\(\int (\frac {c}{(a+b x)^2})^{5/2} \, dx\) [2825]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^2 \sqrt {\frac {c}{(a+b x)^2}}}{4 b (a+b x)^3} \]

[Out]

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

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} \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^2 \sqrt {\frac {c}{(a+b x)^2}}}{4 b (a+b x)^3} \]

[In]

Int[(c/(a + b*x)^2)^(5/2),x]

[Out]

-1/4*(c^2*Sqrt[c/(a + b*x)^2])/(b*(a + b*x)^3)

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 \left (\frac {c}{x^2}\right )^{5/2} \, dx,x,a+b x\right )}{b} \\ & = \frac {\left (c^2 \sqrt {\frac {c}{(a+b x)^2}} (a+b x)\right ) \text {Subst}\left (\int \frac {1}{x^5} \, dx,x,a+b x\right )}{b} \\ & = -\frac {c^2 \sqrt {\frac {c}{(a+b x)^2}}}{4 b (a+b x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {\left (\frac {c}{(a+b x)^2}\right )^{5/2} (a+b x)}{4 b} \]

[In]

Integrate[(c/(a + b*x)^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{2}}\right )^{\frac {5}{2}}}{4 b}\)	\(22\)
default	\(-\frac {\left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{2}}\right )^{\frac {5}{2}}}{4 b}\)	\(22\)
risch	\(-\frac {c^{2} \sqrt {\frac {c}{\left (b x +a \right )^{2}}}}{4 b \left (b x +a \right )^{3}}\)	\(27\)
trager	\(\frac {c^{2} \left (b^{3} x^{3}+4 a \,b^{2} x^{2}+6 a^{2} b x +4 a^{3}\right ) x \sqrt {\frac {c}{b^{2} x^{2}+2 a b x +a^{2}}}}{4 a^{4} \left (b x +a \right )^{3}}\)	\(68\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^{2} \sqrt {\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{4 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 4.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=\begin {cases} - \frac {\left (\frac {c}{\left (a + b x\right )^{2}}\right )^{\frac {5}{2}} \left (\frac {a}{b} + x\right )}{4} & \text {for}\: b \neq 0 \\x \left (\frac {c}{a^{2}}\right )^{\frac {5}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-(c/(a + b*x)**2)**(5/2)*(a/b + x)/4, Ne(b, 0)), (x*(c/a**2)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^{\frac {5}{2}}}{4 \, {\left (b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} b\right )}} \]

[In]

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

[Out]

-1/4*c^(5/2)/(b^5*x^4 + 4*a*b^4*x^3 + 6*a^2*b^3*x^2 + 4*a^3*b^2*x + a^4*b)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right )}{4 \, {\left (b x + a\right )}^{4} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (\frac {c}{(a+b x)^2}\right )^{5/2} \, dx=-\frac {c^2\,\sqrt {\frac {c}{{\left (a+b\,x\right )}^2}}}{4\,b\,{\left (a+b\,x\right )}^3} \]

[In]

int((c/(a + b*x)^2)^(5/2),x)

[Out]

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