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

Optimal. Leaf size=159 \[ \frac{b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}+\frac{3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac{a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3} \]

[Out]

(a^3*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b*x^2)^3 + (a^2*b*x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b
*x^2)^3 + (3*a*b^2*x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(5*(a + b*x^2)^3) + (b^3*x^7*(a^2 + 2*a*b*x^2 + b^2*
x^4)^(3/2))/(7*(a + b*x^2)^3)

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Rubi [A]  time = 0.0332733, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1088, 194} \[ \frac{b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}+\frac{3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac{a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b*x^2)^3 + (a^2*b*x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b
*x^2)^3 + (3*a*b^2*x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(5*(a + b*x^2)^3) + (b^3*x^7*(a^2 + 2*a*b*x^2 + b^2*
x^4)^(3/2))/(7*(a + b*x^2)^3)

Rule 1088

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^p/(b + 2*c*x^2)^(2*p), In
t[(b + 2*c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \int \left (2 a b+2 b^2 x^2\right )^3 \, dx}{\left (2 a b+2 b^2 x^2\right )^3}\\ &=\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \int \left (8 a^3 b^3+24 a^2 b^4 x^2+24 a b^5 x^4+8 b^6 x^6\right ) \, dx}{\left (2 a b+2 b^2 x^2\right )^3}\\ &=\frac{a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac{b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.0122895, size = 59, normalized size = 0.37 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (35 a^2 b x^3+35 a^3 x+21 a b^2 x^5+5 b^3 x^7\right )}{35 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(35*a^3*x + 35*a^2*b*x^3 + 21*a*b^2*x^5 + 5*b^3*x^7))/(35*(a + b*x^2))

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Maple [A]  time = 0.042, size = 56, normalized size = 0.4 \begin{align*}{\frac{x \left ( 5\,{b}^{3}{x}^{6}+21\,a{x}^{4}{b}^{2}+35\,{a}^{2}b{x}^{2}+35\,{a}^{3} \right ) }{35\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*x*(5*b^3*x^6+21*a*b^2*x^4+35*a^2*b*x^2+35*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [A]  time = 1.00336, size = 42, normalized size = 0.26 \begin{align*} \frac{1}{7} \, b^{3} x^{7} + \frac{3}{5} \, a b^{2} x^{5} + a^{2} b x^{3} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.4736, size = 66, normalized size = 0.42 \begin{align*} \frac{1}{7} \, b^{3} x^{7} + \frac{3}{5} \, a b^{2} x^{5} + a^{2} b x^{3} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09579, size = 85, normalized size = 0.53 \begin{align*} \frac{1}{7} \, b^{3} x^{7} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{3}{5} \, a b^{2} x^{5} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{2} b x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{3} x \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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