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

Optimal. Leaf size=107 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3}-\frac{2 a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^3}+\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0409676, antiderivative size = 107, normalized size of antiderivative = 1., 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 = {646, 43} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3}-\frac{2 a \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^3}+\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 43

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])

Rubi steps

\begin{align*} \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int x^2 \left (a b+b^2 x\right )^5 \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{a^2 \left (a b+b^2 x\right )^5}{b^2}-\frac{2 a \left (a b+b^2 x\right )^6}{b^3}+\frac{\left (a b+b^2 x\right )^7}{b^4}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{a^2 (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{6 b^3}-\frac{2 a (a+b x)^6 \sqrt{a^2+2 a b x+b^2 x^2}}{7 b^3}+\frac{(a+b x)^7 \sqrt{a^2+2 a b x+b^2 x^2}}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0195258, size = 77, normalized size = 0.72 \[ \frac{x^3 \sqrt{(a+b x)^2} \left (336 a^3 b^2 x^2+280 a^2 b^3 x^3+210 a^4 b x+56 a^5+120 a b^4 x^4+21 b^5 x^5\right )}{168 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(56*a^5 + 210*a^4*b*x + 336*a^3*b^2*x^2 + 280*a^2*b^3*x^3 + 120*a*b^4*x^4 + 21*b^5*x^5)
)/(168*(a + b*x))

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Maple [A]  time = 0.176, size = 74, normalized size = 0.7 \begin{align*}{\frac{{x}^{3} \left ( 21\,{b}^{5}{x}^{5}+120\,a{b}^{4}{x}^{4}+280\,{a}^{2}{b}^{3}{x}^{3}+336\,{a}^{3}{b}^{2}{x}^{2}+210\,{a}^{4}bx+56\,{a}^{5} \right ) }{168\, \left ( bx+a \right ) ^{5}} \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^2*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/168*x^3*(21*b^5*x^5+120*a*b^4*x^4+280*a^2*b^3*x^3+336*a^3*b^2*x^2+210*a^4*b*x+56*a^5)*((b*x+a)^2)^(5/2)/(b*x
+a)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.58719, size = 126, normalized size = 1.18 \begin{align*} \frac{1}{8} \, b^{5} x^{8} + \frac{5}{7} \, a b^{4} x^{7} + \frac{5}{3} \, a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{5} + \frac{5}{4} \, a^{4} b x^{4} + \frac{1}{3} \, a^{5} x^{3} \end{align*}

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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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**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 = 1.3392, size = 144, normalized size = 1.35 \begin{align*} \frac{1}{8} \, b^{5} x^{8} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{7} \, a b^{4} x^{7} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a^{2} b^{3} x^{6} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{4} \, a^{4} b x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a^{5} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{a^{8} \mathrm{sgn}\left (b x + a\right )}{168 \, b^{3}} \end{align*}

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]

1/8*b^5*x^8*sgn(b*x + a) + 5/7*a*b^4*x^7*sgn(b*x + a) + 5/3*a^2*b^3*x^6*sgn(b*x + a) + 2*a^3*b^2*x^5*sgn(b*x +
a) + 5/4*a^4*b*x^4*sgn(b*x + a) + 1/3*a^5*x^3*sgn(b*x + a) + 1/168*a^8*sgn(b*x + a)/b^3