\(\int \frac {1}{(c (a+b x)^2)^{3/2}} \, dx\) [2806]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {1}{2 b c (a+b x) \sqrt {c (a+b x)^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 15, 30} \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {1}{2 b c (a+b x) \sqrt {c (a+b x)^2}} \]

[In]

Int[(c*(a + b*x)^2)^(-3/2),x]

[Out]

-1/2*1/(b*c*(a + b*x)*Sqrt[c*(a + b*x)^2])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (c x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {(a+b x) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,a+b x\right )}{b c \sqrt {c (a+b x)^2}} \\ & = -\frac {1}{2 b c (a+b x) \sqrt {c (a+b x)^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {a+b x}{2 b \left (c (a+b x)^2\right )^{3/2}} \]

[In]

Integrate[(c*(a + b*x)^2)^(-3/2),x]

[Out]

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

Maple [A] (verified)

Time = 3.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {b x +a}{2 b \left (c \left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(22\)
default	\(-\frac {b x +a}{2 b \left (c \left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(22\)
risch	\(-\frac {1}{2 b c \left (b x +a \right ) \sqrt {c \left (b x +a \right )^{2}}}\)	\(27\)
trager	\(\frac {\left (b x +2 a \right ) x \sqrt {b^{2} c \,x^{2}+2 a b c x +a^{2} c}}{2 a^{2} c^{2} \left (b x +a \right )^{3}}\)	\(46\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (26) = 52\).

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {\sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{2 \, {\left (b^{4} c^{2} x^{3} + 3 \, a b^{3} c^{2} x^{2} + 3 \, a^{2} b^{2} c^{2} x + a^{3} b c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=\begin {cases} - \frac {\frac {a}{b} + x}{2 \left (c \left (a + b x\right )^{2}\right )^{\frac {3}{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (a^{2} c\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-(a/b + x)/(2*(c*(a + b*x)**2)**(3/2)), Ne(b, 0)), (x/(a**2*c)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {1}{2 \, b^{3} c^{\frac {3}{2}} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

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

[Out]

-1/2/(b^3*c^(3/2)*(x + a/b)^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {1}{2 \, {\left (b x + a\right )}^{2} b c^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (c (a+b x)^2\right )^{3/2}} \, dx=-\frac {\sqrt {c\,{\left (a+b\,x\right )}^2}}{2\,b\,c^2\,{\left (a+b\,x\right )}^3} \]

[In]

int(1/(c*(a + b*x)^2)^(3/2),x)

[Out]

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