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

Optimal. Leaf size=37 \[ \frac{x^4}{4 a (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0161654, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 37} \[ \frac{x^4}{4 a (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0147132, size = 55, normalized size = 1.49 \[ \frac{-4 a^2 b x-a^3-6 a b^2 x^2-4 b^3 x^3}{4 b^4 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-a^3 - 4*a^2*b*x - 6*a*b^2*x^2 - 4*b^3*x^3)/(4*b^4*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.175, size = 48, normalized size = 1.3 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 4\,{b}^{3}{x}^{3}+6\,a{b}^{2}{x}^{2}+4\,b{a}^{2}x+{a}^{3} \right ) }{4\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.12336, size = 181, normalized size = 4.89 \begin{align*} -\frac{x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{2 \, a^{2}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{4}} - \frac{a^{3} b}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, a^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{a}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} + \frac{a^{3}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) - 1/4*a^3*b/((b^2)^
(9/2)*(x + a/b)^4) + 2/3*a^2/((b^2)^(7/2)*(x + a/b)^3) - 1/2*a/((b^2)^(5/2)*b*(x + a/b)^2) + 1/2*a^3/((b^2)^(5
/2)*b^3*(x + a/b)^4)

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Fricas [B]  time = 1.64007, size = 154, normalized size = 4.16 \begin{align*} -\frac{4 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3}}{4 \,{\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*b^3*x^3 + 6*a*b^2*x^2 + 4*a^2*b*x + a^3)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^
4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (\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(x**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**3/((a + b*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x