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

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

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {x^4}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 55, normalized size = 1.49 \begin {gather*} \frac {-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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.68, size = 266, normalized size = 7.19 \begin {gather*} \frac {2 \sqrt {b^2} \left (-a^7+a^3 b^4 x^4+4 a^2 b^5 x^5+6 a b^6 x^6+4 b^7 x^7\right )+2 \sqrt {a^2+2 a b x+b^2 x^2} \left (-a^6 b+a^5 b^2 x-a^4 b^3 x^2+a^3 b^4 x^3-2 a^2 b^5 x^4-2 a b^6 x^5-4 b^7 x^6\right )}{\sqrt {b^2} b^5 x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^3-24 a^2 b^4 x-24 a b^5 x^2-8 b^6 x^3\right )+b^5 x^4 \left (8 a^4 b^4+32 a^3 b^5 x+48 a^2 b^6 x^2+32 a b^7 x^3+8 b^8 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-(a^6*b) + a^5*b^2*x - a^4*b^3*x^2 + a^3*b^4*x^3 - 2*a^2*b^5*x^4 - 2*a*b^6*x
^5 - 4*b^7*x^6) + 2*Sqrt[b^2]*(-a^7 + a^3*b^4*x^4 + 4*a^2*b^5*x^5 + 6*a*b^6*x^6 + 4*b^7*x^7))/(b^5*Sqrt[b^2]*x
^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^3 - 24*a^2*b^4*x - 24*a*b^5*x^2 - 8*b^6*x^3) + b^5*x^4*(8*a^4*b^4 +
32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*b^7*x^3 + 8*b^8*x^4))

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fricas [B]  time = 0.42, size = 76, normalized size = 2.05 \begin {gather*} -\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 {gather*}

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

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

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maple [A]  time = 0.04, size = 48, normalized size = 1.30 \begin {gather*} -\frac {\left (b x +a \right ) \left (4 b^{3} x^{3}+6 a \,b^{2} x^{2}+4 a^{2} b x +a^{3}\right )}{4 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \end {gather*}

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 = 1.36, size = 102, normalized size = 2.76 \begin {gather*} -\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}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a^{2}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {a^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

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

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mupad [B]  time = 0.24, size = 128, normalized size = 3.46 \begin {gather*} \frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^4\,{\left (a+b\,x\right )}^5}-\frac {a^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^4}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2}+\frac {3\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^4\,{\left (a+b\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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