\(\int \frac {\sqrt {b x+c x^2}}{x^5} \, dx\) [9]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=-\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac {8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3} \]

[Out]

-2/7*(c*x^2+b*x)^(3/2)/b/x^5+8/35*c*(c*x^2+b*x)^(3/2)/b^2/x^4-16/105*c^2*(c*x^2+b*x)^(3/2)/b^3/x^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3}+\frac {8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^5,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(7*b*x^5) + (8*c*(b*x + c*x^2)^(3/2))/(35*b^2*x^4) - (16*c^2*(b*x + c*x^2)^(3/2))/(10
5*b^3*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}}{7 b x^5}-\frac {(4 c) \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx}{7 b} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac {8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}+\frac {\left (8 c^2\right ) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{35 b^2} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}+\frac {8 c \left (b x+c x^2\right )^{3/2}}{35 b^2 x^4}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (15 b^3+3 b^2 c x-4 b c^2 x^2+8 c^3 x^3\right )}{105 b^3 x^4} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^5,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(15*b^3 + 3*b^2*c*x - 4*b*c^2*x^2 + 8*c^3*x^3))/(105*b^3*x^4)

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54

method	result	size
pseudoelliptic	\(-\frac {2 \left (\frac {8}{15} c^{2} x^{2}-\frac {4}{5} b c x +b^{2}\right ) \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}{7 x^{4} b^{3}}\)	\(40\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (8 c^{2} x^{2}-12 b c x +15 b^{2}\right ) \sqrt {c \,x^{2}+b x}}{105 b^{3} x^{4}}\)	\(44\)
trager	\(-\frac {2 \left (8 c^{3} x^{3}-4 b \,c^{2} x^{2}+3 b^{2} c x +15 b^{3}\right ) \sqrt {c \,x^{2}+b x}}{105 b^{3} x^{4}}\)	\(50\)
risch	\(-\frac {2 \left (c x +b \right ) \left (8 c^{3} x^{3}-4 b \,c^{2} x^{2}+3 b^{2} c x +15 b^{3}\right )}{105 x^{3} \sqrt {x \left (c x +b \right )}\, b^{3}}\)	\(53\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\)	\(67\)

[In]

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

[Out]

-2/7*(8/15*c^2*x^2-4/5*b*c*x+b^2)*(c*x+b)*(x*(c*x+b))^(1/2)/x^4/b^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=-\frac {2 \, {\left (8 \, c^{3} x^{3} - 4 \, b c^{2} x^{2} + 3 \, b^{2} c x + 15 \, b^{3}\right )} \sqrt {c x^{2} + b x}}{105 \, b^{3} x^{4}} \]

[In]

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

[Out]

-2/105*(8*c^3*x^3 - 4*b*c^2*x^2 + 3*b^2*c*x + 15*b^3)*sqrt(c*x^2 + b*x)/(b^3*x^4)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=-\frac {16 \, \sqrt {c x^{2} + b x} c^{3}}{105 \, b^{3} x} + \frac {8 \, \sqrt {c x^{2} + b x} c^{2}}{105 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} c}{35 \, b x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x}}{7 \, x^{4}} \]

[In]

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

[Out]

-16/105*sqrt(c*x^2 + b*x)*c^3/(b^3*x) + 8/105*sqrt(c*x^2 + b*x)*c^2/(b^2*x^2) - 2/35*sqrt(c*x^2 + b*x)*c/(b*x^
3) - 2/7*sqrt(c*x^2 + b*x)/x^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (62) = 124\).

Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=\frac {2 \, {\left (140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} c^{2} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b c^{\frac {3}{2}} + 273 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{2} c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{3} \sqrt {c} + 15 \, b^{4}\right )}}{105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]

[In]

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

[Out]

2/105*(140*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c^(3/2) + 273*(sqrt
(c)*x - sqrt(c*x^2 + b*x))^2*b^2*c + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*sqrt(c) + 15*b^4)/(sqrt(c)*x - sq
rt(c*x^2 + b*x))^7

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx=\frac {8\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^2}-\frac {2\,\sqrt {c\,x^2+b\,x}}{7\,x^4}-\frac {16\,c^3\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}}{35\,b\,x^3} \]

[In]

int((b*x + c*x^2)^(1/2)/x^5,x)

[Out]

(8*c^2*(b*x + c*x^2)^(1/2))/(105*b^2*x^2) - (2*(b*x + c*x^2)^(1/2))/(7*x^4) - (16*c^3*(b*x + c*x^2)^(1/2))/(10
5*b^3*x) - (2*c*(b*x + c*x^2)^(1/2))/(35*b*x^3)