3.370 \(\int x^5 (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac{\left (a+b x^2\right )^{9/2}}{9 b^3}-\frac{2 a \left (a+b x^2\right )^{7/2}}{7 b^3} \]

[Out]

(a^2*(a + b*x^2)^(5/2))/(5*b^3) - (2*a*(a + b*x^2)^(7/2))/(7*b^3) + (a + b*x^2)^(9/2)/(9*b^3)

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Rubi [A]  time = 0.0351328, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{a^2 \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac{\left (a+b x^2\right )^{9/2}}{9 b^3}-\frac{2 a \left (a+b x^2\right )^{7/2}}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^(3/2),x]

[Out]

(a^2*(a + b*x^2)^(5/2))/(5*b^3) - (2*a*(a + b*x^2)^(7/2))/(7*b^3) + (a + b*x^2)^(9/2)/(9*b^3)

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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+b x^2\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b x)^{3/2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^{3/2}}{b^2}-\frac{2 a (a+b x)^{5/2}}{b^2}+\frac{(a+b x)^{7/2}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac{2 a \left (a+b x^2\right )^{7/2}}{7 b^3}+\frac{\left (a+b x^2\right )^{9/2}}{9 b^3}\\ \end{align*}

Mathematica [A]  time = 0.017978, size = 39, normalized size = 0.66 \[ \frac{\left (a+b x^2\right )^{5/2} \left (8 a^2-20 a b x^2+35 b^2 x^4\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^(3/2),x]

[Out]

((a + b*x^2)^(5/2)*(8*a^2 - 20*a*b*x^2 + 35*b^2*x^4))/(315*b^3)

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Maple [A]  time = 0.005, size = 36, normalized size = 0.6 \begin{align*}{\frac{35\,{b}^{2}{x}^{4}-20\,ab{x}^{2}+8\,{a}^{2}}{315\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^(3/2),x)

[Out]

1/315*(b*x^2+a)^(5/2)*(35*b^2*x^4-20*a*b*x^2+8*a^2)/b^3

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57227, size = 126, normalized size = 2.14 \begin{align*} \frac{{\left (35 \, b^{4} x^{8} + 50 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} - 4 \, a^{3} b x^{2} + 8 \, a^{4}\right )} \sqrt{b x^{2} + a}}{315 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

1/315*(35*b^4*x^8 + 50*a*b^3*x^6 + 3*a^2*b^2*x^4 - 4*a^3*b*x^2 + 8*a^4)*sqrt(b*x^2 + a)/b^3

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Sympy [A]  time = 2.07804, size = 109, normalized size = 1.85 \begin{align*} \begin{cases} \frac{8 a^{4} \sqrt{a + b x^{2}}}{315 b^{3}} - \frac{4 a^{3} x^{2} \sqrt{a + b x^{2}}}{315 b^{2}} + \frac{a^{2} x^{4} \sqrt{a + b x^{2}}}{105 b} + \frac{10 a x^{6} \sqrt{a + b x^{2}}}{63} + \frac{b x^{8} \sqrt{a + b x^{2}}}{9} & \text{for}\: b \neq 0 \\\frac{a^{\frac{3}{2}} x^{6}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**(3/2),x)

[Out]

Piecewise((8*a**4*sqrt(a + b*x**2)/(315*b**3) - 4*a**3*x**2*sqrt(a + b*x**2)/(315*b**2) + a**2*x**4*sqrt(a + b
*x**2)/(105*b) + 10*a*x**6*sqrt(a + b*x**2)/63 + b*x**8*sqrt(a + b*x**2)/9, Ne(b, 0)), (a**(3/2)*x**6/6, True)
)

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Giac [B]  time = 2.72273, size = 143, normalized size = 2.42 \begin{align*} \frac{\frac{3 \,{\left (15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2}\right )} a}{b^{2}} + \frac{35 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}}{b^{2}}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/315*(3*(15*(b*x^2 + a)^(7/2) - 42*(b*x^2 + a)^(5/2)*a + 35*(b*x^2 + a)^(3/2)*a^2)*a/b^2 + (35*(b*x^2 + a)^(9
/2) - 135*(b*x^2 + a)^(7/2)*a + 189*(b*x^2 + a)^(5/2)*a^2 - 105*(b*x^2 + a)^(3/2)*a^3)/b^2)/b