\(\int \frac {1}{x (b x+c x^2)^{3/2}} \, dx\) [57]

Optimal result

Integrand size = 17, antiderivative size = 51 \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{3 b x \sqrt {b x+c x^2}}+\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, 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, 627} \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2}{3 b x \sqrt {b x+c x^2}} \]

[In]

Int[1/(x*(b*x + c*x^2)^(3/2)),x]

[Out]

-2/(3*b*x*Sqrt[b*x + c*x^2]) + (8*c*(b + 2*c*x))/(3*b^3*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}{3 b x \sqrt {b x+c x^2}}-\frac {(4 c) \int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b} \\ & = -\frac {2}{3 b x \sqrt {b x+c x^2}}+\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (b+c x) \left (b^2-4 b c x-8 c^2 x^2\right )}{3 b^3 (x (b+c x))^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {2 \left (-8 c^{2} x^{2}-4 b c x +b^{2}\right )}{3 x \sqrt {x \left (c x +b \right )}\, b^{3}}\)	\(35\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-8 c^{2} x^{2}-4 b c x +b^{2}\right )}{3 b^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(39\)
default	\(-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\)	\(44\)
trager	\(-\frac {2 \left (-8 c^{2} x^{2}-4 b c x +b^{2}\right ) \sqrt {c \,x^{2}+b x}}{3 \left (c x +b \right ) b^{3} x^{2}}\)	\(44\)
risch	\(-\frac {2 \left (c x +b \right ) \left (-5 c x +b \right )}{3 b^{3} x \sqrt {x \left (c x +b \right )}}+\frac {2 c^{2} x}{\sqrt {x \left (c x +b \right )}\, b^{3}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (8 \, c^{2} x^{2} + 4 \, b c x - b^{2}\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {16 \, c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {8 \, c}{3 \, \sqrt {c x^{2} + b x} b^{2}} - \frac {2}{3 \, \sqrt {c x^{2} + b x} b x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (-b^2+4\,b\,c\,x+8\,c^2\,x^2\right )}{3\,b^3\,x^2\,\left (b+c\,x\right )} \]

[In]

int(1/(x*(b*x + c*x^2)^(3/2)),x)

[Out]

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