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

Optimal. Leaf size=151 \[ \frac {a b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {3 a^2 b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b^3 x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a^3 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 151, 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 = {646, 43} \[ \frac {b^3 x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {3 a^2 b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^3 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x)) + (3*a^2*b*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*
x)) + (a*b^2*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + (b^3*x^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a
+ b*x))

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 55, normalized size = 0.36 \[ \frac {x^4 \sqrt {(a+b x)^2} \left (35 a^3+84 a^2 b x+70 a b^2 x^2+20 b^3 x^3\right )}{140 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x)^2]*(35*a^3 + 84*a^2*b*x + 70*a*b^2*x^2 + 20*b^3*x^3))/(140*(a + b*x))

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fricas [A]  time = 0.92, size = 35, normalized size = 0.23 \[ \frac {1}{7} \, b^{3} x^{7} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{4} \, a^{3} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 73, normalized size = 0.48 \[ \frac {1}{7} \, b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a^{2} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{7} \mathrm {sgn}\left (b x + a\right )}{140 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7*sgn(b*x + a) + 1/2*a*b^2*x^6*sgn(b*x + a) + 3/5*a^2*b*x^5*sgn(b*x + a) + 1/4*a^3*x^4*sgn(b*x + a)
- 1/140*a^7*sgn(b*x + a)/b^4

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maple [A]  time = 0.05, size = 52, normalized size = 0.34 \[ \frac {\left (20 b^{3} x^{3}+70 a \,b^{2} x^{2}+84 a^{2} b x +35 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{4}}{140 \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*x^4*(20*b^3*x^3+70*a*b^2*x^2+84*a^2*b*x+35*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 1.34, size = 131, normalized size = 0.87 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x^{2}}{7 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4}}{4 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x}{14 \, b^{3}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2}}{70 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3*x/b^3 + 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^2/b^2 - 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*a^4/b^4 - 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x/b^3 + 17/70*(b^2*x^2 + 2*a*b*x + a^2)^
(5/2)*a^2/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*(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 \[ \int x^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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