\(\int \frac {1}{(a+b x)^{5/2}} \, dx\) [355]

Optimal result

Integrand size = 9, antiderivative size = 16 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3 b (a+b x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {32} \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3 b (a+b x)^{3/2}} \]

[In]

Int[(a + b*x)^(-5/2),x]

[Out]

-2/(3*b*(a + b*x)^(3/2))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{3 b (a+b x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3 b (a+b x)^{3/2}} \]

[In]

Integrate[(a + b*x)^(-5/2),x]

[Out]

-2/(3*b*(a + b*x)^(3/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
gosper	\(-\frac {2}{3 b \left (b x +a \right )^{\frac {3}{2}}}\)	\(13\)
derivativedivides	\(-\frac {2}{3 b \left (b x +a \right )^{\frac {3}{2}}}\)	\(13\)
default	\(-\frac {2}{3 b \left (b x +a \right )^{\frac {3}{2}}}\)	\(13\)
trager	\(-\frac {2}{3 b \left (b x +a \right )^{\frac {3}{2}}}\)	\(13\)
pseudoelliptic	\(-\frac {2}{3 b \left (b x +a \right )^{\frac {3}{2}}}\)	\(13\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (12) = 24\).

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2 \, \sqrt {b x + a}}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=- \frac {2}{3 b \left (a + b x\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-2/(3*b*(a + b*x)**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b} \]

[In]

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

[Out]

-2/3/((b*x + a)^(3/2)*b)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b} \]

[In]

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

[Out]

-2/3/((b*x + a)^(3/2)*b)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+b x)^{5/2}} \, dx=-\frac {2}{3\,b\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int(1/(a + b*x)^(5/2),x)

[Out]

-2/(3*b*(a + b*x)^(3/2))