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

Optimal. Leaf size=96 \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^3}-\frac {2 a \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^3}+\frac {a^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {645} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3}-\frac {2 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{5 b^3}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^3) - (2*a*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b
^3) + ((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3)

Rule 645

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

Rubi steps

\begin {align*} \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^2 \left (a b+b^2 x\right )^3}{b^2}-\frac {2 a \left (a b+b^2 x\right )^4}{b^3}+\frac {\left (a b+b^2 x\right )^5}{b^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}-\frac {2 a (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 55, normalized size = 0.57 \begin {gather*} \frac {x^3 \sqrt {(a+b x)^2} \left (20 a^3+45 a^2 b x+36 a b^2 x^2+10 b^3 x^3\right )}{60 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(20*a^3 + 45*a^2*b*x + 36*a*b^2*x^2 + 10*b^3*x^3))/(60*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.46, size = 0, normalized size = 0.00 \begin {gather*} \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 35, normalized size = 0.36 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 73, normalized size = 0.76 \begin {gather*} \frac {1}{6} \, b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a^{2} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{6} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{3}} \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)^(3/2),x, algorithm="giac")

[Out]

1/6*b^3*x^6*sgn(b*x + a) + 3/5*a*b^2*x^5*sgn(b*x + a) + 3/4*a^2*b*x^4*sgn(b*x + a) + 1/3*a^3*x^3*sgn(b*x + a)
+ 1/60*a^6*sgn(b*x + a)/b^3

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maple [A]  time = 0.05, size = 52, normalized size = 0.54 \begin {gather*} \frac {\left (10 b^{3} x^{3}+36 a \,b^{2} x^{2}+45 a^{2} b x +20 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{3}}{60 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*x^3*(10*b^3*x^3+36*a*b^2*x^2+45*a^2*b*x+20*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 1.37, size = 102, normalized size = 1.06 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x}{6 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a}{30 \, b^{3}} \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)^(3/2),x, algorithm="maxima")

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*x/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3/b^3 + 1/6*(b^2*x^2 + 2
*a*b*x + a^2)^(5/2)*x/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \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)^(3/2),x)

[Out]

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

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

[Out]

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

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