3.606 \(\int x^{10} (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac{b^5 x^{21} \sqrt{a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}+\frac{5 a b^4 x^{19} \sqrt{a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac{10 a^2 b^3 x^{17} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )}+\frac{2 a^3 b^2 x^{15} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac{5 a^4 b x^{13} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac{a^5 x^{11} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )} \]

[Out]

(a^5*x^11*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*(a + b*x^2)) + (5*a^4*b*x^13*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(
13*(a + b*x^2)) + (2*a^3*b^2*x^15*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (10*a^2*b^3*x^17*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4])/(17*(a + b*x^2)) + (5*a*b^4*x^19*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(19*(a + b*x^2)) +
(b^5*x^21*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(21*(a + b*x^2))

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Rubi [A]  time = 0.0577635, antiderivative size = 255, normalized size of antiderivative = 1., 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} \[ \frac{b^5 x^{21} \sqrt{a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}+\frac{5 a b^4 x^{19} \sqrt{a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac{10 a^2 b^3 x^{17} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )}+\frac{2 a^3 b^2 x^{15} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac{5 a^4 b x^{13} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac{a^5 x^{11} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^10*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^5*x^11*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*(a + b*x^2)) + (5*a^4*b*x^13*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(
13*(a + b*x^2)) + (2*a^3*b^2*x^15*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (10*a^2*b^3*x^17*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4])/(17*(a + b*x^2)) + (5*a*b^4*x^19*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(19*(a + b*x^2)) +
(b^5*x^21*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(21*(a + b*x^2))

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]

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]

Rubi steps

\begin{align*} \int x^{10} \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^{10} \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^{10}+5 a^4 b^6 x^{12}+10 a^3 b^7 x^{14}+10 a^2 b^8 x^{16}+5 a b^9 x^{18}+b^{10} x^{20}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{a^5 x^{11} \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac{5 a^4 b x^{13} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac{2 a^3 b^2 x^{15} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac{10 a^2 b^3 x^{17} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 \left (a+b x^2\right )}+\frac{5 a b^4 x^{19} \sqrt{a^2+2 a b x^2+b^2 x^4}}{19 \left (a+b x^2\right )}+\frac{b^5 x^{21} \sqrt{a^2+2 a b x^2+b^2 x^4}}{21 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0208419, size = 83, normalized size = 0.33 \[ \frac{x^{11} \sqrt{\left (a+b x^2\right )^2} \left (570570 a^2 b^3 x^6+646646 a^3 b^2 x^4+373065 a^4 b x^2+88179 a^5+255255 a b^4 x^8+46189 b^5 x^{10}\right )}{969969 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^11*Sqrt[(a + b*x^2)^2]*(88179*a^5 + 373065*a^4*b*x^2 + 646646*a^3*b^2*x^4 + 570570*a^2*b^3*x^6 + 255255*a*b
^4*x^8 + 46189*b^5*x^10))/(969969*(a + b*x^2))

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Maple [A]  time = 0.162, size = 80, normalized size = 0.3 \begin{align*}{\frac{{x}^{11} \left ( 46189\,{b}^{5}{x}^{10}+255255\,a{b}^{4}{x}^{8}+570570\,{a}^{2}{b}^{3}{x}^{6}+646646\,{b}^{2}{a}^{3}{x}^{4}+373065\,{a}^{4}b{x}^{2}+88179\,{a}^{5} \right ) }{969969\, \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/969969*x^11*(46189*b^5*x^10+255255*a*b^4*x^8+570570*a^2*b^3*x^6+646646*a^3*b^2*x^4+373065*a^4*b*x^2+88179*a^
5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5

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Maxima [A]  time = 0.997234, size = 77, normalized size = 0.3 \begin{align*} \frac{1}{21} \, b^{5} x^{21} + \frac{5}{19} \, a b^{4} x^{19} + \frac{10}{17} \, a^{2} b^{3} x^{17} + \frac{2}{3} \, a^{3} b^{2} x^{15} + \frac{5}{13} \, a^{4} b x^{13} + \frac{1}{11} \, a^{5} x^{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/21*b^5*x^21 + 5/19*a*b^4*x^19 + 10/17*a^2*b^3*x^17 + 2/3*a^3*b^2*x^15 + 5/13*a^4*b*x^13 + 1/11*a^5*x^11

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Fricas [A]  time = 1.266, size = 144, normalized size = 0.56 \begin{align*} \frac{1}{21} \, b^{5} x^{21} + \frac{5}{19} \, a b^{4} x^{19} + \frac{10}{17} \, a^{2} b^{3} x^{17} + \frac{2}{3} \, a^{3} b^{2} x^{15} + \frac{5}{13} \, a^{4} b x^{13} + \frac{1}{11} \, a^{5} x^{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/21*b^5*x^21 + 5/19*a*b^4*x^19 + 10/17*a^2*b^3*x^17 + 2/3*a^3*b^2*x^15 + 5/13*a^4*b*x^13 + 1/11*a^5*x^11

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{10} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**10*((a + b*x**2)**2)**(5/2), x)

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Giac [A]  time = 1.13374, size = 142, normalized size = 0.56 \begin{align*} \frac{1}{21} \, b^{5} x^{21} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{19} \, a b^{4} x^{19} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{10}{17} \, a^{2} b^{3} x^{17} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{2}{3} \, a^{3} b^{2} x^{15} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{13} \, a^{4} b x^{13} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{11} \, a^{5} x^{11} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/21*b^5*x^21*sgn(b*x^2 + a) + 5/19*a*b^4*x^19*sgn(b*x^2 + a) + 10/17*a^2*b^3*x^17*sgn(b*x^2 + a) + 2/3*a^3*b^
2*x^15*sgn(b*x^2 + a) + 5/13*a^4*b*x^13*sgn(b*x^2 + a) + 1/11*a^5*x^11*sgn(b*x^2 + a)