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

Optimal. Leaf size=119 \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{16 b^3}-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{7 b^3}+\frac{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^3} \]

[Out]

(a^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^3) - (a*(a + b*x^2)^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(7*b^3) + ((a + b*x^2)^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*b^3)

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Rubi [A]  time = 0.0971745, antiderivative size = 119, 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} \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{16 b^3}-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{7 b^3}+\frac{a^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^3) - (a*(a + b*x^2)^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(7*b^3) + ((a + b*x^2)^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*b^3)

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^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \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^2 \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 (\frac{a^2 \left (a b+b^2 x\right )^5}{b^2}-\frac{2 a \left (a b+b^2 x\right )^6}{b^3}+\frac{\left (a b+b^2 x\right )^7}{b^4}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{a^2 \left (a+b x^2\right )^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{12 b^3}-\frac{a \left (a+b x^2\right )^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{7 b^3}+\frac{\left (a+b x^2\right )^7 \sqrt{a^2+2 a b x^2+b^2 x^4}}{16 b^3}\\ \end{align*}

Mathematica [A]  time = 0.021695, size = 83, normalized size = 0.7 \[ \frac{x^6 \sqrt{\left (a+b x^2\right )^2} \left (280 a^2 b^3 x^6+336 a^3 b^2 x^4+210 a^4 b x^2+56 a^5+120 a b^4 x^8+21 b^5 x^{10}\right )}{336 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^6*Sqrt[(a + b*x^2)^2]*(56*a^5 + 210*a^4*b*x^2 + 336*a^3*b^2*x^4 + 280*a^2*b^3*x^6 + 120*a*b^4*x^8 + 21*b^5*
x^10))/(336*(a + b*x^2))

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Maple [A]  time = 0.164, size = 80, normalized size = 0.7 \begin{align*}{\frac{{x}^{6} \left ( 21\,{b}^{5}{x}^{10}+120\,a{b}^{4}{x}^{8}+280\,{a}^{2}{b}^{3}{x}^{6}+336\,{b}^{2}{a}^{3}{x}^{4}+210\,{a}^{4}b{x}^{2}+56\,{a}^{5} \right ) }{336\, \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^5*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/336*x^6*(21*b^5*x^10+120*a*b^4*x^8+280*a^2*b^3*x^6+336*a^3*b^2*x^4+210*a^4*b*x^2+56*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^5*(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 = 1.44155, size = 131, normalized size = 1.1 \begin{align*} \frac{1}{16} \, b^{5} x^{16} + \frac{5}{14} \, a b^{4} x^{14} + \frac{5}{6} \, a^{2} b^{3} x^{12} + a^{3} b^{2} x^{10} + \frac{5}{8} \, a^{4} b x^{8} + \frac{1}{6} \, a^{5} x^{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^5*x^16 + 5/14*a*b^4*x^14 + 5/6*a^2*b^3*x^12 + a^3*b^2*x^10 + 5/8*a^4*b*x^8 + 1/6*a^5*x^6

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

[Out]

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

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Giac [A]  time = 1.13122, size = 140, normalized size = 1.18 \begin{align*} \frac{1}{16} \, b^{5} x^{16} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{14} \, a b^{4} x^{14} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{6} \, a^{2} b^{3} x^{12} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{3} b^{2} x^{10} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{8} \, a^{4} b x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{6} \, a^{5} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^5*x^16*sgn(b*x^2 + a) + 5/14*a*b^4*x^14*sgn(b*x^2 + a) + 5/6*a^2*b^3*x^12*sgn(b*x^2 + a) + a^3*b^2*x^10
*sgn(b*x^2 + a) + 5/8*a^4*b*x^8*sgn(b*x^2 + a) + 1/6*a^5*x^6*sgn(b*x^2 + a)