\(\int \frac {\sqrt {b x+c x^2}}{x^4} \, dx\) [8]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3} \]

[Out]

-2/5*(c*x^2+b*x)^(3/2)/b/x^4+4/15*c*(c*x^2+b*x)^(3/2)/b^2/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=\frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(5*b*x^4) + (4*c*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

Rule 664

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

Rule 672

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}-\frac {(2 c) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{5 b} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (3 b^2+b c x-2 c^2 x^2\right )}{15 b^2 x^3} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(3*b^2 + b*c*x - 2*c^2*x^2))/(15*b^2*x^3)

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(-\frac {2 \left (c x +b \right ) \sqrt {x \left (c x +b \right )}\, \left (-\frac {2 c x}{3}+b \right )}{5 x^{3} b^{2}}\)	\(29\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-2 c x +3 b \right ) \sqrt {c \,x^{2}+b x}}{15 b^{2} x^{3}}\)	\(33\)
trager	\(-\frac {2 \left (-2 c^{2} x^{2}+b c x +3 b^{2}\right ) \sqrt {c \,x^{2}+b x}}{15 b^{2} x^{3}}\)	\(38\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\)	\(41\)
risch	\(-\frac {2 \left (c x +b \right ) \left (-2 c^{2} x^{2}+b c x +3 b^{2}\right )}{15 x^{2} \sqrt {x \left (c x +b \right )}\, b^{2}}\)	\(41\)

[In]

int((c*x^2+b*x)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=\frac {2 \, {\left (2 \, c^{2} x^{2} - b c x - 3 \, b^{2}\right )} \sqrt {c x^{2} + b x}}{15 \, b^{2} x^{3}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^4,x, algorithm="fricas")

[Out]

2/15*(2*c^2*x^2 - b*c*x - 3*b^2)*sqrt(c*x^2 + b*x)/(b^2*x^3)

Sympy [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{x^{4}}\, dx \]

[In]

integrate((c*x**2+b*x)**(1/2)/x**4,x)

[Out]

Integral(sqrt(x*(b + c*x))/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=\frac {4 \, \sqrt {c x^{2} + b x} c^{2}}{15 \, b^{2} x} - \frac {2 \, \sqrt {c x^{2} + b x} c}{15 \, b x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x}}{5 \, x^{3}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^4,x, algorithm="maxima")

[Out]

4/15*sqrt(c*x^2 + b*x)*c^2/(b^2*x) - 2/15*sqrt(c*x^2 + b*x)*c/(b*x^2) - 2/5*sqrt(c*x^2 + b*x)/x^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (40) = 80\).

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.23 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{\frac {3}{2}} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + 3 \, b^{3}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^4,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*c^(3/2) + 25*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c + 15*(sqrt(c)*x
- sqrt(c*x^2 + b*x))*b^2*sqrt(c) + 3*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^5

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,b^2+b\,c\,x-2\,c^2\,x^2\right )}{15\,b^2\,x^3} \]

[In]

int((b*x + c*x^2)^(1/2)/x^4,x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*b^2 - 2*c^2*x^2 + b*c*x))/(15*b^2*x^3)