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

Optimal. Leaf size=61 \[ \frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {640, 609} \begin {gather*} \frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}-\frac {a \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx}{b}\\ &=-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 77, normalized size = 1.26 \begin {gather*} \frac {x^2 \sqrt {(a+b x)^2} \left (21 a^5+70 a^4 b x+105 a^3 b^2 x^2+84 a^2 b^3 x^3+35 a b^4 x^4+6 b^5 x^5\right )}{42 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[(a + b*x)^2]*(21*a^5 + 70*a^4*b*x + 105*a^3*b^2*x^2 + 84*a^2*b^3*x^3 + 35*a*b^4*x^4 + 6*b^5*x^5))/(4
2*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.69, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 57, normalized size = 0.93 \begin {gather*} \frac {1}{7} \, b^{5} x^{7} + \frac {5}{6} \, a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{5} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{2} \, a^{5} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 107, normalized size = 1.75 \begin {gather*} \frac {1}{7} \, b^{5} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, a b^{4} x^{6} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{4} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{7} \mathrm {sgn}\left (b x + a\right )}{42 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^5*x^7*sgn(b*x + a) + 5/6*a*b^4*x^6*sgn(b*x + a) + 2*a^2*b^3*x^5*sgn(b*x + a) + 5/2*a^3*b^2*x^4*sgn(b*x +
a) + 5/3*a^4*b*x^3*sgn(b*x + a) + 1/2*a^5*x^2*sgn(b*x + a) - 1/42*a^7*sgn(b*x + a)/b^2

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maple [A]  time = 0.04, size = 74, normalized size = 1.21 \begin {gather*} \frac {\left (6 b^{5} x^{5}+35 a \,b^{4} x^{4}+84 a^{2} b^{3} x^{3}+105 a^{3} b^{2} x^{2}+70 a^{4} b x +21 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x^{2}}{42 \left (b x +a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*x^2*(6*b^5*x^5+35*a*b^4*x^4+84*a^2*b^3*x^3+105*a^3*b^2*x^2+70*a^4*b*x+21*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5

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maxima [A]  time = 1.34, size = 75, normalized size = 1.23 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2}}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{7 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(x*((a + b*x)**2)**(5/2), x)

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