\(\int \frac {1}{x^2 (b x+c x^2)^{3/2}} \, dx\) [58]

Optimal result

Integrand size = 17, antiderivative size = 77 \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}+\frac {4 c}{5 b^2 x \sqrt {b x+c x^2}}-\frac {16 c^2 (b+2 c x)}{5 b^4 \sqrt {b x+c x^2}} \]

[Out]

-2/5/b/x^2/(c*x^2+b*x)^(1/2)+4/5*c/b^2/x/(c*x^2+b*x)^(1/2)-16/5*c^2*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, 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, 627} \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {16 c^2 (b+2 c x)}{5 b^4 \sqrt {b x+c x^2}}+\frac {4 c}{5 b^2 x \sqrt {b x+c x^2}}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}} \]

[In]

Int[1/(x^2*(b*x + c*x^2)^(3/2)),x]

[Out]

-2/(5*b*x^2*Sqrt[b*x + c*x^2]) + (4*c)/(5*b^2*x*Sqrt[b*x + c*x^2]) - (16*c^2*(b + 2*c*x))/(5*b^4*Sqrt[b*x + c*
x^2])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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}{5 b x^2 \sqrt {b x+c x^2}}-\frac {(6 c) \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{5 b} \\ & = -\frac {2}{5 b x^2 \sqrt {b x+c x^2}}+\frac {4 c}{5 b^2 x \sqrt {b x+c x^2}}+\frac {\left (8 c^2\right ) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 b^2} \\ & = -\frac {2}{5 b x^2 \sqrt {b x+c x^2}}+\frac {4 c}{5 b^2 x \sqrt {b x+c x^2}}-\frac {16 c^2 (b+2 c x)}{5 b^4 \sqrt {b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )}{5 b^4 x^2 \sqrt {x (b+c x)}} \]

[In]

Integrate[1/(x^2*(b*x + c*x^2)^(3/2)),x]

[Out]

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

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(-\frac {2 \left (16 c^{3} x^{3}+8 b \,c^{2} x^{2}-2 b^{2} c x +b^{3}\right )}{5 x^{2} \sqrt {x \left (c x +b \right )}\, b^{4}}\)	\(46\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (16 c^{3} x^{3}+8 b \,c^{2} x^{2}-2 b^{2} c x +b^{3}\right )}{5 x \,b^{4} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(53\)
trager	\(-\frac {2 \left (16 c^{3} x^{3}+8 b \,c^{2} x^{2}-2 b^{2} c x +b^{3}\right ) \sqrt {c \,x^{2}+b x}}{5 \left (c x +b \right ) b^{4} x^{3}}\)	\(55\)
risch	\(-\frac {2 \left (c x +b \right ) \left (11 c^{2} x^{2}-3 b c x +b^{2}\right )}{5 b^{4} x^{2} \sqrt {x \left (c x +b \right )}}-\frac {2 c^{3} x}{\sqrt {x \left (c x +b \right )}\, b^{4}}\)	\(59\)
default	\(-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\)	\(70\)

[In]

int(1/x^2/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (16 \, c^{3} x^{3} + 8 \, b c^{2} x^{2} - 2 \, b^{2} c x + b^{3}\right )} \sqrt {c x^{2} + b x}}{5 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}} \]

[In]

integrate(1/x^2/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/x**2/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(x**2*(x*(b + c*x))**(3/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {32 \, c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {16 \, c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {4 \, c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {2}{5 \, \sqrt {c x^{2} + b x} b x^{2}} \]

[In]

integrate(1/x^2/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*x^2), x)

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (b^3-2\,b^2\,c\,x+8\,b\,c^2\,x^2+16\,c^3\,x^3\right )}{5\,b^4\,x^3\,\left (b+c\,x\right )} \]

[In]

int(1/(x^2*(b*x + c*x^2)^(3/2)),x)

[Out]

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