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

Optimal. Leaf size=67 \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2}-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^2} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 640, 609} \[ \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{7/2}}{14 b^2}-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 609

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 640

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 1111

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*} \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {1}{2} \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 1.24 \[ \frac {x^4 \sqrt {\left (a+b x^2\right )^2} \left (21 a^5+70 a^4 b x^2+105 a^3 b^2 x^4+84 a^2 b^3 x^6+35 a b^4 x^8+6 b^5 x^{10}\right )}{84 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x^2)^2]*(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
))/(84*(a + b*x^2))

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fricas [A]  time = 1.16, size = 56, normalized size = 0.84 \[ \frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.20, size = 67, normalized size = 1.00 \[ \frac {1}{84} \, {\left (6 \, b^{5} x^{14} + 35 \, a b^{4} x^{12} + 84 \, a^{2} b^{3} x^{10} + 105 \, a^{3} b^{2} x^{8} + 70 \, a^{4} b x^{6} + 21 \, a^{5} x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 80, normalized size = 1.19 \[ \frac {\left (6 b^{5} x^{10}+35 a \,b^{4} x^{8}+84 a^{2} b^{3} x^{6}+105 a^{3} b^{2} x^{4}+70 a^{4} b \,x^{2}+21 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{4}}{84 \left (b \,x^{2}+a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.25, size = 56, normalized size = 0.84 \[ \frac {1}{14} \, b^{5} x^{14} + \frac {5}{12} \, a b^{4} x^{12} + a^{2} b^{3} x^{10} + \frac {5}{4} \, a^{3} b^{2} x^{8} + \frac {5}{6} \, a^{4} b x^{6} + \frac {1}{4} \, a^{5} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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