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

Optimal. Leaf size=79 \[ \frac{b x^8 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac{a x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )} \]

[Out]

(a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (b*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^
2))

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Rubi [A]  time = 0.0643209, antiderivative size = 79, 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} \[ \frac{b x^8 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac{a x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(a*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (b*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*(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^5 \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \sqrt{a^2+2 a b x+b^2 x^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 ) \, dx,x,x^2\right )}{2 \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 b x^2+b^2 x^3\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )}\\ &=\frac{a x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac{b x^8 \sqrt{a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0142518, size = 39, normalized size = 0.49 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (4 a x^6+3 b x^8\right )}{24 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(4*a*x^6 + 3*b*x^8))/(24*(a + b*x^2))

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Maple [A]  time = 0.172, size = 36, normalized size = 0.5 \begin{align*}{\frac{{x}^{6} \left ( 3\,b{x}^{2}+4\,a \right ) }{24\,b{x}^{2}+24\,a}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*((b*x^2+a)^2)^(1/2),x)

[Out]

1/24*x^6*(3*b*x^2+4*a)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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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*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68822, size = 31, normalized size = 0.39 \begin{align*} \frac{1}{8} \, b x^{8} + \frac{1}{6} \, a x^{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/8*b*x^8 + 1/6*a*x^6

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Sympy [A]  time = 0.093207, size = 12, normalized size = 0.15 \begin{align*} \frac{a x^{6}}{6} + \frac{b x^{8}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*((b*x**2+a)**2)**(1/2),x)

[Out]

a*x**6/6 + b*x**8/8

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Giac [A]  time = 1.14404, size = 39, normalized size = 0.49 \begin{align*} \frac{1}{8} \, b x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{1}{6} \, a 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*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/8*b*x^8*sgn(b*x^2 + a) + 1/6*a*x^6*sgn(b*x^2 + a)