\(\int \frac {\sqrt {b x+c x^2}}{x^6} \, dx\) [10]

Optimal result

Integrand size = 17, antiderivative size = 100 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}+\frac {32 c^3 \left (b x+c x^2\right )^{3/2}}{315 b^4 x^3} \]

[Out]

-2/9*(c*x^2+b*x)^(3/2)/b/x^6+4/21*c*(c*x^2+b*x)^(3/2)/b^2/x^5-16/105*c^2*(c*x^2+b*x)^(3/2)/b^3/x^4+32/315*c^3*
(c*x^2+b*x)^(3/2)/b^4/x^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\frac {32 c^3 \left (b x+c x^2\right )^{3/2}}{315 b^4 x^3}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6} \]

[In]

Int[Sqrt[b*x + c*x^2]/x^6,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(9*b*x^6) + (4*c*(b*x + c*x^2)^(3/2))/(21*b^2*x^5) - (16*c^2*(b*x + c*x^2)^(3/2))/(10
5*b^3*x^4) + (32*c^3*(b*x + c*x^2)^(3/2))/(315*b^4*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}}{9 b x^6}-\frac {(2 c) \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx}{3 b} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}+\frac {\left (8 c^2\right ) \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx}{21 b^2} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}-\frac {\left (16 c^3\right ) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{105 b^3} \\ & = -\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}+\frac {4 c \left (b x+c x^2\right )^{3/2}}{21 b^2 x^5}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2}}{105 b^3 x^4}+\frac {32 c^3 \left (b x+c x^2\right )^{3/2}}{315 b^4 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (35 b^4+5 b^3 c x-6 b^2 c^2 x^2+8 b c^3 x^3-16 c^4 x^4\right )}{315 b^4 x^5} \]

[In]

Integrate[Sqrt[b*x + c*x^2]/x^6,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(35*b^4 + 5*b^3*c*x - 6*b^2*c^2*x^2 + 8*b*c^3*x^3 - 16*c^4*x^4))/(315*b^4*x^5)

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-16 c^{3} x^{3}+24 b \,c^{2} x^{2}-30 b^{2} c x +35 b^{3}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{4} x^{5}}\)	\(55\)
pseudoelliptic	\(\frac {2 \left (16 c^{4} x^{4}-8 b \,c^{3} x^{3}+6 b^{2} c^{2} x^{2}-5 b^{3} c x -35 b^{4}\right ) \sqrt {x \left (c x +b \right )}}{315 x^{5} b^{4}}\)	\(59\)
trager	\(-\frac {2 \left (-16 c^{4} x^{4}+8 b \,c^{3} x^{3}-6 b^{2} c^{2} x^{2}+5 b^{3} c x +35 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{4} x^{5}}\)	\(61\)
risch	\(-\frac {2 \left (c x +b \right ) \left (-16 c^{4} x^{4}+8 b \,c^{3} x^{3}-6 b^{2} c^{2} x^{2}+5 b^{3} c x +35 b^{4}\right )}{315 x^{4} \sqrt {x \left (c x +b \right )}\, b^{4}}\)	\(64\)
default	\(-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{9 b \,x^{6}}-\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )}{3 b}\)	\(93\)

[In]

int((c*x^2+b*x)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

-2/315*(c*x+b)*(-16*c^3*x^3+24*b*c^2*x^2-30*b^2*c*x+35*b^3)*(c*x^2+b*x)^(1/2)/b^4/x^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\frac {2 \, {\left (16 \, c^{4} x^{4} - 8 \, b c^{3} x^{3} + 6 \, b^{2} c^{2} x^{2} - 5 \, b^{3} c x - 35 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{315 \, b^{4} x^{5}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^6,x, algorithm="fricas")

[Out]

2/315*(16*c^4*x^4 - 8*b*c^3*x^3 + 6*b^2*c^2*x^2 - 5*b^3*c*x - 35*b^4)*sqrt(c*x^2 + b*x)/(b^4*x^5)

Sympy [F]

\[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{x^{6}}\, dx \]

[In]

integrate((c*x**2+b*x)**(1/2)/x**6,x)

[Out]

Integral(sqrt(x*(b + c*x))/x**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\frac {32 \, \sqrt {c x^{2} + b x} c^{4}}{315 \, b^{4} x} - \frac {16 \, \sqrt {c x^{2} + b x} c^{3}}{315 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{2} + b x} c^{2}}{105 \, b^{2} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} c}{63 \, b x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x}}{9 \, x^{5}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^6,x, algorithm="maxima")

[Out]

32/315*sqrt(c*x^2 + b*x)*c^4/(b^4*x) - 16/315*sqrt(c*x^2 + b*x)*c^3/(b^3*x^2) + 4/105*sqrt(c*x^2 + b*x)*c^2/(b
^2*x^3) - 2/63*sqrt(c*x^2 + b*x)*c/(b*x^4) - 2/9*sqrt(c*x^2 + b*x)/x^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\frac {2 \, {\left (630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} c^{\frac {5}{2}} + 1764 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b c^{2} + 1995 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{2} c^{\frac {3}{2}} + 1125 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{3} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{4} \sqrt {c} + 35 \, b^{5}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

[In]

integrate((c*x^2+b*x)^(1/2)/x^6,x, algorithm="giac")

[Out]

2/315*(630*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*c^(5/2) + 1764*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b*c^2 + 1995*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^3*b^2*c^(3/2) + 1125*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b^3*c + 315*(sqrt(c)*x - s
qrt(c*x^2 + b*x))*b^4*sqrt(c) + 35*b^5)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9

Mupad [B] (verification not implemented)

Time = 9.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx=\frac {4\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^3}-\frac {2\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {16\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x^2}+\frac {32\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^4\,x}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}}{63\,b\,x^4} \]

[In]

int((b*x + c*x^2)^(1/2)/x^6,x)

[Out]

(4*c^2*(b*x + c*x^2)^(1/2))/(105*b^2*x^3) - (2*(b*x + c*x^2)^(1/2))/(9*x^5) - (16*c^3*(b*x + c*x^2)^(1/2))/(31
5*b^3*x^2) + (32*c^4*(b*x + c*x^2)^(1/2))/(315*b^4*x) - (2*c*(b*x + c*x^2)^(1/2))/(63*b*x^4)