\(\int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx\) [51]

Optimal result

Integrand size = 17, antiderivative size = 126 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {b x+c x^2}}{9 b x^5}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {32 c^2 \sqrt {b x+c x^2}}{105 b^3 x^3}+\frac {128 c^3 \sqrt {b x+c x^2}}{315 b^4 x^2}-\frac {256 c^4 \sqrt {b x+c x^2}}{315 b^5 x} \]

[Out]

-2/9*(c*x^2+b*x)^(1/2)/b/x^5+16/63*c*(c*x^2+b*x)^(1/2)/b^2/x^4-32/105*c^2*(c*x^2+b*x)^(1/2)/b^3/x^3+128/315*c^
3*(c*x^2+b*x)^(1/2)/b^4/x^2-256/315*c^4*(c*x^2+b*x)^(1/2)/b^5/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {672, 664} \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=-\frac {256 c^4 \sqrt {b x+c x^2}}{315 b^5 x}+\frac {128 c^3 \sqrt {b x+c x^2}}{315 b^4 x^2}-\frac {32 c^2 \sqrt {b x+c x^2}}{105 b^3 x^3}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {2 \sqrt {b x+c x^2}}{9 b x^5} \]

[In]

Int[1/(x^5*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[b*x + c*x^2])/(9*b*x^5) + (16*c*Sqrt[b*x + c*x^2])/(63*b^2*x^4) - (32*c^2*Sqrt[b*x + c*x^2])/(105*b^3
*x^3) + (128*c^3*Sqrt[b*x + c*x^2])/(315*b^4*x^2) - (256*c^4*Sqrt[b*x + c*x^2])/(315*b^5*x)

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 \sqrt {b x+c x^2}}{9 b x^5}-\frac {(8 c) \int \frac {1}{x^4 \sqrt {b x+c x^2}} \, dx}{9 b} \\ & = -\frac {2 \sqrt {b x+c x^2}}{9 b x^5}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}+\frac {\left (16 c^2\right ) \int \frac {1}{x^3 \sqrt {b x+c x^2}} \, dx}{21 b^2} \\ & = -\frac {2 \sqrt {b x+c x^2}}{9 b x^5}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {32 c^2 \sqrt {b x+c x^2}}{105 b^3 x^3}-\frac {\left (64 c^3\right ) \int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx}{105 b^3} \\ & = -\frac {2 \sqrt {b x+c x^2}}{9 b x^5}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {32 c^2 \sqrt {b x+c x^2}}{105 b^3 x^3}+\frac {128 c^3 \sqrt {b x+c x^2}}{315 b^4 x^2}+\frac {\left (128 c^4\right ) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{315 b^4} \\ & = -\frac {2 \sqrt {b x+c x^2}}{9 b x^5}+\frac {16 c \sqrt {b x+c x^2}}{63 b^2 x^4}-\frac {32 c^2 \sqrt {b x+c x^2}}{105 b^3 x^3}+\frac {128 c^3 \sqrt {b x+c x^2}}{315 b^4 x^2}-\frac {256 c^4 \sqrt {b x+c x^2}}{315 b^5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (35 b^4-40 b^3 c x+48 b^2 c^2 x^2-64 b c^3 x^3+128 c^4 x^4\right )}{315 b^5 x^5} \]

[In]

Integrate[1/(x^5*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(35*b^4 - 40*b^3*c*x + 48*b^2*c^2*x^2 - 64*b*c^3*x^3 + 128*c^4*x^4))/(315*b^5*x^5)

