3.149 \(\int x^4 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=151 \[ \frac{b^3 x^8 \sqrt{a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac{3 a b^2 x^7 \sqrt{a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac{a^2 b x^6 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{a^3 x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0415061, antiderivative size = 151, 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} \[ \frac{b^3 x^8 \sqrt{a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac{3 a b^2 x^7 \sqrt{a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac{a^2 b x^6 \sqrt{a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac{a^3 x^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0141467, size = 55, normalized size = 0.36 \[ \frac{x^5 \sqrt{(a+b x)^2} \left (140 a^2 b x+56 a^3+120 a b^2 x^2+35 b^3 x^3\right )}{280 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x^5*Sqrt[(a + b*x)^2]*(56*a^3 + 140*a^2*b*x + 120*a*b^2*x^2 + 35*b^3*x^3))/(280*(a + b*x))

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Maple [A]  time = 0.182, size = 52, normalized size = 0.3 \begin{align*}{\frac{{x}^{5} \left ( 35\,{b}^{3}{x}^{3}+120\,a{b}^{2}{x}^{2}+140\,{a}^{2}bx+56\,{a}^{3} \right ) }{280\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/280*x^5*(35*b^3*x^3+120*a*b^2*x^2+140*a^2*b*x+56*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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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^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73161, size = 80, normalized size = 0.53 \begin{align*} \frac{1}{8} \, b^{3} x^{8} + \frac{3}{7} \, a b^{2} x^{7} + \frac{1}{2} \, a^{2} b x^{6} + \frac{1}{5} \, a^{3} x^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x**4*((a + b*x)**2)**(3/2), x)

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Giac [A]  time = 1.19534, size = 99, normalized size = 0.66 \begin{align*} \frac{1}{8} \, b^{3} x^{8} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{7} \, a b^{2} x^{7} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{2} b x^{6} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, a^{3} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{a^{8} \mathrm{sgn}\left (b x + a\right )}{280 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/8*b^3*x^8*sgn(b*x + a) + 3/7*a*b^2*x^7*sgn(b*x + a) + 1/2*a^2*b*x^6*sgn(b*x + a) + 1/5*a^3*x^5*sgn(b*x + a)
+ 1/280*a^8*sgn(b*x + a)/b^5