\(\int x^3 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [150]

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a^3 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {3 a^2 b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^3 x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 151, 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} \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a b^2 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {3 a^2 b x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b^3 x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {a^3 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[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 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 )^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*}

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^4 \left (35 a^3+84 a^2 b x+70 a b^2 x^2+20 b^3 x^3\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{140 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

(x^4*(35*a^3 + 84*a^2*b*x + 70*a*b^2*x^2 + 20*b^3*x^3)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(1
40*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.34

method	result	size
gosper	\(\frac {x^{4} \left (20 b^{3} x^{3}+70 a \,b^{2} x^{2}+84 a^{2} b x +35 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (b x +a \right )^{3}}\)	\(52\)
default	\(\frac {x^{4} \left (20 b^{3} x^{3}+70 a \,b^{2} x^{2}+84 a^{2} b x +35 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (b x +a \right )^{3}}\)	\(52\)
risch	\(\frac {a^{3} x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}+\frac {3 a^{2} b \,x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 \left (b x +a \right )}+\frac {a \,b^{2} x^{6} \sqrt {\left (b x +a \right )^{2}}}{2 b x +2 a}+\frac {b^{3} x^{7} \sqrt {\left (b x +a \right )^{2}}}{7 b x +7 a}\)	\(100\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.23 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{7} \, b^{3} x^{7} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{4} \, a^{3} x^{4} \]

[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

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.22 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{6}}{140 b^{4}} + \frac {a^{5} x}{140 b^{3}} - \frac {a^{4} x^{2}}{140 b^{2}} + \frac {a^{3} x^{3}}{140 b} + \frac {17 a^{2} x^{4}}{70} + \frac {5 a b x^{5}}{14} + \frac {b^{2} x^{6}}{7}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} - \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \left (a^{2}\right )^{\frac {3}{2}}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**6/(140*b**4) + a**5*x/(140*b**3) - a**4*x**2/(140*b**2) + a**
3*x**3/(140*b) + 17*a**2*x**4/70 + 5*a*b*x**5/14 + b**2*x**6/7), Ne(b**2, 0)), ((-a**6*(a**2 + 2*a*b*x)**(5/2)
/5 + 3*a**4*(a**2 + 2*a*b*x)**(7/2)/7 - a**2*(a**2 + 2*a*b*x)**(9/2)/3 + (a**2 + 2*a*b*x)**(11/2)/11)/(8*a**4*
b**4), Ne(a*b, 0)), (x**4*(a**2)**(3/2)/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x^{2}}{7 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4}}{4 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x}{14 \, b^{3}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2}}{70 \, b^{4}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.48 \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{7} \, b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a^{2} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{7} \mathrm {sgn}\left (b x + a\right )}{140 \, b^{4}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

[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)