/2), x)

Maxima [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x}}{x^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^2 + b*x)/x^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=\frac {\frac {c^{2} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {\sqrt {c x + b} c}{x}}{c} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b x+c x^2}}{x^{5/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}}{x^{5/2}} \,d x \]

[In]

int((b*x + c*x^2)^(1/2)/x^(5/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/x^(5/2), x)