3.586 \(\int x^{11} (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=255 \[ \frac{b^5 x^{22} \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac{a b^4 x^{20} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^{18} \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac{5 a^3 b^2 x^{16} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac{5 a^4 b x^{14} \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac{a^5 x^{12} \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )} \]

[Out]

(a^5*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*(a + b*x^2)) + (5*a^4*b*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(
14*(a + b*x^2)) + (5*a^3*b^2*x^16*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^2)) + (5*a^2*b^3*x^18*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4])/(9*(a + b*x^2)) + (a*b^4*x^20*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*(a + b*x^2)) + (b^5*
x^22*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(22*(a + b*x^2))

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Rubi [A]  time = 0.15898, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \[ \frac{b^5 x^{22} \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}+\frac{a b^4 x^{20} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^{18} \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac{5 a^3 b^2 x^{16} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac{5 a^4 b x^{14} \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac{a^5 x^{12} \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(a^5*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*(a + b*x^2)) + (5*a^4*b*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(
14*(a + b*x^2)) + (5*a^3*b^2*x^16*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^2)) + (5*a^2*b^3*x^18*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4])/(9*(a + b*x^2)) + (a*b^4*x^20*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*(a + b*x^2)) + (b^5*
x^22*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(22*(a + b*x^2))

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])

Rule 646

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[(d + e*x)^m*(b/2 + c*x)^(2*p), 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]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^5 \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 x^5 \left (a b+b^2 x\right )^5 \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \left (a^5 b^5 x^5+5 a^4 b^6 x^6+10 a^3 b^7 x^7+10 a^2 b^8 x^8+5 a b^9 x^9+b^{10} x^{10}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{a^5 x^{12} \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )}+\frac{5 a^4 b x^{14} \sqrt{a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac{5 a^3 b^2 x^{16} \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^{18} \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac{a b^4 x^{20} \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{b^5 x^{22} \sqrt{a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0206497, size = 83, normalized size = 0.33 \[ \frac{x^{12} \sqrt{\left (a+b x^2\right )^2} \left (3080 a^2 b^3 x^6+3465 a^3 b^2 x^4+1980 a^4 b x^2+462 a^5+1386 a b^4 x^8+252 b^5 x^{10}\right )}{5544 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x^12*Sqrt[(a + b*x^2)^2]*(462*a^5 + 1980*a^4*b*x^2 + 3465*a^3*b^2*x^4 + 3080*a^2*b^3*x^6 + 1386*a*b^4*x^8 + 2
52*b^5*x^10))/(5544*(a + b*x^2))

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Maple [A]  time = 0.163, size = 80, normalized size = 0.3 \begin{align*}{\frac{{x}^{12} \left ( 252\,{b}^{5}{x}^{10}+1386\,a{b}^{4}{x}^{8}+3080\,{a}^{2}{b}^{3}{x}^{6}+3465\,{b}^{2}{a}^{3}{x}^{4}+1980\,{a}^{4}b{x}^{2}+462\,{a}^{5} \right ) }{5544\, \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^11*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/5544*x^12*(252*b^5*x^10+1386*a*b^4*x^8+3080*a^2*b^3*x^6+3465*a^3*b^2*x^4+1980*a^4*b*x^2+462*a^5)*((b*x^2+a)^
2)^(5/2)/(b*x^2+a)^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46215, size = 140, normalized size = 0.55 \begin{align*} \frac{1}{22} \, b^{5} x^{22} + \frac{1}{4} \, a b^{4} x^{20} + \frac{5}{9} \, a^{2} b^{3} x^{18} + \frac{5}{8} \, a^{3} b^{2} x^{16} + \frac{5}{14} \, a^{4} b x^{14} + \frac{1}{12} \, a^{5} x^{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/22*b^5*x^22 + 1/4*a*b^4*x^20 + 5/9*a^2*b^3*x^18 + 5/8*a^3*b^2*x^16 + 5/14*a^4*b*x^14 + 1/12*a^5*x^12

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{11} \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**11*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**11*((a + b*x**2)**2)**(5/2), x)

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Giac [A]  time = 1.12745, size = 142, normalized size = 0.56 \begin{align*} \frac{1}{22} \, b^{5} x^{22} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{4} \, a b^{4} x^{20} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{9} \, a^{2} b^{3} x^{18} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{8} \, a^{3} b^{2} x^{16} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{14} \, a^{4} b x^{14} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{12} \, a^{5} x^{12} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/22*b^5*x^22*sgn(b*x^2 + a) + 1/4*a*b^4*x^20*sgn(b*x^2 + a) + 5/9*a^2*b^3*x^18*sgn(b*x^2 + a) + 5/8*a^3*b^2*x
^16*sgn(b*x^2 + a) + 5/14*a^4*b*x^14*sgn(b*x^2 + a) + 1/12*a^5*x^12*sgn(b*x^2 + a)