\(\int x^3 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\) [590]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2} \]

[Out]

-1/12*a*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^(5/2)/b^2+1/14*(b^2*x^4+2*a*b*x^2+a^2)^(7/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 654, 623} \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2}-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^2} \]

[In]

Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

-1/12*(a*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/b^2 + (a^2 + 2*a*b*x^2 + b^2*x^4)^(7/2)/(14*b^2)

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^2\right ) \\ & = \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2}-\frac {a \text {Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(135\) vs. \(2(67)=134\).

Time = 0.66 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.01 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^4 \left (21 a^5+70 a^4 b x^2+105 a^3 b^2 x^4+84 a^2 b^3 x^6+35 a b^4 x^8+6 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{84 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^4*(21*a^5 + 70*a^4*b*x^2 + 105*a^3*b^2*x^4 + 84*a^2*b^3*x^6 + 35*a*b^4*x^8 + 6*b^5*x^10)*(Sqrt[a^2]*b*x^2 +
a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(84*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99

method	result	size
pseudoelliptic	\(\frac {\left (\frac {2}{7} x^{10} b^{5}+\frac {5}{3} a \,x^{8} b^{4}+4 a^{2} x^{6} b^{3}+5 a^{3} x^{4} b^{2}+\frac {10}{3} x^{2} a^{4} b +a^{5}\right ) x^{4} \operatorname {csgn}\left (b \,x^{2}+a \right )}{4}\)	\(66\)
gosper	\(\frac {x^{4} \left (6 x^{10} b^{5}+35 a \,x^{8} b^{4}+84 a^{2} x^{6} b^{3}+105 a^{3} x^{4} b^{2}+70 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{4} \left (6 x^{10} b^{5}+35 a \,x^{8} b^{4}+84 a^{2} x^{6} b^{3}+105 a^{3} x^{4} b^{2}+70 x^{2} a^{4} b +21 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{84 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{4}}{4 b \,x^{2}+4 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{6}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{8}}{4 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{10}}{b \,x^{2}+a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{12}}{12 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{14}}{14 b \,x^{2}+14 a}\)	\(177\)

[In]

int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(2/7*x^10*b^5+5/3*a*x^8*b^4+4*a^2*x^6*b^3+5*a^3*x^4*b^2+10/3*x^2*a^4*b+a^5)*x^4*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/14*b^5*x^14 + 5/12*a*b^4*x^12 + a^2*b^3*x^10 + 5/4*a^3*b^2*x^8 + 5/6*a^4*b*x^6 + 1/4*a^5*x^4

Sympy [F]

\[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

[In]

integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**3*((a + b*x**2)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/14*b^5*x^14 + 5/12*a*b^4*x^12 + a^2*b^3*x^10 + 5/4*a^3*b^2*x^8 + 5/6*a^4*b*x^6 + 1/4*a^5*x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{84} \, {\left (6 \, b^{5} x^{14} + 35 \, a b^{4} x^{12} + 84 \, a^{2} b^{3} x^{10} + 105 \, a^{3} b^{2} x^{8} + 70 \, a^{4} b x^{6} + 21 \, a^{5} x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/84*(6*b^5*x^14 + 35*a*b^4*x^12 + 84*a^2*b^3*x^10 + 105*a^3*b^2*x^8 + 70*a^4*b*x^6 + 21*a^5*x^4)*sgn(b*x^2 +
a)

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

[In]

int(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)