\(\int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [138]

Optimal result

Integrand size = 24, antiderivative size = 71 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {a x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[In]

Int[x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 45

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 660

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^4 (5 a+4 b x) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{20 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

Integrate[x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x^4*(5*a + 4*b*x)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(20*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a
+ b*x)^2]))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.42

method	result	size
gosper	\(\frac {x^{4} \left (4 b x +5 a \right ) \sqrt {\left (b x +a \right )^{2}}}{20 b x +20 a}\)	\(30\)
risch	\(\frac {a \,x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}+\frac {b \,x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}\)	\(46\)
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (-4 b^{3} x^{3}+3 a \,b^{2} x^{2}-2 a^{2} b x +a^{3}\right )}{20 b^{4}}\)	\(47\)

[In]

int(x^3*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/20*x^4*(4*b*x+5*a)*((b*x+a)^2)^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.18 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} \]

[In]

integrate(x^3*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*b*x^5 + 1/4*a*x^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (46) = 92\).

Time = 1.34 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.25 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{4}} + \frac {a^{3} x}{20 b^{3}} - \frac {a^{2} x^{2}}{20 b^{2}} + \frac {a x^{3}}{20 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*((b*x+a)**2)**(1/2),x)

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3*x/(20*b**3) - a**2*x**2/(20*b**2) + a*x**3
/(20*b) + x**4/5), Ne(b**2, 0)), ((-a**6*(a**2 + 2*a*b*x)**(3/2)/3 + 3*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 3*a**2
*(a**2 + 2*a*b*x)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8*a**4*b**4), Ne(a*b, 0)), (x**4*sqrt(a**2)/4, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (45) = 90\).

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.85 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} x}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}}{2 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x}{20 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}}{20 \, b^{4}} \]

[In]

integrate(x^3*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*x/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*x^2/b^2 - 1/2*sqrt(b^2*x^2
+ 2*a*b*x + a^2)*a^4/b^4 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*x/b^3 + 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)
*a^2/b^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.55 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a x^{4} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{5} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{4}} \]

[In]

integrate(x^3*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*b*x^5*sgn(b*x + a) + 1/4*a*x^4*sgn(b*x + a) - 1/20*a^5*sgn(b*x + a)/b^4

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30 \[ \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{20\,b^4} \]

[In]

int(x^3*((a + b*x)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(4*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) - a^4 + 9*a^2*b^2*x^2 + 8*a^3*b*x - 7*a*
b*x*(a^2 + b^2*x^2 + 2*a*b*x)))/(20*b^4)