3.5.2 \(\int x^4 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=167 \[ \frac {a b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {a^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 167, 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^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {a b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {a^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(a^3*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*(a + b*x^2)) + (3*a^2*b*x^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*(
a + b*x^2)) + (a*b^2*x^9*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (b^3*x^11*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4])/(11*(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^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^4 \left (a b+b^2 x^2\right )^3 \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^3 b^3 x^4+3 a^2 b^4 x^6+3 a b^5 x^8+b^6 x^{10}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {a^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {a b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \begin {gather*} \frac {x^5 \sqrt {\left (a+b x^2\right )^2} \left (231 a^3+495 a^2 b x^2+385 a b^2 x^4+105 b^3 x^6\right )}{1155 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^5*Sqrt[(a + b*x^2)^2]*(231*a^3 + 495*a^2*b*x^2 + 385*a*b^2*x^4 + 105*b^3*x^6))/(1155*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 5.88, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (231 a^3 x^5+495 a^2 b x^7+385 a b^2 x^9+105 b^3 x^{11}\right )}{1155 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(Sqrt[(a + b*x^2)^2]*(231*a^3*x^5 + 495*a^2*b*x^7 + 385*a*b^2*x^9 + 105*b^3*x^11))/(1155*(a + b*x^2))

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fricas [A]  time = 0.83, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{11} \, b^{3} x^{11} + \frac {1}{3} \, a b^{2} x^{9} + \frac {3}{7} \, a^{2} b x^{7} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/11*b^3*x^11 + 1/3*a*b^2*x^9 + 3/7*a^2*b*x^7 + 1/5*a^3*x^5

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giac [A]  time = 0.18, size = 67, normalized size = 0.40 \begin {gather*} \frac {1}{11} \, b^{3} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a b^{2} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{7} \, a^{2} b x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{5} \, a^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/11*b^3*x^11*sgn(b*x^2 + a) + 1/3*a*b^2*x^9*sgn(b*x^2 + a) + 3/7*a^2*b*x^7*sgn(b*x^2 + a) + 1/5*a^3*x^5*sgn(b
*x^2 + a)

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \begin {gather*} \frac {\left (105 b^{3} x^{6}+385 a \,b^{2} x^{4}+495 a^{2} b \,x^{2}+231 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x^{5}}{1155 \left (b \,x^{2}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/1155*x^5*(105*b^3*x^6+385*a*b^2*x^4+495*a^2*b*x^2+231*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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maxima [A]  time = 1.39, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{11} \, b^{3} x^{11} + \frac {1}{3} \, a b^{2} x^{9} + \frac {3}{7} \, a^{2} b x^{7} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/11*b^3*x^11 + 1/3*a*b^2*x^9 + 3/7*a^2*b*x^7 + 1/5*a^3*x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**4*((a + b*x**2)**2)**(3/2), x)

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