3.587 \(\int x^9 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=201 \[ \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^9}{20 b^5}-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^8}{9 b^5}+\frac {3 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{8 b^5}+\frac {a^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^5}-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{7 b^5} \]

[Out]

1/12*a^4*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^5-2/7*a^3*(b*x^2+a)^6*((b*x^2+a)^2)^(1/2)/b^5+3/8*a^2*(b*x^2+a)^7*(
(b*x^2+a)^2)^(1/2)/b^5-2/9*a*(b*x^2+a)^8*((b*x^2+a)^2)^(1/2)/b^5+1/20*(b*x^2+a)^9*((b*x^2+a)^2)^(1/2)/b^5

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Rubi [A]  time = 0.13, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1111, 645} \[ \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^9}{20 b^5}-\frac {2 a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^8}{9 b^5}+\frac {3 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{8 b^5}-\frac {2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{7 b^5}+\frac {a^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^5} \]

Antiderivative was successfully verified.

[In]

Int[x^9*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^4*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^5) - (2*a^3*(a + b*x^2)^6*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4])/(7*b^5) + (3*a^2*(a + b*x^2)^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*b^5) - (2*a*(a + b*x^2)^8*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(9*b^5) + ((a + b*x^2)^9*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(20*b^5)

Rule 645

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

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^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^4 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a^4 \left (a b+b^2 x\right )^5}{b^4}-\frac {4 a^3 \left (a b+b^2 x\right )^6}{b^5}+\frac {6 a^2 \left (a b+b^2 x\right )^7}{b^6}-\frac {4 a \left (a b+b^2 x\right )^8}{b^7}+\frac {\left (a b+b^2 x\right )^9}{b^8}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {a^4 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^5}-\frac {2 a^3 \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 b^5}+\frac {3 a^2 \left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 b^5}-\frac {2 a \left (a+b x^2\right )^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 b^5}+\frac {\left (a+b x^2\right )^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{20 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.41 \[ \frac {x^{10} \sqrt {\left (a+b x^2\right )^2} \left (252 a^5+1050 a^4 b x^2+1800 a^3 b^2 x^4+1575 a^2 b^3 x^6+700 a b^4 x^8+126 b^5 x^{10}\right )}{2520 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^10*Sqrt[(a + b*x^2)^2]*(252*a^5 + 1050*a^4*b*x^2 + 1800*a^3*b^2*x^4 + 1575*a^2*b^3*x^6 + 700*a*b^4*x^8 + 12
6*b^5*x^10))/(2520*(a + b*x^2))

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fricas [A]  time = 0.73, size = 57, normalized size = 0.28 \[ \frac {1}{20} \, b^{5} x^{20} + \frac {5}{18} \, a b^{4} x^{18} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{12} \, a^{4} b x^{12} + \frac {1}{10} \, a^{5} x^{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/20*b^5*x^20 + 5/18*a*b^4*x^18 + 5/8*a^2*b^3*x^16 + 5/7*a^3*b^2*x^14 + 5/12*a^4*b*x^12 + 1/10*a^5*x^10

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giac [A]  time = 0.16, size = 105, normalized size = 0.52 \[ \frac {1}{20} \, b^{5} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{18} \, a b^{4} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{2} b^{3} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{3} b^{2} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{12} \, a^{4} b x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{10} \, a^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/20*b^5*x^20*sgn(b*x^2 + a) + 5/18*a*b^4*x^18*sgn(b*x^2 + a) + 5/8*a^2*b^3*x^16*sgn(b*x^2 + a) + 5/7*a^3*b^2*
x^14*sgn(b*x^2 + a) + 5/12*a^4*b*x^12*sgn(b*x^2 + a) + 1/10*a^5*x^10*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.40 \[ \frac {\left (126 b^{5} x^{10}+700 a \,b^{4} x^{8}+1575 a^{2} b^{3} x^{6}+1800 a^{3} b^{2} x^{4}+1050 a^{4} b \,x^{2}+252 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{10}}{2520 \left (b \,x^{2}+a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/2520*x^10*(126*b^5*x^10+700*a*b^4*x^8+1575*a^2*b^3*x^6+1800*a^3*b^2*x^4+1050*a^4*b*x^2+252*a^5)*((b*x^2+a)^2
)^(5/2)/(b*x^2+a)^5

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maxima [A]  time = 1.43, size = 57, normalized size = 0.28 \[ \frac {1}{20} \, b^{5} x^{20} + \frac {5}{18} \, a b^{4} x^{18} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{12} \, a^{4} b x^{12} + \frac {1}{10} \, a^{5} x^{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/20*b^5*x^20 + 5/18*a*b^4*x^18 + 5/8*a^2*b^3*x^16 + 5/7*a^3*b^2*x^14 + 5/12*a^4*b*x^12 + 1/10*a^5*x^10

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^9\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^9*(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 \[ \int x^{9} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**9*((a + b*x**2)**2)**(5/2), x)

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