3.2807 \(\int \frac {1}{(c (a+b x)^2)^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ -\frac {1}{4 b c^2 (a+b x)^3 \sqrt {c (a+b x)^2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ -\frac {1}{4 b c^2 (a+b x)^3 \sqrt {c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x)^2)^(-5/2),x]

[Out]

-1/(4*b*c^2*(a + b*x)^3*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 247

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.83 \[ -\frac {a+b x}{4 b \left (c (a+b x)^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(a + b*x)^2)^(-5/2),x]

[Out]

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

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fricas [B]  time = 0.82, size = 97, normalized size = 3.23 \[ -\frac {\sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \, {\left (b^{6} c^{3} x^{5} + 5 \, a b^{5} c^{3} x^{4} + 10 \, a^{2} b^{4} c^{3} x^{3} + 10 \, a^{3} b^{3} c^{3} x^{2} + 5 \, a^{4} b^{2} c^{3} x + a^{5} b c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(b^2*c*x^2 + 2*a*b*c*x + a^2*c)/(b^6*c^3*x^5 + 5*a*b^5*c^3*x^4 + 10*a^2*b^4*c^3*x^3 + 10*a^3*b^3*c^3*
x^2 + 5*a^4*b^2*c^3*x + a^5*b*c^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.00, size = 22, normalized size = 0.73 \[ -\frac {b x +a}{4 \left (\left (b x +a \right )^{2} c \right )^{\frac {5}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x+a)^2*c)^(5/2),x)

[Out]

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

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maxima [A]  time = 0.56, size = 17, normalized size = 0.57 \[ -\frac {1}{4 \, b^{5} c^{\frac {5}{2}} {\left (x + \frac {a}{b}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(b^5*c^(5/2)*(x + a/b)^4)

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mupad [B]  time = 1.28, size = 26, normalized size = 0.87 \[ -\frac {\sqrt {c\,{\left (a+b\,x\right )}^2}}{4\,b\,c^3\,{\left (a+b\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*(a + b*x)^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(a + b*x)**2)**(-5/2), x)

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