3.2830 \(\int \frac{1}{(\frac{c}{(a+b x)^2})^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac{(a+b x)^5}{6 b c^2 \sqrt{\frac{c}{(a+b x)^2}}} \]

[Out]

(a + b*x)^5/(6*b*c^2*Sqrt[c/(a + b*x)^2])

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Rubi [A]  time = 0.0090109, antiderivative size = 30, normalized size of antiderivative = 1., 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} \[ \frac{(a+b x)^5}{6 b c^2 \sqrt{\frac{c}{(a+b x)^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^2)^(-5/2),x]

[Out]

(a + b*x)^5/(6*b*c^2*Sqrt[c/(a + b*x)^2])

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]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (\frac{c}{(a+b x)^2}\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (\frac{c}{x^2}\right )^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x^5 \, dx,x,a+b x\right )}{b c^2 \sqrt{\frac{c}{(a+b x)^2}} (a+b x)}\\ &=\frac{(a+b x)^5}{6 b c^2 \sqrt{\frac{c}{(a+b x)^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0167713, size = 25, normalized size = 0.83 \[ \frac{a+b x}{6 b \left (\frac{c}{(a+b x)^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^2)^(-5/2),x]

[Out]

(a + b*x)/(6*b*(c/(a + b*x)^2)^(5/2))

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Maple [B]  time = 0.003, size = 73, normalized size = 2.4 \begin{align*}{\frac{x \left ({b}^{5}{x}^{5}+6\,a{b}^{4}{x}^{4}+15\,{a}^{2}{b}^{3}{x}^{3}+20\,{a}^{3}{b}^{2}{x}^{2}+15\,{a}^{4}bx+6\,{a}^{5} \right ) }{6\, \left ( bx+a \right ) ^{5}} \left ({\frac{c}{ \left ( bx+a \right ) ^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(b^5*x^5+6*a*b^4*x^4+15*a^2*b^3*x^3+20*a^3*b^2*x^2+15*a^4*b*x+6*a^5)/(b*x+a)^5/(c/(b*x+a)^2)^(5/2)

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Maxima [B]  time = 1.28996, size = 80, normalized size = 2.67 \begin{align*} \frac{b^{5} x^{6} + 6 \, a b^{4} x^{5} + 15 \, a^{2} b^{3} x^{4} + 20 \, a^{3} b^{2} x^{3} + 15 \, a^{4} b x^{2} + 6 \, a^{5} x}{6 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b^5*x^6 + 6*a*b^4*x^5 + 15*a^2*b^3*x^4 + 20*a^3*b^2*x^3 + 15*a^4*b*x^2 + 6*a^5*x)/c^(5/2)

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Fricas [B]  time = 1.20083, size = 193, normalized size = 6.43 \begin{align*} \frac{{\left (b^{6} x^{7} + 7 \, a b^{5} x^{6} + 21 \, a^{2} b^{4} x^{5} + 35 \, a^{3} b^{3} x^{4} + 35 \, a^{4} b^{2} x^{3} + 21 \, a^{5} b x^{2} + 6 \, a^{6} x\right )} \sqrt{\frac{c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b^6*x^7 + 7*a*b^5*x^6 + 21*a^2*b^4*x^5 + 35*a^3*b^3*x^4 + 35*a^4*b^2*x^3 + 21*a^5*b*x^2 + 6*a^6*x)*sqrt(c
/(b^2*x^2 + 2*a*b*x + a^2))/c^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\frac{c}{\left (a + b x\right )^{2}}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\frac{c}{{\left (b x + a\right )}^{2}}\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c/(b*x + a)^2)^(-5/2), x)