3.5.33 \(\int x^{12} (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac {b^5 x^{23} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 \left (a+b x^2\right )}+\frac {5 a b^4 x^{21} \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{19} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac {a^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {a^4 b x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 255, 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^{23} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 \left (a+b x^2\right )}+\frac {5 a b^4 x^{21} \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{19} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )}+\frac {a^4 b x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^12*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^5*x^13*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*(a + b*x^2)) + (a^4*b*x^15*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*
(a + b*x^2)) + (10*a^3*b^2*x^17*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*(a + b*x^2)) + (10*a^2*b^3*x^19*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4])/(19*(a + b*x^2)) + (5*a*b^4*x^21*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(21*(a + b*x^2)) + (
b^5*x^23*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(23*(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^{12} \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^{12} \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^{12}+5 a^4 b^6 x^{14}+10 a^3 b^7 x^{16}+10 a^2 b^8 x^{18}+5 a b^9 x^{20}+b^{10} x^{22}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {a^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {a^4 b x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{19} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac {5 a b^4 x^{21} \sqrt {a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}+\frac {b^5 x^{23} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 \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^{13} \sqrt {\left (a+b x^2\right )^2} \left (156009 a^5+676039 a^4 b x^2+1193010 a^3 b^2 x^4+1067430 a^2 b^3 x^6+482885 a b^4 x^8+88179 b^5 x^{10}\right )}{2028117 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^12*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^13*Sqrt[(a + b*x^2)^2]*(156009*a^5 + 676039*a^4*b*x^2 + 1193010*a^3*b^2*x^4 + 1067430*a^2*b^3*x^6 + 482885*
a*b^4*x^8 + 88179*b^5*x^10))/(2028117*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 19.39, size = 83, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (156009 a^5 x^{13}+676039 a^4 b x^{15}+1193010 a^3 b^2 x^{17}+1067430 a^2 b^3 x^{19}+482885 a b^4 x^{21}+88179 b^5 x^{23}\right )}{2028117 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^12*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(Sqrt[(a + b*x^2)^2]*(156009*a^5*x^13 + 676039*a^4*b*x^15 + 1193010*a^3*b^2*x^17 + 1067430*a^2*b^3*x^19 + 4828
85*a*b^4*x^21 + 88179*b^5*x^23))/(2028117*(a + b*x^2))

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fricas [A]  time = 1.07, size = 57, normalized size = 0.22 \begin {gather*} \frac {1}{23} \, b^{5} x^{23} + \frac {5}{21} \, a b^{4} x^{21} + \frac {10}{19} \, a^{2} b^{3} x^{19} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {1}{3} \, a^{4} b x^{15} + \frac {1}{13} \, a^{5} x^{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/23*b^5*x^23 + 5/21*a*b^4*x^21 + 10/19*a^2*b^3*x^19 + 10/17*a^3*b^2*x^17 + 1/3*a^4*b*x^15 + 1/13*a^5*x^13

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giac [A]  time = 0.16, size = 105, normalized size = 0.41 \begin {gather*} \frac {1}{23} \, b^{5} x^{23} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{21} \, a b^{4} x^{21} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{19} \, a^{2} b^{3} x^{19} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{17} \, a^{3} b^{2} x^{17} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a^{4} b x^{15} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{13} \, a^{5} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/23*b^5*x^23*sgn(b*x^2 + a) + 5/21*a*b^4*x^21*sgn(b*x^2 + a) + 10/19*a^2*b^3*x^19*sgn(b*x^2 + a) + 10/17*a^3*
b^2*x^17*sgn(b*x^2 + a) + 1/3*a^4*b*x^15*sgn(b*x^2 + a) + 1/13*a^5*x^13*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 80, normalized size = 0.31 \begin {gather*} \frac {\left (88179 b^{5} x^{10}+482885 a \,b^{4} x^{8}+1067430 a^{2} b^{3} x^{6}+1193010 a^{3} b^{2} x^{4}+676039 a^{4} b \,x^{2}+156009 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x^{13}}{2028117 \left (b \,x^{2}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/2028117*x^13*(88179*b^5*x^10+482885*a*b^4*x^8+1067430*a^2*b^3*x^6+1193010*a^3*b^2*x^4+676039*a^4*b*x^2+15600
9*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5

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maxima [A]  time = 1.28, size = 57, normalized size = 0.22 \begin {gather*} \frac {1}{23} \, b^{5} x^{23} + \frac {5}{21} \, a b^{4} x^{21} + \frac {10}{19} \, a^{2} b^{3} x^{19} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {1}{3} \, a^{4} b x^{15} + \frac {1}{13} \, a^{5} x^{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/23*b^5*x^23 + 5/21*a*b^4*x^21 + 10/19*a^2*b^3*x^19 + 10/17*a^3*b^2*x^17 + 1/3*a^4*b*x^15 + 1/13*a^5*x^13

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{12}\,{\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^12*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(x^12*(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^{12} \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**12*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**12*((a + b*x**2)**2)**(5/2), x)

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