3.3.2 \(\int \frac {x^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [202]

Optimal. Leaf size=107 \[ -\frac {a^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \begin {gather*} -\frac {a^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*a^2/(b^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*a)/(3*b^3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]) - 1/(2*b^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 660

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^2}{\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^2}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {a^2}{b^7 (a+b x)^5}-\frac {2 a}{b^7 (a+b x)^4}+\frac {1}{b^7 (a+b x)^3}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {a^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 44, normalized size = 0.41 \begin {gather*} \frac {-a^2-4 a b x-6 b^2 x^2}{12 b^3 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-a^2 - 4*a*b*x - 6*b^2*x^2)/(12*b^3*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.51, size = 37, normalized size = 0.35

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (6 b^{2} x^{2}+4 a b x +a^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(37\)
default	\(-\frac {\left (b x +a \right ) \left (6 b^{2} x^{2}+4 a b x +a^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(37\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {x^{2}}{2 b}-\frac {a x}{3 b^{2}}-\frac {a^{2}}{12 b^{3}}\right )}{\left (b x +a \right )^{5}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(b*x+a)*(6*b^2*x^2+4*a*b*x+a^2)/b^3/((b*x+a)^2)^(5/2)

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Maxima [A]
time = 0.27, size = 47, normalized size = 0.44 \begin {gather*} -\frac {1}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.51, size = 65, normalized size = 0.61 \begin {gather*} -\frac {6 \, b^{2} x^{2} + 4 \, a b x + a^{2}}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*b^2*x^2 + 4*a*b*x + a^2)/(b^7*x^4 + 4*a*b^6*x^3 + 6*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\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**2/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

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

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Giac [A]
time = 0.57, size = 37, normalized size = 0.35 \begin {gather*} -\frac {6 \, b^{2} x^{2} + 4 \, a b x + a^{2}}{12 \, {\left (b x + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*b^2*x^2 + 4*a*b*x + a^2)/((b*x + a)^4*b^3*sgn(b*x + a))

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Mupad [B]
time = 0.24, size = 47, normalized size = 0.44 \begin {gather*} -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^2+4\,a\,b\,x+6\,b^2\,x^2\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^2 + 6*b^2*x^2 + 4*a*b*x))/(12*b^3*(a + b*x)^5)

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