3.2836 \(\int \frac{1}{(\frac{c}{(a+b x)^3})^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac{2 (a+b x)^7}{17 b c^2 \sqrt{\frac{c}{(a+b x)^3}}} \]

[Out]

(2*(a + b*x)^7)/(17*b*c^2*Sqrt[c/(a + b*x)^3])

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Rubi [A]  time = 0.0095543, 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} \[ \frac{2 (a+b x)^7}{17 b c^2 \sqrt{\frac{c}{(a+b x)^3}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^3)^(-5/2),x]

[Out]

(2*(a + b*x)^7)/(17*b*c^2*Sqrt[c/(a + b*x)^3])

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)^3}\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (\frac{c}{x^3}\right )^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x^{15/2} \, dx,x,a+b x\right )}{b c^2 \sqrt{\frac{c}{(a+b x)^3}} (a+b x)^{3/2}}\\ &=\frac{2 (a+b x)^7}{17 b c^2 \sqrt{\frac{c}{(a+b x)^3}}}\\ \end{align*}

Mathematica [A]  time = 0.0144372, size = 25, normalized size = 0.83 \[ \frac{2 (a+b x)}{17 b \left (\frac{c}{(a+b x)^3}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^3)^(-5/2),x]

[Out]

(2*(a + b*x))/(17*b*(c/(a + b*x)^3)^(5/2))

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Maple [A]  time = 0.001, size = 22, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{17\,b} \left ({\frac{c}{ \left ( bx+a \right ) ^{3}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/(b*x+a)^3)^(5/2),x)

[Out]

2/17*(b*x+a)/b/(c/(b*x+a)^3)^(5/2)

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Maxima [A]  time = 1.26531, size = 36, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (b \sqrt{c} x + a \sqrt{c}\right )}{\left (b x + a\right )}^{\frac{15}{2}}}{17 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a)^3)^(5/2),x, algorithm="maxima")

[Out]

2/17*(b*sqrt(c)*x + a*sqrt(c))*(b*x + a)^(15/2)/(b*c^3)

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Fricas [B]  time = 1.21817, size = 317, normalized size = 10.57 \begin{align*} \frac{2 \,{\left (b^{10} x^{10} + 10 \, a b^{9} x^{9} + 45 \, a^{2} b^{8} x^{8} + 120 \, a^{3} b^{7} x^{7} + 210 \, a^{4} b^{6} x^{6} + 252 \, a^{5} b^{5} x^{5} + 210 \, a^{6} b^{4} x^{4} + 120 \, a^{7} b^{3} x^{3} + 45 \, a^{8} b^{2} x^{2} + 10 \, a^{9} b x + a^{10}\right )} \sqrt{\frac{c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{17 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a)^3)^(5/2),x, algorithm="fricas")

[Out]

2/17*(b^10*x^10 + 10*a*b^9*x^9 + 45*a^2*b^8*x^8 + 120*a^3*b^7*x^7 + 210*a^4*b^6*x^6 + 252*a^5*b^5*x^5 + 210*a^
6*b^4*x^4 + 120*a^7*b^3*x^3 + 45*a^8*b^2*x^2 + 10*a^9*b*x + a^10)*sqrt(c/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x +
a^3))/(b*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 )^{3}}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c/(b*x+a)**3)**(5/2),x)

[Out]

Integral((c/(a + b*x)**3)**(-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 )}^{3}}\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)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((c/(b*x + a)^3)^(-5/2), x)