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

Optimal. Leaf size=252 \[ \frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 252, 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 = {1112, 270} \begin {gather*} \frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (a^4*b*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b
*x^2) + (10*a^3*b^2*x^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*(a + b*x^2)) + (10*a^2*b^3*x^9*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])/(9*(a + b*x^2)) + (5*a*b^4*x^11*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*(a + b*x^2)) + (b^5*x^13*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*(a + b*x^2))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^2 \left (a b+b^2 x^2\right )^5 \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^5 b^5 x^2+5 a^4 b^6 x^4+10 a^3 b^7 x^6+10 a^2 b^8 x^8+5 a b^9 x^{10}+b^{10} x^{12}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.33 \begin {gather*} \frac {x^3 \sqrt {\left (a+b x^2\right )^2} \left (3003 a^5+9009 a^4 b x^2+12870 a^3 b^2 x^4+10010 a^2 b^3 x^6+4095 a b^4 x^8+693 b^5 x^{10}\right )}{9009 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x^2)^2]*(3003*a^5 + 9009*a^4*b*x^2 + 12870*a^3*b^2*x^4 + 10010*a^2*b^3*x^6 + 4095*a*b^4*x^8 +
693*b^5*x^10))/(9009*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 6.75, size = 83, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (3003 a^5 x^3+9009 a^4 b x^5+12870 a^3 b^2 x^7+10010 a^2 b^3 x^9+4095 a b^4 x^{11}+693 b^5 x^{13}\right )}{9009 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(3003*a^5*x^3 + 9009*a^4*b*x^5 + 12870*a^3*b^2*x^7 + 10010*a^2*b^3*x^9 + 4095*a*b^4*x^11
+ 693*b^5*x^13))/(9009*(a + b*x^2))

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fricas [A]  time = 0.99, size = 56, normalized size = 0.22 \begin {gather*} \frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^5*x^13 + 5/11*a*b^4*x^11 + 10/9*a^2*b^3*x^9 + 10/7*a^3*b^2*x^7 + a^4*b*x^5 + 1/3*a^5*x^3

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giac [A]  time = 0.18, size = 104, normalized size = 0.41 \begin {gather*} \frac {1}{13} \, b^{5} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{11} \, a b^{4} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a^{5} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^5*x^13*sgn(b*x^2 + a) + 5/11*a*b^4*x^11*sgn(b*x^2 + a) + 10/9*a^2*b^3*x^9*sgn(b*x^2 + a) + 10/7*a^3*b^2
*x^7*sgn(b*x^2 + a) + a^4*b*x^5*sgn(b*x^2 + a) + 1/3*a^5*x^3*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.32 \begin {gather*} \frac {\left (693 b^{5} x^{10}+4095 a \,b^{4} x^{8}+10010 a^{2} b^{3} x^{6}+12870 a^{3} b^{2} x^{4}+9009 a^{4} b \,x^{2}+3003 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{3}}{9009 \left (b \,x^{2}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9009*x^3*(693*b^5*x^10+4095*a*b^4*x^8+10010*a^2*b^3*x^6+12870*a^3*b^2*x^4+9009*a^4*b*x^2+3003*a^5)*((b*x^2+a
)^2)^(5/2)/(b*x^2+a)^5

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maxima [A]  time = 1.36, size = 56, normalized size = 0.22 \begin {gather*} \frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^5*x^13 + 5/11*a*b^4*x^11 + 10/9*a^2*b^3*x^9 + 10/7*a^3*b^2*x^7 + a^4*b*x^5 + 1/3*a^5*x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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 \begin {gather*} \int x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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