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.47

method	result	size
pseudoelliptic	\(-\frac {2 \left (128 c^{4} x^{4}-64 b \,c^{3} x^{3}+48 b^{2} c^{2} x^{2}-40 b^{3} c x +35 b^{4}\right ) \sqrt {x \left (c x +b \right )}}{315 b^{5} x^{5}}\)	\(59\)
trager	\(-\frac {2 \left (128 c^{4} x^{4}-64 b \,c^{3} x^{3}+48 b^{2} c^{2} x^{2}-40 b^{3} c x +35 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{315 b^{5} x^{5}}\)	\(61\)
risch	\(-\frac {2 \left (c x +b \right ) \left (128 c^{4} x^{4}-64 b \,c^{3} x^{3}+48 b^{2} c^{2} x^{2}-40 b^{3} c x +35 b^{4}\right )}{315 b^{5} x^{4} \sqrt {x \left (c x +b \right )}}\)	\(64\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (128 c^{4} x^{4}-64 b \,c^{3} x^{3}+48 b^{2} c^{2} x^{2}-40 b^{3} c x +35 b^{4}\right )}{315 x^{4} b^{5} \sqrt {c \,x^{2}+b x}}\)	\(66\)
default	\(-\frac {2 \sqrt {c \,x^{2}+b x}}{9 b \,x^{5}}-\frac {8 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{7 b \,x^{4}}-\frac {6 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{5 b \,x^{3}}-\frac {4 c \left (-\frac {2 \sqrt {c \,x^{2}+b x}}{3 b \,x^{2}}+\frac {4 c \sqrt {c \,x^{2}+b x}}{3 b^{2} x}\right )}{5 b}\right )}{7 b}\right )}{9 b}\)	\(119\)

[In]

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

[Out]

-2/315*(128*c^4*x^4-64*b*c^3*x^3+48*b^2*c^2*x^2-40*b^3*c*x+35*b^4)/b^5/x^5*(x*(c*x+b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=-\frac {2 \, {\left (128 \, c^{4} x^{4} - 64 \, b c^{3} x^{3} + 48 \, b^{2} c^{2} x^{2} - 40 \, b^{3} c x + 35 \, b^{4}\right )} \sqrt {c x^{2} + b x}}{315 \, b^{5} x^{5}} \]

[In]

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

[Out]

-2/315*(128*c^4*x^4 - 64*b*c^3*x^3 + 48*b^2*c^2*x^2 - 40*b^3*c*x + 35*b^4)*sqrt(c*x^2 + b*x)/(b^5*x^5)

Sympy [F]

\[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=\int \frac {1}{x^{5} \sqrt {x \left (b + c x\right )}}\, dx \]

[In]

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

[Out]

Integral(1/(x**5*sqrt(x*(b + c*x))), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=-\frac {256 \, \sqrt {c x^{2} + b x} c^{4}}{315 \, b^{5} x} + \frac {128 \, \sqrt {c x^{2} + b x} c^{3}}{315 \, b^{4} x^{2}} - \frac {32 \, \sqrt {c x^{2} + b x} c^{2}}{105 \, b^{3} x^{3}} + \frac {16 \, \sqrt {c x^{2} + b x} c}{63 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x}}{9 \, b x^{5}} \]

[In]

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

[Out]

-256/315*sqrt(c*x^2 + b*x)*c^4/(b^5*x) + 128/315*sqrt(c*x^2 + b*x)*c^3/(b^4*x^2) - 32/105*sqrt(c*x^2 + b*x)*c^
2/(b^3*x^3) + 16/63*sqrt(c*x^2 + b*x)*c/(b^2*x^4) - 2/9*sqrt(c*x^2 + b*x)/(b*x^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=\frac {2 \, {\left (1008 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} c^{2} + 1680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b c^{\frac {3}{2}} + 1080 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{2} c + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{3} \sqrt {c} + 35 \, b^{4}\right )}}{315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{9}} \]

[In]

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

[Out]

2/315*(1008*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*c^2 + 1680*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b*c^(3/2) + 1080*(s
qrt(c)*x - sqrt(c*x^2 + b*x))^2*b^2*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^3*sqrt(c) + 35*b^4)/(sqrt(c)*x -
sqrt(c*x^2 + b*x))^9

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \sqrt {b x+c x^2}} \, dx=\frac {128\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^4\,x^2}-\frac {32\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^3\,x^3}-\frac {2\,\sqrt {c\,x^2+b\,x}}{9\,b\,x^5}-\frac {256\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^5\,x}+\frac {16\,c\,\sqrt {c\,x^2+b\,x}}{63\,b^2\,x^4} \]

[In]

int(1/(x^5*(b*x + c*x^2)^(1/2)),x)

[Out]

(128*c^3*(b*x + c*x^2)^(1/2))/(315*b^4*x^2) - (32*c^2*(b*x + c*x^2)^(1/2))/(105*b^3*x^3) - (2*(b*x + c*x^2)^(1
/2))/(9*b*x^5) - (256*c^4*(b*x + c*x^2)^(1/2))/(315*b^5*x) + (16*c*(b*x + c*x^2)^(1/2))/(63*b^2*x^4)